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\textcolorblueAdvanced Algebra HW2 \textcolorblue2022.3.16
1.2.1 four subspaces
Prove or find counter examples.
- For four subspaces, if any three of them are linearly independent,
then the four subspaces are linearly independent.
- If subspaces V1,V2 are
linearly independent, and V1,V3,V4 are linearly independent, and V2,V3,V4 are linearly
independent, then all four subspaces are linearly independent.
- If V1,V2 are linearly
independent, and V3,V4 are
linearly independent, and V1+V2,V3+V4 are linearly independent, then all four subspaces
are linearly independent.
(1) Construct four subspaces
below. It is obvious that any three of them are linearly independent,
but four subspaces together are linearly dependent. (dimR3=3) V1={[k00]∣k∈R},V2={[0k0]∣k∈R},V3={[00k]∣k∈R},V4={[kkk]∣k∈R} (2) Construct four
subspaces below. We can prove that each of V1,V2 and V1,V3,V4 and V2,V3,V4 are linearly independent.
V1={[k00]∣k∈R},V2={[0k0]∣k∈R},V3={[k0k]∣k∈R},V4={[0kk]∣k∈R} However, pick some special vectors from these subspaces and
its linear combination are zero 1⋅[100]+(−1)⋅[010]+(−1)[101]+1⋅[011]=→0 (3) Reduction to
absurdity, assume that four real numbers that not all of them is zero
a1→v1+a2→v2+a3→v3+a4→v4=0 We can proof that a1→v1+a2→v2≠0,
otherwise according to the independence of V1,V2 and
V3,V4 a1=a2=0,a3=a4=0, so a1→v1+a2→v2≠0,a3→v3+a4→v4≠0
But a1→v1+a2→v2∈V1+V2,a3→v3+a4→v4∈V3+V4, so linear
combination of a1→v1+a2→v2 and
a3→v3+a4→v4 can
add up to 0, which is contrary to
the independence of V1+V2.V3+V4
All in all, the four subspaces must be linearly independence
1.2.2 decomposition of
transpose
Let V be the space of n×n real matrices. Let T:V→V be the transpose
operation, i.e., T sends A to AT for each A∈V. Find a non-trivial T-invariant decomposition of V, and find the corresponding block form
of T. (Here we use real matrices
for your convenience, but the statement is totally fine for complex
matrices and conjugate transpose.)
Set S={A∣A=AT,A∈Mn×n}, this decomposition is invariant. Because after
transposing any symmetric matrix, the matrix remains itself.
Set S′={A∣A=−AT,A∈Mn×n}, A=−AT⟹AT=−A=−(AT)T, so any antisymmetric matrix's transpose
is antisymmetric.
So decompose the linear map of T
into S and S′, dimS=n(n+1)2,dimS′=n(n−1)2
Because any n×n matrix
B=B+BT2+B−BT2 ,
so transpose T can be
decomposed
into S and S′ two block form.
A→(A+AT200A−AT2)
so the corresponding block form of T is (I00−I)
1.2.3 ultimate subspaces
Let p(x) be any polynomial, and
define p(A) in the obvious manner.
E.g., if p(x)= x2+2x+3, then p(A)=A2+2A+3I. We fix some n×n matrix A.
- If AB=BA, show that Ker(B),Ran(B) are both A-invariant subspaces.
- Prove that Ap(A)=p(A)A.
- Conclude that N∞(A−λI),R∞(A−λI) are both A-invariant for any λ∈C.
(1) For any vector →v∈Ker(B),B→v=→0, so B(A→v)=BA→v=AB→v=0,A→v∈Ker(B)
And for any vector →v∈Ran(B),B→x=→v, so A→v=A(B→v)=AB→v=BA→v=B(A→v)
A→v∈Ran(B). So
Ker(B) and Ran(B) are both A−invariant subspaces
(2) Similar to polynomial, set
p(A)=n∑i=0aiAn, so calculate Ap(A)
Ap(A)=An∑i=0aiAn=n∑i=0aiAn+1=(n∑i=0aiAn)A=p(A)A (3) Due to limited
dimension n of A, N∞(A−λI) and R∞(A−λI) are limited
combinations
of Ker(A−λI)k
and Ran(A−λI)k .
According to the conclusion from (1) and (2)
A(A−λI)=(A−λI)A,A(A−λI)2=(A−λI)2A,⋯,A(A−λI)k=(A−λI)kA
so ∀ k∈Z , Ker(A−λI)k and Ran(A−λI)k are all A−invariant. Add them all together
So, N∞(A−λI),R∞(A−λI) are both A-invariant for any λ∈C.
1.2.4 interchangeability and common
eigenvector
Note that any linear map must have at least one eigenvector. (You may
try to prove this yourself, but it is not part of this homework.) You
may use this fact freely in this problem. Fix any two n×n square matrices A,B. Suppose AB=BA.
- If W is an A-invariant
subspace, show that A has an
eigenvector in W.
- Show that Ker(A−λI) is always B-invariant
for all λ∈C.
(Hint: Last problem.)
- Show that A,B has a common
eigenvector. (Hint: Last two sub-problem.)
(1) Construct a new linear map
from W to W W⟼W:→w⟼A→w According to the fact that any linear map must have at least
one eigenvector, A has an eigenvector in W
(2) For any vector →v from Ker(A−λI), (A−λI)→v=0,A→v=λ→v, use AB=BA
Because (A−λI)(B→v)=(AB−λB)→v=(BA−λB)→v=B(A−λI)→v=0
So Ker(A−λI) is
B−invariant for all λ∈C
(3) According to (1) and (2), there exists at least one
eigenvector →v in Ker(A−λI) that
B→v=λB→v And
the vector →v also satisfies
that A→v=λA→v
So if A and B are interchangeable, they must have
common eigenvector.
\textcolorblueAdvanced Algebra HW3 \textcolorblue2022.3.24
Find a basis in the following vector space so that the linear map
involved will be in Jordan normal form. Also find the Jordan normal
form.
- V=C2 is a real
vector space, and A:V→V that sends [xy] to [ˉx−ℜ(y)(1+i)ℑ(x)−y] is a real linear map. (Here ˉx means the complex conjugate of a
complex number x, and ℜ(x),ℑ(x) means the real part and
the imaginary part of a complex number x.)
- V=P4, the real vector space
space of all real polynomials of degree at most 4. And A:V→V is a linear map such
that A(p(x))=p′(x)+p(0)+p′(0)x2 for each polynomial p∈P4.
- A=[a1a2a3a4]. Be careful here.
Maybe we have many possibilities for its Jordan normal form depending on
the values of a1,a2,a3,a4.
(1) A1=(10−100−10001−10010−1)=(1210004004000401)(10000−11000−10000−1)(1210004004000401)−1
(2) A:a0+a1x+a2x2+a3x3+a4x4⟼(a0+a1)+2a2x+(a1+2a3)x2+4a4x3
(1100000200010300000400000)=J(100000√200000−√2000000100000)J−1J=(1√2+11−√21212011−1200√22−√220−60004000001) (3) J=(J1,4OOJ2,3),where Ji,j={(√aiaj00−√aiaj)ai≠0,aj≠0(0000)ai=0,aj≠0 or ai≠0,aj=0(0100)ai=aj=0
1.3.2 partitions of
interger
A partition of integer n is a
way to write n as a sum of other
positive integers, say 5=2+2+1. If
you always order the summands from large to small, you end up with a dot
diagram, where each column represent an integer: [⋅⋅⋅⋅⋅]. Similarly, 7=4+2+1 should be represented as [⋅⋅⋅⋅⋅⋅⋅]
If the Jordan normal form of an n×n nilpotent matrix A is diag (Ja1,Ja2,…,Jak), then we have a partition of integer n=a1+…+ak. However, we also
have a partition of integer n=[dimKer(A)]+[dimKer(A2)−dimKer(A)]+[dimKer(A3)−dimKer(A2)]+… where we
treat the content of each bracket as a positive integer. Can you find a
relation between the two dot diagrams?
A partition of integer n=a1+…+ak is called
self-conjugate if, for the matrix A=diag(Ja1,Ja2,…,Jak), the two dot diagrams you obtained
above are the same. Show that, for a fixed integer n, the number of
self-conjugate partition of n is
equal to the number of partition of n into distinct odd positive integers.
(Hint: For a self-conjugate dot diagram, count the total number of dots
that are either in the first column or in the first row or in both. Is
this always odd?)
Suppose a 4 by 4 matrix A is
nilpotent and upper trianguler, and all (i,j) entries for i<j are
chosen randomly and uniformly in the interval [−1,1]. What are the probabilities that
its Jordan canonical form corresponds to the partitions 4=4,4=3+1,4=2+2,4=2+1+1,4=1+1+1+1
?
(1) I can find that the sequence
of each bracket' number is not incremental, which is just like the dot
graph. Put [dimKer(A)],[dimKer(A2)−dimKer(A)],[dimKer(A3)−dimKer(A2)],… like the dot
graph. According to the 'killing chain', the number of each row's dots
of the dot graph is just a1,a2,a3,…
(2) The self-conjugate of the
partition is just like a flying wing. Like this [⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅]
If rudely call this as a matrix A, because of the condition, we have
A=AT.
Consider the outermost corner, the total number of the dots is n+n−1=2n−1∈odd
And deprive each corner, the newest outermost corner also satisfies
the odd condition.
So every self-conjugate can be correspondence to odd-partition. And
for each odd-partition, we can construct the matrix one corner by one
corner.
In conclusion, the two numbers are the same.
(3) A is nilpotent and its dimension is 4, so A4=O. Consider the diagonal
elements.
A4(i,i)=(A(i,i))4=0, so A is like A=(0ranranran00ranran000ran0000).
It is obvious that all the
eigenvalues are 0. Consider
Ker(A),Ker(A2),Ker(A3),Ke(A4),
dimension of each is nearly
1,2,3,4 , since each ran is randomly chosen from [−1,1]. The probability 4=1+1+1+1
is 1, others are 0.
\textcolorblueAdvanced Algebra HW6 \textcolorblue2022.4.29
1.6.1 Vandermode matrix
Let V be the space of real
polynomials of degree less than n.
So dimV=n. Then for
each a∈R, the
evaluation eva is a
dual vector.
For any real numbers a1,…,an∈R, consider the map L:V→Rn such that L(p)=[p(a1)⋮p(an)].
- Write out the matrix for L
under the basis 1,x,…,xn−1 for V and the
standard basis for Rn.
(Do you know the name for this matrix?)
- Prove that L is invertible if
and only if a1,…,an
are distinct. (If you can name the matrix L, then you may use its determinant
formula without proof.)
- Show that eva1,…,evan form a basis for V∗ if and only if all a1,…,an are distinct.
- Set n=3. Find polynomials p1,p2,p3 such that pi(j)=δij for i,j∈{−1,0,1}.
- Set n=4, and consider ev−2,ev−1,ev0,ev1,ev2∈V∗.
Since dimV∗=4,
these must be linearly dependent. Find a non-trivial linear combination
of these which is zero.
(1) L=[1a1⋯an−111a2⋯an−12⋮⋮1an⋯an−1n]Vandermode matrix
Calculate its inverse and we can get standard basis using Lagrange
interpolation. (1a1⋯an−111a2⋯an−12⋮⋮1an⋯an−1n)(x0x1x2⋮xn−1)=(y0y1y2⋮yn−1)(1a1⋯an−111a2⋯an−12⋮⋮1an⋯an−1n)−1(y0y1y2⋮yn−1)=(x0x1x2⋮xn−1) Construct polynomial f(a)=∑iyi∏j≠ia−ajai−aj So f(ai)=xi
So (V−1)ij is
the coefficient of ∏k≠ia−akai−ak at xj−1 , which is (V−1)ij=[xj−1]∏k≠ia−akai−ak (V−1)ij=(−1)j+1∑0≤p1<⋯<pn−ζ;p1,p2,⋯pn−j≠ixp1xp2⋯xpn−j∏0≤k<n;k≠i(xk−xi) The column of the (V−1) is the standard basis of V
(2) det
(3) e v_{a_{1}}, \cdots, ev_{a_n} form a
basis for V^{*} \Longleftrightarrow e
v_{a_{1}}, e v_{a_{2}}, \cdots, ev_{a_n}, evan are linearly
indepenctent.
The matrix \left(\begin{array}{c}e
v_{a_{1}} \\ e v_{a_{2}} \\ \vdots \\ e
v_{a_{n}}\end{array}\right) is invertible, which means L is invertible. According to \small(2), a_i are distinct.
(4) Pick original basis \{1,x,x^2\} So \alpha_{1}=(1,-1,1) \quad \alpha_{2}=(1,0,0) \quad
\alpha_{3}=(1,1,1)
A=\left(\begin{array}{l}\alpha_{1} \\ \alpha_{2} \\
\alpha_{3}\end{array}\right)=\left(\begin{array}{ccc}1 & -1 & 1
\\ 1 & 0 & 0 \\ 1 & 1 & 1\end{array}\right) \quad
A^{-1}=\left(\begin{array}{ccc}0 & 1 & 0 \\ -\frac{1}{2} & 0
& \frac{1}{2} \\ \frac{1}{2} & -1 &
\frac{1}{2}\end{array}\right)
So p_{1}=-\dfrac{1}{2}
x+\dfrac{1}{2} x^{2} \quad p_{2}=1-x^{2} \quad p_{3}=\dfrac{1}{2}
x+\dfrac{1}{2} x^{2}
(5) Set p(x)=ax^3+bx^2+cx+d and m e v_{-2}+n e v_{-1}+p e v_{0}+q e v_{1}+r e
v_{2}=0
\begin{gathered}
e v_{-2}=-8 a+4 b-2 c+d \quad e v_{-1}=-a+b-c+d \quad e v_{0}=d\\ \quad
e v_{1}=a+b+c+d \quad e v_{2}=8 a+4 b+2 c+d\\
\left(\begin{array}{ccccc}-8 & -1 & 0 & 1 & 8 \\ 4 &
1 & 0 & 1 & 4 \\ -2 & -1 & 0 & 1 & 2 \\ 1
& 1 & 1 & 1 &1\end{array}\right)\left(\begin{array}{l}m
\\ n \\ p \\ q \\ r\end{array}\right)=\boldsymbol 0
\end{gathered}
Solve that e v_{-2}-4 e v_{-1}+6 e
v_{0}-4 e v_{1}+e v_{2}=0
1.6.2\ \small \mbox{dual vector in
polynomials}
Let V be the space of real
polynomials of degree less than 3.
Which of the following is a dual vector? Prove it or show why not.
- p \mapsto \operatorname{ev}_{5}((x+1)
p(x)).
- p \mapsto \lim _{x \rightarrow \infty}
\dfrac{p(x)}{x}.
- p \mapsto \lim _{x \rightarrow \infty}
\dfrac{p(x)}{x^{2}}.
- p \mapsto p(3)
p^{\prime}(4).
- p \mapsto
\operatorname{deg}(p), the degree of the polynomial p.
(1) Yes. ①\ e_{s}((x+1) p(x))=6 p(5). that is the
map from V to \mathbb{R}. ② For \forall\ p, q \in V and
\forall m, n \in \mathbb{R}, L(m p+n q)=6
m p(5)+6 n p(5)=m L(p)+n L(q), so its bilinear.
(2) No. Sometimes the limit
doeen't exist when \mbox{deg}(p)\geq
2
(3) Yes. This vector is just a
'taking the coefficient of x^2',
which is bilinear.
(4) No. For instance, p(x)=x^{2}+2, q(x)=-2 x, \quad L(p)=11 \times 8=88
\quad L(q)=12
L(p+q)=5 \times 6=30 . \quad L(p+q) \neq
L(p)+L(q). So it's not a dual vector.
(5) No. For instance, p(x)=x^{2}+x, q(x)=-x^{2}+1,
L(p)=\operatorname{deg}(p)=2, L(q)=\operatorname{deg}(q)=2
L(p+q)=\operatorname{deg}(x+1)=1 .
L(p)+L(q) \neq L(p+q). So it's not a clual vector
1.6.3\ \small \mbox{directional
derivative}
Fix a differentiable function f:
\mathbb{R}^{2} \rightarrow \mathbb{R}, and fix a point \boldsymbol{p} \in \mathbb{R}^{2}. For
any vector \boldsymbol{v} \in
\mathbb{R}^{2}, then the directional derivative of f at \boldsymbol{p} in the direction of \boldsymbol{v} is defined as \nabla_{\boldsymbol{v}} f:=\lim _{t \rightarrow 0}
\dfrac{f(\boldsymbol{p}+t \boldsymbol{v})-f(\boldsymbol{p})}{t} . S
\mathrm{Show} that the map \nabla
f: \boldsymbol{v} \mapsto \nabla_{\boldsymbol{v}}(f) is a dual
vector in \left(\mathbb{R}^{2}\right)^{*}, i.e., a
row vector. Also, what are its "coordinates" under the standard dual
basis? (Remark: In calculus, we write \nabla
f as a column vector for historical reasons. By all means, from
a mathematical perspective, the correct way to write \nabla f is to write it as a row vector,
as illlustrated in this problem. (But don't annoy your calculus teachers
though.... In your calculus class, you use whatever notation your
calculus teacher told you.) (Extra Remark: If we use row vector, then
the evaluation of \nabla f at \boldsymbol{v} is purely linear, and no
inner product structure is needed, which is sweet. But if we HAVE TO
write \nabla f as a column vector
(for historical reason), then we would have to do a dot product between
\nabla f and \boldsymbol{v}, which now requires an
inner product structure. That is an unnecessary dependence on an extra
structure that actually should have no influence.)
Set \vec{v}=\begin{pmatrix}a\\b\end{pmatrix},
since the function f is
differentiable, \nabla_{\vec{v}}
f=\dfrac{\partial f(\vec{v})}{\partial x} \cdot
\dfrac{a}{\sqrt{a^{2}+b^{2}}}+\dfrac{\partial f(\vec{v})}{\partial y}
\dfrac{b}{\sqrt{a^{2}+b}}
So the map \nabla f is a map
from \left(\mathbb{R}^{2}\right)^{*} to \mathbb{R}.And for \forall diffential functions f, g, \forall m, n \in \mathbb{R}, we
have:
\begin{aligned}
\nabla_{\vec{v}}(m f+n g) &=\frac{\partial[m f(\vec{v})+n
g(\vec{v})]}{\partial x} \cdot
\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{\partial[m f(\vec{v})+n
g(\vec{v})]}{\partial y} \cdot \frac{b}{\sqrt{a^{2}+b^{2}}} \\
&=m \nabla_{\vec{v}} f+n \nabla_{\vec{v}} g
\end{aligned}
So the map \nabla f is
bilinear, i.e, is a dual vector in \left(\mathbb{R}^{2}\right)^{*}.
Since x-axis and y-axis are perpendicular, the standard
dual basis is \dfrac{\partial f}{\partial
x} and \dfrac{\partial f}{\partial
y} . And the coordinates are \dfrac{\vec{v} \cdot
\hat{x}}{\|\vec{v}\|} and \dfrac{\vec{v} \cdot
\hat{y}}{\|\vec{v}\|}
1.6.4
Consider a linear map L: V \rightarrow
W and its dual map L^{*}: W^{*}
\rightarrow V^{*}. Prove the following.
- \operatorname{Ker}\left(L^{*}\right) is
exactly the collection of dual vectors in W^{*} that kills \operatorname{Ran}(L).
- \operatorname{Ran}\left(L^{*}\right) is
exactly the collection of dual vectors in V^{*} that kills \operatorname{Ker}(L).
(1) First, all the elements in
\ker(L^*)\in W^{*}, so just prove
\forall \ \vec{v}\in
\ker(L^*),\vec{\omega}\in \mbox{Ran}(L),\vec{v}^{T}\cdot
\vec{\omega}=0
L^{*}\vec{v}=0,L^{*}\vec{v}\cdot\vec{x}=0=\vec{v}^{T}(L\vec{x}),
set L\vec{x}=\vec{\omega} So \vec{v}^{T}\cdot \vec{\omega}=\langle
\vec{v},\vec{\omega}\rangle=0
(2) First, all the elements in
\mbox{Ran}(L^{*})\in V^{*} ,so just
prove \forall \ \vec{v}\in
\mbox{Ran}(L^*),\vec{\omega}\in \mbox{Ker}(L),\vec{v}^{T}\cdot
\vec{\omega}=0
L\vec{\omega}=0,L\vec{\omega}\cdot\vec{x}=0=\vec{\omega}^{T}(L^{*}\vec{x}),
set L^{*}\vec{x}=\vec{v}, So \vec{\omega}^{T}\cdot \vec{v}=\langle
\vec{\omega},\vec{v}\rangle=0
\large\textcolor{blue}{\mbox{Advanced Algebra }
\small \mathbb{HW}\mathrm{7}}\ \ \ \ \ \
_\textcolor{blue}{2022.5.3}
1.7.1\ \small \mbox{bra map and Riesz
map}
On the space \mathbb{R}^{n}, we
fix a symmetric positive-definite matrix A, and define (\boldsymbol{v},
\boldsymbol{w})=\boldsymbol{v}^{\mathrm{T}}A\boldsymbol{w}
- Show that this is an inner product.
- The Riesz map (inverse of the bra map) from V^{*} to V would send a row vector \boldsymbol{v}^{\mathrm{T}} to what?
- The bra map from V to V^{*} would send a vector \boldsymbol{v} to what?
- The dual of the Riesz map from V^{*} to V would send a row vector \boldsymbol{v}^{\mathrm{T}} to what?
(1) Easy to prove.
(2) The bra map means \forall \ \langle \boldsymbol{v}|\in
\mathcal{B},s.t.\langle
\boldsymbol{v}|:\boldsymbol{\omega\longmapsto}\langle\boldsymbol{v},\boldsymbol{\omega}\rangle=\boldsymbol{v}^{\mathrm{T}}A\boldsymbol{w}
So the inverse of \langle
\boldsymbol{v}| is make \boldsymbol{v}^{\mathrm{T}}A
to \boldsymbol {v}, which is
from V^{*} to V. Set \boldsymbol{u}=\boldsymbol {v}^{T}A, so
\boldsymbol{v}^{T}=\boldsymbol{u}A^{-1}
then transpose it we can get \boldsymbol{v}=(A^{-1})^{T}\boldsymbol{u^{T}}
So if the input is \boldsymbol{v}^{T}\in
V^{*}, the output is (A^{-1})^{T}(\boldsymbol
v^{T})^{T}=(A^{-1})^{T}\boldsymbol{v}\in V
(3) Obviously, it sends \boldsymbol{v} to \boldsymbol{v}^{T}A
(4) I guess the result is the
same with \small(2), which sends
\boldsymbol {v}^{T} to (A^{-1})^{T}\boldsymbol{v}
1.7.2\ \small \mbox{What is a
derivative}
The discussions in this problem holds for all manifolds M. But for simplicities sake, suppose
M=\mathbb{R}^{3} for this
problem.
Let V be the space of all
analytic functions from M to \mathbb{R}. Here analytic means f(x, y, z) is a infinite polynomial
series (its Taylor expansion) with variables x, y, z. Approximately f(x, y, z)=a_{0}+a_{1} x+a_{2} y+ a_{3} z+a_{4} x^{2}+a_{5} x y+a_{6} x z+a_{7}
y^{2}+\ldots, and things should converge always.
Then a dual vector v \in V^{*}
is said to be a "derivation at \boldsymbol{p} \in M^{\prime \prime} if
it satisfy the following Leibniz rule (or product rule):
v(f g)=f(\boldsymbol{p}) v(g)+g(\boldsymbol{p}) v(f) .
(Note the similarity with your traditional product rule (f g)^{\prime}(x)=f(x) g^{\prime}(x)+g(x)
f^{\prime}(x).) Prove the following:
- Constant functions in V must be
sent to zero by all derivations at any point.
- Let x, y, z \in V be the
coordinate function. Suppose \boldsymbol{p}=\left[\begin{array}{l}p_{1} \\
p_{2} \\ p_{3}\end{array}\right], then for any derivation v at \boldsymbol{p}, then we have v\left(\left(x-p_{1}\right)
f\right)=f(\boldsymbol{p}) v(x), v\left(\left(y-p_{2}\right)
f\right)=f(\boldsymbol{p}) v(y) and v\left(\left(z-p_{3}\right)
f\right)=f(\boldsymbol{p}) v(z).
- Let x, y, z \in V be the
coordinate function. Suppose \boldsymbol{p}=\left[\begin{array}{l}p_{1} \\
p_{2} \\ p_{3}\end{array}\right], then for any derivation v at \boldsymbol{p}, then we have v\left(\left(x-p_{1}\right)^{a}\left(y-p_{2}\right)^{b}\left(z-p_{3}\right)^{c}\right)=0
for any non-negative integers a, b,
c such that a+b+c>1.
- Let x, y, z \in V be the
coordinate function. Suppose \boldsymbol{p}=\left[\begin{array}{l}p_{1} \\
p_{2} \\ p_{3}\end{array}\right], then for any derivation v at \boldsymbol{p}, v(f)= \dfrac{\partial f}{\partial x}(\boldsymbol{p})
v(x)+\dfrac{\partial f}{\partial y}(\boldsymbol{p}) v(y)+\dfrac{\partial
f}{\partial z}(\boldsymbol{p}) v(z). (Hint: use the Taylor
expansion of f at \left.\boldsymbol{p} .\right)
- Any derivation v at \boldsymbol{p} must be exactly the
directional derivative operator \nabla_{\boldsymbol{v}} where \boldsymbol{v}=\left[\begin{array}{l}v(x) \\ v(y)
\\ v(z)\end{array}\right]. (Remark: So, algebraically speaking,
tangent vectors are exactly derivations, i.e., things that satisfy the
Leibniz rule.)
(1) Set f(\vec{p})\equiv a, because v(f g)=f(\boldsymbol{p}) v(g)+g(\boldsymbol{p})
v(f) and v\in V^{*} is
linear, v(fg)=av(g)+g(\boldsymbol{p})v(f)
Set g=f, so v(ff)=2av(f)=v(af)=av(f) If a=0, v(g)=v(0)=v(gg)=0+0=0 If a\neq 0
av(f)=0, and v(f)=0 So constant functions in V are all sent to 0
(2) Calculate v((x-p_1)f)=f(\boldsymbol
{p})v(x-p_1)+(x-p_1)(\boldsymbol{p})v(f) And x(\boldsymbol{p})=p_1,p_1(\boldsymbol{p})=p_1
So (x-p_1)(\boldsymbol{p})=0 For
constant \boldsymbol{p}, v(\boldsymbol{p}=0) So v((x-p_1)f)=f(\boldsymbol{p})(v(x)-v(p_1))=f(\boldsymbol{p})v(x)
(3) As a,b,c are all integers, if a,b,c<1, the sum of them is a+b+c\leq 0, contradiction
So at least one intergers is \geq
1, without loss of generalization, assume a\geq 1
v\left(\left(x-p_{1}\right)^{a}\left(y-p_{2}\right)^{b}\left(z-p_{3}\right)^{c}\right)=v(x-p_1)((x-p_1)^{a-1}(y-p_2)^b(z-p_3)^c)(\boldsymbol{p})
And if a> 1, (x-p_1)^{a-1}(\boldsymbol{p})=(x-p_1)^{a-2}(x-p_1)(\boldsymbol{p})=0
if a=1, then b+c>0
at least one integer in b and
c \geq
1, without of generalization, assume b\geq 1. Then
(y-p_2)^{b}(\boldsymbol{p})=(y-p_2)^{b-1}(y-p_2)(\boldsymbol{p})=0
So v\left(\left(x-p_{1}\right)^{a}\left(y-p_{2}\right)^{b}\left(z-p_{3}\right)^{c}\right)=0
(4) Use Taylor expansion for
f, we have
f(x, y, z)=f\left(p_{1}, p_{2}, p_{3}\right)+\frac{\partial f}{\partial
x}(\boldsymbol{p})\left(x-p_{1}\right)+\frac{\partial f}{\partial
y}(\boldsymbol{p})\left(y-p_{2}\right)+\frac{\partial f}{\partial
z}(\boldsymbol{p})\left(z-p_{3}\right)+o(|\boldsymbol{r}-\boldsymbol{p}|)
According to (3), the
remainder o(|\boldsymbol{r}-\boldsymbol{p}|)=\left(x-p_{1}\right)^{a}\left(y-p_{2}\right)^{b}\left(z-p_{3}\right)^{c},a+b+c>1
So v(o(|\boldsymbol{r}-\boldsymbol{p}|))=0
For constant number v sends to
0. Take v function to the Taylor expansion
v(f)=\dfrac{\partial f}{\partial x}(\boldsymbol{p}) v(x)+\dfrac{\partial
f}{\partial y}(\boldsymbol{p}) v(y)+\dfrac{\partial f}{\partial
z}(\boldsymbol{p}) v(z)
Which shows the complete differential at \boldsymbol{p}
(5) Just calculate the
directional derivative \nabla_{\boldsymbol{v}} where \boldsymbol{v}=\left[\begin{array}{l}v(x) \\ v(y)
\\ v(z)\end{array}\right]
\begin{aligned}
\nabla_{v} f &=\lim _{t \rightarrow 0^{+}} \frac{f(\boldsymbol{p}+t
\boldsymbol{v})-f(\boldsymbol{p})}{t} \\
&=\lim _{t \rightarrow 0^{+}}
\dfrac{f(\boldsymbol{p})+\dfrac{\partial f}{\partial x}(\boldsymbol{p})
v(x) t+\dfrac{\partial f}{\partial y}(\boldsymbol{p}) v(y)
t+\dfrac{\partial f}{\partial z}(\boldsymbol{p}) v(z)
t+o(\|\boldsymbol{v}\| t)-f(\boldsymbol{p})}{t} \\
&=\frac{\partial f}{\partial x}(\boldsymbol{p}) v(x)+\frac{\partial
f}{\partial y}(\boldsymbol{p}) v(y)+\frac{\partial f}{\partial
z}(\boldsymbol{p}) v(z) \\
&=v(f)
\end{aligned}
So derivative in calculus is just a vector in the dual space
of R^{n} which suits Leibniz
rule
1.7.3\ \small \mbox{What is a vector
field}
The discussions in this problem holds for all manifolds M. But for simplicities sake, suppose
M=\mathbb{R}^{3} for this problem.
Let V be the space of all analytic
functions from M to \mathbb{R} as usual. We say X: V \rightarrow V is a vector field on
X if X(f g)=f X(g)+g X(f), i.e., the Leibniz
rule again! Prove the following:
Show that X_{\boldsymbol{p}}: V
\rightarrow \mathbb{R} such that X_{\boldsymbol{p}}(f)=(X(f))(\boldsymbol{p})
is a derivation at \boldsymbol{p}.
(Hence X is indeed a vector field,
since it is the same as picking a tangent vector at each
point.)
Note that each f on M induces a covector field \mathrm{d} f. Then at each point \boldsymbol p, the cotangent vector \mathrm{d} f and the tangent vector X would evaluate to some number. So \mathrm{d} f(X) is a function M \rightarrow \mathbb{R}. Show that \mathrm{d} f(X)=X(f), i.e., the two are
the same. (Hint: just use definitions and calculate directly.)
If X, Y: V \rightarrow V are
vector fields, then note that X \circ Y: V
\rightarrow V might not be a vector field. (Leibniz rule might
fail.) However, show that X \circ Y-Y \circ
X is always a vector field.
On a related note, show that if A,
B are skew-symmetric matrices, then A B-B A is still skewsymmetric.
(Skew-symmetric matrices actually corresponds to certain vector fields
on the manifold of orthogonal matrices. So this is no
coincidence.)
(1) X_{\boldsymbol{p}}(fg)=(X(fg))(\boldsymbol{p})=(fX(g)+gX(f))(\boldsymbol{p})=f(\boldsymbol{p})X(g)(\boldsymbol{p})+g(\boldsymbol{p})X(f)(\boldsymbol{p})
=f(\boldsymbol{p})X_{\boldsymbol{p}}(g)+g(\boldsymbol{p})X_{\boldsymbol{p}}(f)
So it suits the definition of derivative at \boldsymbol{p}
(2) X_{\boldsymbol{p}}:V\rightarrow R So
X_{\boldsymbol{p}}\in V^{*} X_{p}=\left[\begin{array}{c}
X_{\boldsymbol{p}}(X) &
X_{p}(Y) &
X_{p}(Z)
\end{array}\right]
df(X)(\boldsymbol{p})=df_{\boldsymbol{p}
}(X_{\boldsymbol{p}}) where df=\left[\begin{array}{c}
\dfrac{\partial f}{\partial x}(\boldsymbol{p})\\
\dfrac{\partial f}{\partial y}(\boldsymbol{p})\\
\dfrac{\partial f}{\partial z}(\boldsymbol{p})\\
\end{array}\right] combine them together then we have
From \small (1.4) (X(f))(\boldsymbol{p})=X_{p}(f)=\dfrac{\partial
f}{\partial x}(\boldsymbol{p}) X_{\boldsymbol{p}}(X)+\dfrac{\partial
f}{\partial y}(\boldsymbol{p}) X_{\boldsymbol{p}}(Y)+\dfrac{\partial
f}{\partial z}(\boldsymbol{p}) X_{\boldsymbol{p}}(Z)
=\left[\begin{array}{c}
X_{\boldsymbol{p}}(X) &
X_{p}(Y) &
X_{p}(Z)
\end{array}\right]\left[\begin{array}{c}
\dfrac{\partial f}{\partial x}(\boldsymbol{p})\\
\dfrac{\partial f}{\partial y}(\boldsymbol{p})\\
\dfrac{\partial f}{\partial z}(\boldsymbol{p})\\
\end{array}\right]=df(X)
(3) Just unfold the fomula
\begin{gathered}
(X\circ Y-Y\circ X)(fg)=X\circ Y(fg)-Y\circ X(fg)=X\circ
(fY(g)+gY(f))-Y\circ (fX(g)+gX(f))\\
=\small f (X\circ Y)g+X(f)Y(g)+g (X\circ Y)f+X(g)Y(f)-(f (Y\circ
X)g+Y(f)X(g)+g (Y\circ X)f+Y(g)X(f))\\
=f(X\circ Y-Y\circ X)g+g(X\circ Y-Y\circ X)f
\end{gathered}
So X\circ Y-Y \circ X
suits the Leibniz rule, which is a vector field.
(4) Calculate directly we can
prove (AB-BA)^{T}=-(AB-BA)
\begin{aligned}
(A B-B A)^{T} =(A B)^{T}-(B A)^{T}
=(B^{T} A^{T}-A^{T} B^{T}) \\
=(-B)(-A)-(-A)(-B) =BA-AB=-(A B-B A)
\end{aligned}
For all the positive orthogonal matrices \mathcal{Q} \forall \ Q\in \mathcal{Q},\det(Q)=1. At
one matrix, its tangent vector
(Q+A)^{T}(Q+A)=I=(Q^{T}+A^{T})(Q+A)=I\Longrightarrow
A=-A^{T},||A||\to 0
are all Skew-symmetric matrices. According to \small (3), AB-BA is also a Skew-symmetric matrix
\Large \mathbf{If\ your\ life\ is\
tense,\ it\ could\ be\ a\ tensor. }
\large\textcolor{blue}{\mbox{Advanced Algebra }
\small \mathbb{HW}\mathrm{8}}\ \ \ \ \ \
_\textcolor{blue}{2022.5.9}
1.8.1\ \small \mbox{Elementary layer operations for
tensors}
Note that, for "2D" matrices we have row and column operations, and
the two kinds of operations corresponds to the two dimensions of the
array.
For simplicity, let M be a 2 \times 2 \times 2 "3D matrix". Then we
have "row layer operations", "column layer operations", "horizontal
layer operations". The three kinds corresponds to the three dimensions
of the array. We interpret this as a multilinear map M: \mathbb{R}^{2} \times \mathbb{R}^{2} \times
\mathbb{R}^{2} \rightarrow \mathbb{R}. Let \left(\left(\mathbb{R}^{2}\right)^{*}\right)^{\otimes
3} be the space of all multilinear maps from \mathbb{R}^{2} \times \mathbb{R}^{2} \times
\mathbb{R}^{2} to \mathbb{R}.
- Given \alpha, \beta, \gamma
\in\left(\mathbb{R}^{2}\right)^{*}, what is the (i, j, k)-entry of the "3D matrix" \alpha \otimes \beta \otimes \gamma in
terms of the coordinates of \alpha, \beta,
\gamma ? Here \alpha \otimes \beta
\otimes \gamma is the multilinear map sending (\boldsymbol{u}, \boldsymbol{v},
\boldsymbol{w}) to the real number \alpha(\boldsymbol{u}) \beta(\boldsymbol{v})
\gamma(\boldsymbol{w}).
- Let E be an elementary matrix.
Then we can send \alpha \otimes \beta
\otimes \gamma to (\alpha E)
\otimes \beta \otimes \gamma. Why can this be extended to a
linear map M_{E}:\left(\left(\mathbb{R}^{2}\right)^{*}\right)^{\otimes
3} \rightarrow\left(\left(\mathbb{R}^{2}\right)^{*}\right)^{\otimes
3} ? (This gives a formula for the "elementary layer
operations" on "3D matrices", where the three kinds of layer operations
corresponds to applying E to the
three arguments respectively.)
- Show that elementary layer operations preserve rank. Here we say
M has rank r if r is the smallest possible integer such
that M can be written as the linear
combination of r "rank one" maps,
i.e., maps of the kind \alpha \otimes \beta
\otimes \gamma for some \alpha,
\beta, \gamma \in\left(\mathbb{R}^{2}\right)^{*}.
- Show that, if some "2D" layer matrix of a "3D matrix" has rank r,
then the 3 D matrix has rank at
least r.
- Let M be made of two layers,
\left[\begin{array}{ll}1 & 0 \\ 0 &
1\end{array}\right] and \left[\begin{array}{ll}0 & 1 \\ 1 &
0\end{array}\right]. Find its rank.
- (Read only) Despite some practical interests, finding the tensor
rank in general is NOT easy. In fact, it is NP-complete just for
3-tensors over finite field. Furthermore, a tensor with all real entries
might have different real rank and complex rank.
(1) According to the symmetry of
dot product \langle\alpha,u\rangle=\langle
u,\alpha\rangle we have equation \alpha^Tu=u^T\alpha So
\alpha^{T} u \beta^{T} v \gamma^{T}
\omega=u^{T} \alpha \beta^{T} v \gamma^{T} \omega=u^{T}\left[\alpha
\beta^{T} v \gamma_{1} \quad \partial \beta^{T} v \gamma_{2}\right]
\omega Compare to [u^TA_1v\quad
u^TA_2v]\omega hence
A_{1}=\gamma_{1}\left(\begin{array}{ll}\alpha_{1}
\beta_{1} & \alpha_{1} \beta_{2} \\ \alpha_{2} \beta_{1} &
\alpha_{2}
\beta_{2}\end{array}\right)=\gamma_1\alpha\beta^T,A_{2}=\gamma_{2}\left(\begin{array}{ll}\alpha_{1}
\beta_{1} & \alpha_{1} \beta_{2} \\ \alpha_{2} \beta_{1} &
\alpha_{2} \beta_{2}\end{array}\right)=\gamma_2\alpha\beta^T .
So A_{ijk}=\gamma_i\alpha_j\beta_k
(2) M_E sends tensor M=[[A_1,A_2]]=[[\gamma_1\alpha
\beta^T,\gamma_2\alpha \beta^T]] to M'=[[\gamma_1\alpha E \beta^T,\gamma_2\alpha
E\beta^T]]
For \alpha _1,\alpha_2 in (\mathbb{R}^2)^* as one part in ((\mathbb{R}^2)^*)^{\otimes3}
M_{k\alpha_1+\mu\alpha_2}=[[\gamma_1(k\alpha_1+\mu\alpha_2)
\beta^T,\gamma_2(k\alpha_1+\mu\alpha_2) \beta^T]]=kM_{\alpha _1}+\mu
M_{\alpha _2}\in ((\mathbb{R}^2)^*)^{\otimes3}
So for \alpha ,\beta is
linear. and for \gamma_1,\gamma_2
in (\mathbb{R}^2)^* as one part in
((\mathbb{R}^2)^*)^{\otimes3}, set
\gamma_{11,12} as the component of
\gamma_1
\gamma_{21,22} as the component
of \gamma_2, So
M_{k\gamma_1+\mu\gamma_2}=[[(k\gamma_{11}+\mu\gamma_{21})\alpha\beta^T,(k\gamma_{12}+\mu\gamma_{22})\alpha
\beta^T]]=kM_{\gamma _1}+\mu M_{\gamma _2}\in
((\mathbb{R}^2)^*)^{\otimes3}
So M_E is a linear map,
three operations at \alpha,\beta,\gamma
(3) Suppose M=\displaystyle \sum_{i=1}^rM_{base(i)}
if we operate elementary layer operations for M, the right hand side
is also in "rank one" maps, so r'\leq
r. And if r'<r,
i.e., M'=\displaystyle
\sum_{i=1}^{r'}M_{base(i)} As elementary have its
inverse
operate the inverse of elementary operation, and we have
contradiction, so r'=r
(4) According to \mbox{SVD}, the minimum number of
decomposing a matrix into rank-1
matrixes equals to rank
So if 2D matrix needs at least
r rank-1 matrixes to make up, since every
rank-1 maps in \alpha \otimes \beta \otimes \gamma
contains two rank-1 matrix, so
the 3D matrix also needs at least
r rank-1 tensors to make up
(5) For A_1=\left[\begin{array}{ll}1 & 0 \\ 0 &
1\end{array}\right] its rank is 2, so r(M)\geq 2. Besides, construct two
rank-1 tensors
M_1=[[\dfrac{1}{2}\begin{bmatrix}1 & 1 \\ 1 &
1\end{bmatrix},\dfrac{1}{2}\begin{bmatrix}1 & 1 \\ 1 &
1\end{bmatrix}]],
M_2=[[\dfrac{1}{2}\begin{bmatrix}1 & -1 \\ -1 &
1\end{bmatrix},-\dfrac{1}{2}\begin{bmatrix}1 & -1 \\ -1 &
1\end{bmatrix}]]
And M=M_1+M_2, so its
rank is 2

1.8.2\
\small \mbox{i+j+k rank-3 tensor}
Let M be a 3 \times 3 \times 3 "3D matrix" whose
(i, j, k)-entry is i+j+k. We interpret this as a multilinear
map M: \mathbb{R}^{3} \times \mathbb{R}^{3}
\times \mathbb{R}^{3} \rightarrow \mathbb{R}.
- Let \boldsymbol{v}=\left[\begin{array}{l}x \\ y \\
z\end{array}\right], then M(\boldsymbol{v}, \boldsymbol{v},
\boldsymbol{v}) is a polynomial in x, y, z. What is this polynomial?
- Let \sigma:\{1,2,3\}
\rightarrow\{1,2,3\} be any bijection. Show that M\left(\boldsymbol{v}_{1}, \boldsymbol{v}_{2},
\boldsymbol{v}_{3}\right)=M\left(\boldsymbol{v}_{\sigma(1)},
\boldsymbol{v}_{\sigma(2)}, \boldsymbol{v}_{\sigma(3)}\right).
(Hint: brute force works. But alternatively, try find the (i, j, k) entry of the multilinear map
M^{\sigma}, a map that sends \left(\boldsymbol{v}_{1}, \boldsymbol{v}_{2},
\boldsymbol{v}_{3}\right) to M\left(\boldsymbol{v}_{\sigma(1)},
\boldsymbol{v}_{\sigma(2)},
\boldsymbol{v}_{\sigma(3)}\right).)
- Show that the rank r of M is at least 2 and at most 3. (It is
actually exactly three.)
- (Read only) Any study of polynomial of degree d on n variables is equivalent to the study of
some symmetric d tensor on \mathbb{R}^{n}.
(1) M=[[\begin{bmatrix}3 & 4 & 5 \\ 4 & 5
& 6\\5& 6&7\end{bmatrix},\begin{bmatrix}4 & 5 & 6 \\
5 & 6 & 7\\6& 7&8\end{bmatrix},\begin{bmatrix}5 & 6
& 7 \\ 6 & 7 & 8\\7&
8&9\end{bmatrix}]]=[[A_1,A_2,A_3]] And M(\boldsymbol v,\boldsymbol v,\boldsymbol
v) where \boldsymbol{v}=\begin{pmatrix}x\\y\\z\end{pmatrix}
=[\boldsymbol v^TA_1\boldsymbol v\quad
\boldsymbol v^TA_2\boldsymbol v\quad \boldsymbol v^TA_3\boldsymbol
v]\boldsymbol v=[\boldsymbol v^T\begin{bmatrix}3 & 4 & 5 \\ 4
& 5 & 6\\5& 6&7\end{bmatrix}\boldsymbol v\quad
\boldsymbol v^T\begin{bmatrix}4 & 5 & 6 \\ 5 & 6 &
7\\6& 7&8\end{bmatrix}\boldsymbol v\quad \boldsymbol
v^T\begin{bmatrix}5 & 6 & 7 \\ 6 & 7 & 8\\7&
8&9\end{bmatrix}\boldsymbol v]\boldsymbol v
Calculate it by \mbox{Mathematica}, the result is p(x,y,z)=3 (x+y+z)^2 (x+2 y+3 z)

(2) let a linear map from \mathbb{R}^{3} \times \mathbb{R}^{3} \times
\mathbb{R}^{3} to \mathbb{R} sends \left(\boldsymbol{v}_{1}, \boldsymbol{v}_{2},
\boldsymbol{v}_{3}\right) to M\left(\boldsymbol{v}_{\sigma(1)},
\boldsymbol{v}_{\sigma(2)},
\boldsymbol{v}_{\sigma(3)}\right)
Obviously, it's multi-linear since for \boldsymbol{v}_i the evaluation result is
linear no matter which position \boldsymbol{v}_i is.
And this map have a tensor such M'=[[A_1',A_2',A_3']].
Specialise \boldsymbol{v_i} to get
value of A_{1}',A_{2}',A_{3}'
Set \boldsymbol{b}_1=\begin{pmatrix}1\\0\\0\end{pmatrix},\boldsymbol{b}_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\boldsymbol{b}_3=\begin{pmatrix}0\\0\\1\end{pmatrix}
let (\boldsymbol{v_1},\boldsymbol{v_2},\boldsymbol{v_3})=(\boldsymbol{b_i},\boldsymbol{b_j},\boldsymbol{b_k})
where 1\leq i,j,k\leq 3
and they can be the same. Put one of condition into the map so
A_{k(ij)}'=A_{\sigma^{-1}(i)(\sigma^{-1}(j)\sigma^{-1}(k))}=\sigma^{-1}(i)+\sigma^{-1}(j)+\sigma^{-1}(k)=i+j+k
So for k=1,2,3 A_{k}'=A_k, which implies that M'=M, so the equation is proved.
A brute try (failed):
It is obvious that swaping \boldsymbol{v}_{i},\boldsymbol{v}_{j} at
most twice can make \boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3
$$ \boldsymbol{v}_{\sigma(1)},
\boldsymbol{v}_{\sigma(2)}, \boldsymbol{v}_{\sigma(3)}
If swap \boldsymbol{v}_{1},\boldsymbol{v}_{2},
the value \boldsymbol v_1^TA_i\boldsymbol
v_2=\boldsymbol v_2^TA_i\boldsymbol v_1,i=1,2,3, since A_i=A_i^T so the result stays the
same.
And if we swap \boldsymbol{v}_{2},\boldsymbol{v}_{3},
[\boldsymbol v_1^TA_1\boldsymbol v_2\quad
\boldsymbol v_1^TA_2\boldsymbol v_2\quad \boldsymbol v_1^TA_3\boldsymbol
v_2]\boldsymbol v_3=\boldsymbol v_1^TA_1\boldsymbol
v_2v_{3x}+\boldsymbol v_1^TA_2\boldsymbol v_2v_{3y}+\boldsymbol
v_1^TA_3\boldsymbol v_2v_{3z}
(3) According to 1.8.1(4), since \mbox{rank}(A_i)=2, so r\geq 2. Then just construct a reasonable
combination
I guess \alpha=\begin{pmatrix}1\\1\\-1\end{pmatrix},\beta=\begin{pmatrix}1\\-1\\1\end{pmatrix},\gamma=\begin{pmatrix}-1\\1\\1\end{pmatrix}
so the rank-1 matrix like
\begin{bmatrix}a&b&c\\a&b&c\\-a&-b&-c\end{bmatrix},\begin{bmatrix}d&e&f\\-d&-e&-f\\d&e&f\end{bmatrix},\begin{bmatrix}-g&-h&-i\\g&h&i\\g&h&i\end{bmatrix},
And the linear combinations of these three matrixes are \begin{bmatrix}3 & 4 & 5 \\ 4 & 5
& 6\\5& 6&7\end{bmatrix},\begin{bmatrix}4 & 5 & 6 \\
5 & 6 & 7\\6& 7&8\end{bmatrix},\begin{bmatrix}5 & 6
& 7 \\ 6 & 7 & 8\\7& 8&9\end{bmatrix}
which transfers to three nine - dimensional equations, out of my hand
ability, the coefficient matrix is
A=\left(
\begin{array}{ccccccccc}
1 & 0 & 0 & 1 & 0 & 0 & -1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & -1 \\
0 & 1 & 0 & 0 & -1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & -1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1 & 0 & 0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & -1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & -1 & 0 & 0 & 1 & 0 & 0 & 1 \\
\end{array}
\right),A\vec{x}=\vec{b}_{i}=\begin{pmatrix}3\\4\\5\\4\\5\\6\\5\\6\\7\end{pmatrix},\begin{pmatrix}4\\5\\6\\5\\6\\7\\6\\7\\8\end{pmatrix},\begin{pmatrix}5\\6\\7\\6\\7\\8\\7\\8\\9\end{pmatrix}
And the accurate solution is
\begin{gathered}
\begin{bmatrix}3 & 4 & 5 \\ 4 & 5 & 6\\5&
6&7\end{bmatrix}=\begin{bmatrix}3.5&4.5&5.5\\3.5&4.5&5.5\\-3.5&-4.5&-5.5\end{bmatrix}+
\begin{bmatrix}4&5&6\\-4&-5&-6\\4&5&6\end{bmatrix}+\begin{bmatrix}-4.5&-5.5&-6.5\\4.5&5.5&6.5\\4.5&5.5&6.5\end{bmatrix}\\
\begin{bmatrix}4 & 5 & 6 \\ 5 & 6 & 7\\6&
7&8\end{bmatrix}=\begin{bmatrix}4.5&5.5&6.5\\4.5&5.5&6.5\\-4.5&-5.5&-6.5\end{bmatrix}+
\begin{bmatrix}5&6&7\\-5&-6&-7\\5&6&7\end{bmatrix}+\begin{bmatrix}-5.5&-6.5&-7.5\\5.5&6.5&7.5\\5.5&6.5&7.5\end{bmatrix}\\
\begin{bmatrix}5 & 6 & 7 \\ 6 & 7 & 8\\7&
8&9\end{bmatrix}=\begin{bmatrix}5.5&6.5&7.5\\5.5&6.5&7.5\\-5.5&-6.5&-7.5\end{bmatrix}+
\begin{bmatrix}6&7&8\\-6&-7&-8\\6&7&8\end{bmatrix}+\begin{bmatrix}-6.5&-7.5&-8.5\\6.5&7.5&8.5\\6.5&7.5&8.5\end{bmatrix}\\
\end{gathered}
So it can be decomposed into three rank-1 tensors. So r\leq 3