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Multilinear transformations being determined by their values on basis elements
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The following is stated in the book Analysis on Manifolds by James Munkres
Just as is the case with linear transformations, a multilinear transformation
is entirely determined once one knows its values on basis elements. That
we now prove.
And then he gives the following lemma.
Lemma 26.2 Let $a_1, dots, a_n$ be a basis for $V$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$fleft(a_i_1, dots, a_i_kright) = gleft(a_i_1, dots, a_i_kright) $$ for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$ then $f=g$.
Now what I don't understand is why to we even need the following in the above lemma
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Because $V^k$ has dimension $k cdot n$ since $V$ has dimension $n$, and has as basis elements beginalign*&(a_1, 0, dots, 0), dots, &(a_n,0, dots, 0), \
&(0, a_1, dots, 0), dots, &(0, a_n, dots, 0), \
&. &. \
&. &. \
&. &. \
&(0, 0, dots, a_1), dots, &(0, 0, dots, a_n)
endalign*
So if $mathcalB$ was the set of basis elements of $V^k$ above then I'd say that the following proposed lemma would make more sense
Proposed Lemma: Let $V$ be a vector space of dimension $n$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$f(alpha) = g(alpha)$$ for every $alpha in mathcalB$ where $mathcalB$ is a basis for $V^k$, then $f= g$
Furthermore the same part of Lemma 26.2
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Taken literally gives $n^k$ possible $k$-tuples, which would correspond to checking to see if the values of $f$ and $g$ agree on $n^k$ basis elements, which confuses me since $dim(V^k) = kn$
I'm sure that Lemma 26.2 must be correct and I'm just making some error somewhere, if so could someone please point out what that error is.
linear-algebra vector-spaces linear-transformations tensors
asked Jul 20 at 18:09
Perturbative
3,51411039
add a comment |Â
up vote
1
down vote
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The following is stated in the book Analysis on Manifolds by James Munkres
Just as is the case with linear transformations, a multilinear transformation
is entirely determined once one knows its values on basis elements. That
we now prove.
And then he gives the following lemma.
Lemma 26.2 Let $a_1, dots, a_n$ be a basis for $V$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$fleft(a_i_1, dots, a_i_kright) = gleft(a_i_1, dots, a_i_kright) $$ for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$ then $f=g$.
Now what I don't understand is why to we even need the following in the above lemma
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Because $V^k$ has dimension $k cdot n$ since $V$ has dimension $n$, and has as basis elements beginalign*&(a_1, 0, dots, 0), dots, &(a_n,0, dots, 0), \
&(0, a_1, dots, 0), dots, &(0, a_n, dots, 0), \
&. &. \
&. &. \
&. &. \
&(0, 0, dots, a_1), dots, &(0, 0, dots, a_n)
endalign*
So if $mathcalB$ was the set of basis elements of $V^k$ above then I'd say that the following proposed lemma would make more sense
Proposed Lemma: Let $V$ be a vector space of dimension $n$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$f(alpha) = g(alpha)$$ for every $alpha in mathcalB$ where $mathcalB$ is a basis for $V^k$, then $f= g$
Furthermore the same part of Lemma 26.2
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Taken literally gives $n^k$ possible $k$-tuples, which would correspond to checking to see if the values of $f$ and $g$ agree on $n^k$ basis elements, which confuses me since $dim(V^k) = kn$
I'm sure that Lemma 26.2 must be correct and I'm just making some error somewhere, if so could someone please point out what that error is.
linear-algebra vector-spaces linear-transformations tensors
asked Jul 20 at 18:09
Perturbative
3,51411039
take a look at how this construction looks for $V=Bbb R^1$ or $V=Bbb R^2$ or $V=Bbb R^3$ with $k=1$ or $=2$ or $k=3$
– janmarqz
Jul 20 at 18:33
@janmarqz Okay so take the case $V = mathbbR^2$ and $k=2$, suppose $f, g : V^2 = mathbbR^4 to mathbbR$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\ f(e_1, e_2) = g(e_1, e_2);\ f(e_2, e_1) = g(e_2, e_1); \ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $mathbbR^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $mathbbR^4$
– Perturbative
Jul 20 at 19:48
So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$
– Perturbative
Jul 20 at 19:58
for bilinear maps $Bbb R^2timesBbb R^2toBbb R$ (which is a vector space) you only need the 4 basic "vectors": $$left(beginarraycc1&0\0&0endarrayright)$$ $$left(beginarraycc0&1\0&0endarrayright)$$ $$left(beginarraycc0&0\1&0endarrayright)$$ $$left(beginarraycc0&0\0&1endarrayright)$$
– janmarqz
Jul 20 at 20:23
@janmarqz But that's not what the theorem above asserts..
– Perturbative
Jul 20 at 21:29
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The following is stated in the book Analysis on Manifolds by James Munkres
Just as is the case with linear transformations, a multilinear transformation
is entirely determined once one knows its values on basis elements. That
we now prove.
And then he gives the following lemma.
Lemma 26.2 Let $a_1, dots, a_n$ be a basis for $V$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$fleft(a_i_1, dots, a_i_kright) = gleft(a_i_1, dots, a_i_kright) $$ for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$ then $f=g$.
Now what I don't understand is why to we even need the following in the above lemma
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Because $V^k$ has dimension $k cdot n$ since $V$ has dimension $n$, and has as basis elements beginalign*&(a_1, 0, dots, 0), dots, &(a_n,0, dots, 0), \
&(0, a_1, dots, 0), dots, &(0, a_n, dots, 0), \
&. &. \
&. &. \
&. &. \
&(0, 0, dots, a_1), dots, &(0, 0, dots, a_n)
endalign*
So if $mathcalB$ was the set of basis elements of $V^k$ above then I'd say that the following proposed lemma would make more sense
Proposed Lemma: Let $V$ be a vector space of dimension $n$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$f(alpha) = g(alpha)$$ for every $alpha in mathcalB$ where $mathcalB$ is a basis for $V^k$, then $f= g$
Furthermore the same part of Lemma 26.2
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Taken literally gives $n^k$ possible $k$-tuples, which would correspond to checking to see if the values of $f$ and $g$ agree on $n^k$ basis elements, which confuses me since $dim(V^k) = kn$
I'm sure that Lemma 26.2 must be correct and I'm just making some error somewhere, if so could someone please point out what that error is.
linear-algebra vector-spaces linear-transformations tensors
asked Jul 20 at 18:09
Perturbative
3,51411039
The following is stated in the book Analysis on Manifolds by James Munkres
Just as is the case with linear transformations, a multilinear transformation
is entirely determined once one knows its values on basis elements. That
we now prove.
And then he gives the following lemma.
Lemma 26.2 Let $a_1, dots, a_n$ be a basis for $V$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$fleft(a_i_1, dots, a_i_kright) = gleft(a_i_1, dots, a_i_kright) $$ for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$ then $f=g$.
Now what I don't understand is why to we even need the following in the above lemma
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Because $V^k$ has dimension $k cdot n$ since $V$ has dimension $n$, and has as basis elements beginalign*&(a_1, 0, dots, 0), dots, &(a_n,0, dots, 0), \
&(0, a_1, dots, 0), dots, &(0, a_n, dots, 0), \
&. &. \
&. &. \
&. &. \
&(0, 0, dots, a_1), dots, &(0, 0, dots, a_n)
endalign*
So if $mathcalB$ was the set of basis elements of $V^k$ above then I'd say that the following proposed lemma would make more sense
Proposed Lemma: Let $V$ be a vector space of dimension $n$. If $f, g : V^k to mathbbR$ are $k$-tensors on $V$ and if $$f(alpha) = g(alpha)$$ for every $alpha in mathcalB$ where $mathcalB$ is a basis for $V^k$, then $f= g$
Furthermore the same part of Lemma 26.2
"for every $k$-tuple $I = (i_1, dots, i_k)$ of integers from the set $1, dots, n$"
Taken literally gives $n^k$ possible $k$-tuples, which would correspond to checking to see if the values of $f$ and $g$ agree on $n^k$ basis elements, which confuses me since $dim(V^k) = kn$
I'm sure that Lemma 26.2 must be correct and I'm just making some error somewhere, if so could someone please point out what that error is.
linear-algebra vector-spaces linear-transformations tensors
asked Jul 20 at 18:09
Perturbative
3,51411039
asked Jul 20 at 18:09
Perturbative
3,51411039
asked Jul 20 at 18:09
Perturbative
3,51411039
asked Jul 20 at 18:09
Perturbative
3,51411039
3,51411039
take a look at how this construction looks for $V=Bbb R^1$ or $V=Bbb R^2$ or $V=Bbb R^3$ with $k=1$ or $=2$ or $k=3$
– janmarqz
Jul 20 at 18:33
@janmarqz Okay so take the case $V = mathbbR^2$ and $k=2$, suppose $f, g : V^2 = mathbbR^4 to mathbbR$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\ f(e_1, e_2) = g(e_1, e_2);\ f(e_2, e_1) = g(e_2, e_1); \ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $mathbbR^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $mathbbR^4$
– Perturbative
Jul 20 at 19:48
So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$
– Perturbative
Jul 20 at 19:58
for bilinear maps $Bbb R^2timesBbb R^2toBbb R$ (which is a vector space) you only need the 4 basic "vectors": $$left(beginarraycc1&0\0&0endarrayright)$$ $$left(beginarraycc0&1\0&0endarrayright)$$ $$left(beginarraycc0&0\1&0endarrayright)$$ $$left(beginarraycc0&0\0&1endarrayright)$$
– janmarqz
Jul 20 at 20:23
@janmarqz But that's not what the theorem above asserts..
– Perturbative
Jul 20 at 21:29
add a comment |Â
take a look at how this construction looks for $V=Bbb R^1$ or $V=Bbb R^2$ or $V=Bbb R^3$ with $k=1$ or $=2$ or $k=3$
– janmarqz
Jul 20 at 18:33
@janmarqz Okay so take the case $V = mathbbR^2$ and $k=2$, suppose $f, g : V^2 = mathbbR^4 to mathbbR$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\ f(e_1, e_2) = g(e_1, e_2);\ f(e_2, e_1) = g(e_2, e_1); \ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $mathbbR^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $mathbbR^4$
– Perturbative
Jul 20 at 19:48
So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$
– Perturbative
Jul 20 at 19:58
for bilinear maps $Bbb R^2timesBbb R^2toBbb R$ (which is a vector space) you only need the 4 basic "vectors": $$left(beginarraycc1&0\0&0endarrayright)$$ $$left(beginarraycc0&1\0&0endarrayright)$$ $$left(beginarraycc0&0\1&0endarrayright)$$ $$left(beginarraycc0&0\0&1endarrayright)$$
– janmarqz
Jul 20 at 20:23
@janmarqz But that's not what the theorem above asserts..
– Perturbative
Jul 20 at 21:29
take a look at how this construction looks for $V=Bbb R^1$ or $V=Bbb R^2$ or $V=Bbb R^3$ with $k=1$ or $=2$ or $k=3$
– janmarqz
Jul 20 at 18:33
take a look at how this construction looks for $V=Bbb R^1$ or $V=Bbb R^2$ or $V=Bbb R^3$ with $k=1$ or $=2$ or $k=3$
– janmarqz
Jul 20 at 18:33
@janmarqz Okay so take the case $V = mathbbR^2$ and $k=2$, suppose $f, g : V^2 = mathbbR^4 to mathbbR$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\ f(e_1, e_2) = g(e_1, e_2);\ f(e_2, e_1) = g(e_2, e_1); \ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $mathbbR^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $mathbbR^4$
– Perturbative
Jul 20 at 19:48
@janmarqz Okay so take the case $V = mathbbR^2$ and $k=2$, suppose $f, g : V^2 = mathbbR^4 to mathbbR$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\ f(e_1, e_2) = g(e_1, e_2);\ f(e_2, e_1) = g(e_2, e_1); \ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $mathbbR^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $mathbbR^4$
– Perturbative
Jul 20 at 19:48
So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$
– Perturbative
Jul 20 at 19:58
So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$
– Perturbative
Jul 20 at 19:58
for bilinear maps $Bbb R^2timesBbb R^2toBbb R$ (which is a vector space) you only need the 4 basic "vectors": $$left(beginarraycc1&0\0&0endarrayright)$$ $$left(beginarraycc0&1\0&0endarrayright)$$ $$left(beginarraycc0&0\1&0endarrayright)$$ $$left(beginarraycc0&0\0&1endarrayright)$$
– janmarqz
Jul 20 at 20:23
for bilinear maps $Bbb R^2timesBbb R^2toBbb R$ (which is a vector space) you only need the 4 basic "vectors": $$left(beginarraycc1&0\0&0endarrayright)$$ $$left(beginarraycc0&1\0&0endarrayright)$$ $$left(beginarraycc0&0\1&0endarrayright)$$ $$left(beginarraycc0&0\0&1endarrayright)$$
– janmarqz
Jul 20 at 20:23
@janmarqz But that's not what the theorem above asserts..
– Perturbative
Jul 20 at 21:29
@janmarqz But that's not what the theorem above asserts..
– Perturbative
Jul 20 at 21:29
add a comment |Â
1 Answer
1
active
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votes
up vote
2
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accepted
The dimension of $underbraceVotimesdotsotimes V_k text times = bigotimes^k V$ is $n^k$, even though the dimension of $V^k$ is $kn$. We're talking about multilinear maps on $V^k$, not linear maps. (Note that $n=k=2$ is a bad example to pick, since then $kn = n^k$. :))
EDIT: I should comment that the vector space of multilinear maps on $V^k$ is isomorphic to $big(bigotimes^k Vbig)^*$, but dimensions are the same.
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
1
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The dimension of $underbraceVotimesdotsotimes V_k text times = bigotimes^k V$ is $n^k$, even though the dimension of $V^k$ is $kn$. We're talking about multilinear maps on $V^k$, not linear maps. (Note that $n=k=2$ is a bad example to pick, since then $kn = n^k$. :))
EDIT: I should comment that the vector space of multilinear maps on $V^k$ is isomorphic to $big(bigotimes^k Vbig)^*$, but dimensions are the same.
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
1
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
add a comment |Â
up vote
2
down vote
accepted
The dimension of $underbraceVotimesdotsotimes V_k text times = bigotimes^k V$ is $n^k$, even though the dimension of $V^k$ is $kn$. We're talking about multilinear maps on $V^k$, not linear maps. (Note that $n=k=2$ is a bad example to pick, since then $kn = n^k$. :))
EDIT: I should comment that the vector space of multilinear maps on $V^k$ is isomorphic to $big(bigotimes^k Vbig)^*$, but dimensions are the same.
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
1
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The dimension of $underbraceVotimesdotsotimes V_k text times = bigotimes^k V$ is $n^k$, even though the dimension of $V^k$ is $kn$. We're talking about multilinear maps on $V^k$, not linear maps. (Note that $n=k=2$ is a bad example to pick, since then $kn = n^k$. :))
EDIT: I should comment that the vector space of multilinear maps on $V^k$ is isomorphic to $big(bigotimes^k Vbig)^*$, but dimensions are the same.
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
The dimension of $underbraceVotimesdotsotimes V_k text times = bigotimes^k V$ is $n^k$, even though the dimension of $V^k$ is $kn$. We're talking about multilinear maps on $V^k$, not linear maps. (Note that $n=k=2$ is a bad example to pick, since then $kn = n^k$. :))
EDIT: I should comment that the vector space of multilinear maps on $V^k$ is isomorphic to $big(bigotimes^k Vbig)^*$, but dimensions are the same.
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
edited Jul 21 at 22:24
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
answered Jul 21 at 0:29
Ted Shifrin
59.5k44386
59.5k44386
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
1
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
add a comment |Â
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
1
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
Thanks for your answer Ted! Just one follow up question, you're saying then that Munkres is talking about basis elements from $bigotimes^k V$ and not $V^k$, correct?
– Perturbative
Jul 22 at 13:33
1
1
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Right. That's why we're considering the $n^k$ $k$-tuples $(a_i_1,dots,a_i_k)$.
– Ted Shifrin
Jul 22 at 14:57
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
Okay thanks again Ted! :)
– Perturbative
Jul 22 at 16:58
add a comment |Â
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take a look at how this construction looks for $V=Bbb R^1$ or $V=Bbb R^2$ or $V=Bbb R^3$ with $k=1$ or $=2$ or $k=3$
– janmarqz
Jul 20 at 18:33
@janmarqz Okay so take the case $V = mathbbR^2$ and $k=2$, suppose $f, g : V^2 = mathbbR^4 to mathbbR$ are two $2$-tensors, by the above theorem we need to show $$f(e_1, e_1) = g(e_1, e_1);\ f(e_1, e_2) = g(e_1, e_2);\ f(e_2, e_1) = g(e_2, e_1); \ f(e_2, e_2) = g(e_2, e_2);$$ for basis elements $e_1 = (1, 0)$ and $e_2 = (0, 1)$ of $mathbbR^2$. But for example $(e_1, e_1) = (1, 0, 1, 0)$ is not a basis element of $mathbbR^4$
– Perturbative
Jul 20 at 19:48
So I take it that Munkres is referring to basis elements of $V$ as opposed to $V^k$
– Perturbative
Jul 20 at 19:58
for bilinear maps $Bbb R^2timesBbb R^2toBbb R$ (which is a vector space) you only need the 4 basic "vectors": $$left(beginarraycc1&0\0&0endarrayright)$$ $$left(beginarraycc0&1\0&0endarrayright)$$ $$left(beginarraycc0&0\1&0endarrayright)$$ $$left(beginarraycc0&0\0&1endarrayright)$$
– janmarqz
Jul 20 at 20:23
@janmarqz But that's not what the theorem above asserts..
– Perturbative
Jul 20 at 21:29