Proving that a matrix is symmetric if it can be expressed as a spectral decomposition
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If ${u_1, cdots, u_n}$ is an orthonormal basis for $mathbb{R}^n$, and if $A$ can be expressed as
$$A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$$
then $A$ is symmetric and has eigenvalues $c_1, cdots, c_n$.
I'm trying to prove this. Here's what I have so far.
I figure I need to show:
$A$ is symmetric. I can achieve this by showing that $A$ has an orthonormal set of $n$ eigenvectors (or equivalently, that $A$ is orthogonally diagonalizable). If $P$ orthogonally diagonalizes $A$ then $D = P^TAP equiv A = PDP^T$. $(PDP^T) = (PDP^T)^T$ by trivial manipulations, knowing that $D$ is diagonal and thus $D^T = D$.
$c_1, cdots, c_n$ are the eigenvalues of $A$.
I think both of these would be satisfied if I could show that $c_1u_1u_1^T + cdots + c_nu_nu_n^T$ was equivalent to $PDP^T$ for the orthogonal matrix $P$ and a $D$ such that $D_{ij} = begin{cases}0 & i neq j\c_i & i = jend{cases}$.
If $P$ was an orthogonal matrix such that $P = begin{bmatrix} u_1 & cdots & u_nend{bmatrix}$ where $u_j$ was an eigenvector of $A$ then it would also be a basis for $mathbb{R}^n$, since we'd have $n$ linearly independent vectors. If I had this then I believe you can do the tedious matrix multiplication and get $PDP^T$ given the $D$ defined above and receive $A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$. Then I'd be done.
But to me the question implies that any orthonormal basis for $mathbb{R}^n$ would satisfy this. Perhaps I need to show that if $A$ can be expressed with those basis vectors then those basis vectors must be the eigenvectors of $A$. I'm kind of stuck on this part though!
Edit: To be clear: I have outlined here my approach to the proof and what I know to be true. I'm ultimately stuck on how to prove the quoted question. I am asking how one can prove this.
This is exercise 7.2.26 of Anton and Rorres' Elementary Linear Algebra, 11th ed.
linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices
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up vote
3
down vote
favorite
If ${u_1, cdots, u_n}$ is an orthonormal basis for $mathbb{R}^n$, and if $A$ can be expressed as
$$A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$$
then $A$ is symmetric and has eigenvalues $c_1, cdots, c_n$.
I'm trying to prove this. Here's what I have so far.
I figure I need to show:
$A$ is symmetric. I can achieve this by showing that $A$ has an orthonormal set of $n$ eigenvectors (or equivalently, that $A$ is orthogonally diagonalizable). If $P$ orthogonally diagonalizes $A$ then $D = P^TAP equiv A = PDP^T$. $(PDP^T) = (PDP^T)^T$ by trivial manipulations, knowing that $D$ is diagonal and thus $D^T = D$.
$c_1, cdots, c_n$ are the eigenvalues of $A$.
I think both of these would be satisfied if I could show that $c_1u_1u_1^T + cdots + c_nu_nu_n^T$ was equivalent to $PDP^T$ for the orthogonal matrix $P$ and a $D$ such that $D_{ij} = begin{cases}0 & i neq j\c_i & i = jend{cases}$.
If $P$ was an orthogonal matrix such that $P = begin{bmatrix} u_1 & cdots & u_nend{bmatrix}$ where $u_j$ was an eigenvector of $A$ then it would also be a basis for $mathbb{R}^n$, since we'd have $n$ linearly independent vectors. If I had this then I believe you can do the tedious matrix multiplication and get $PDP^T$ given the $D$ defined above and receive $A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$. Then I'd be done.
But to me the question implies that any orthonormal basis for $mathbb{R}^n$ would satisfy this. Perhaps I need to show that if $A$ can be expressed with those basis vectors then those basis vectors must be the eigenvectors of $A$. I'm kind of stuck on this part though!
Edit: To be clear: I have outlined here my approach to the proof and what I know to be true. I'm ultimately stuck on how to prove the quoted question. I am asking how one can prove this.
This is exercise 7.2.26 of Anton and Rorres' Elementary Linear Algebra, 11th ed.
linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices
1
I assume that the last term should be $c_nu_nu_n^T$ not $c_nu_n+u_n^T$?
– jgon
Nov 21 at 0:21
@jgon Correct, fixed.
– Emily Horsman
Nov 21 at 0:40
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If ${u_1, cdots, u_n}$ is an orthonormal basis for $mathbb{R}^n$, and if $A$ can be expressed as
$$A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$$
then $A$ is symmetric and has eigenvalues $c_1, cdots, c_n$.
I'm trying to prove this. Here's what I have so far.
I figure I need to show:
$A$ is symmetric. I can achieve this by showing that $A$ has an orthonormal set of $n$ eigenvectors (or equivalently, that $A$ is orthogonally diagonalizable). If $P$ orthogonally diagonalizes $A$ then $D = P^TAP equiv A = PDP^T$. $(PDP^T) = (PDP^T)^T$ by trivial manipulations, knowing that $D$ is diagonal and thus $D^T = D$.
$c_1, cdots, c_n$ are the eigenvalues of $A$.
I think both of these would be satisfied if I could show that $c_1u_1u_1^T + cdots + c_nu_nu_n^T$ was equivalent to $PDP^T$ for the orthogonal matrix $P$ and a $D$ such that $D_{ij} = begin{cases}0 & i neq j\c_i & i = jend{cases}$.
If $P$ was an orthogonal matrix such that $P = begin{bmatrix} u_1 & cdots & u_nend{bmatrix}$ where $u_j$ was an eigenvector of $A$ then it would also be a basis for $mathbb{R}^n$, since we'd have $n$ linearly independent vectors. If I had this then I believe you can do the tedious matrix multiplication and get $PDP^T$ given the $D$ defined above and receive $A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$. Then I'd be done.
But to me the question implies that any orthonormal basis for $mathbb{R}^n$ would satisfy this. Perhaps I need to show that if $A$ can be expressed with those basis vectors then those basis vectors must be the eigenvectors of $A$. I'm kind of stuck on this part though!
Edit: To be clear: I have outlined here my approach to the proof and what I know to be true. I'm ultimately stuck on how to prove the quoted question. I am asking how one can prove this.
This is exercise 7.2.26 of Anton and Rorres' Elementary Linear Algebra, 11th ed.
linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices
If ${u_1, cdots, u_n}$ is an orthonormal basis for $mathbb{R}^n$, and if $A$ can be expressed as
$$A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$$
then $A$ is symmetric and has eigenvalues $c_1, cdots, c_n$.
I'm trying to prove this. Here's what I have so far.
I figure I need to show:
$A$ is symmetric. I can achieve this by showing that $A$ has an orthonormal set of $n$ eigenvectors (or equivalently, that $A$ is orthogonally diagonalizable). If $P$ orthogonally diagonalizes $A$ then $D = P^TAP equiv A = PDP^T$. $(PDP^T) = (PDP^T)^T$ by trivial manipulations, knowing that $D$ is diagonal and thus $D^T = D$.
$c_1, cdots, c_n$ are the eigenvalues of $A$.
I think both of these would be satisfied if I could show that $c_1u_1u_1^T + cdots + c_nu_nu_n^T$ was equivalent to $PDP^T$ for the orthogonal matrix $P$ and a $D$ such that $D_{ij} = begin{cases}0 & i neq j\c_i & i = jend{cases}$.
If $P$ was an orthogonal matrix such that $P = begin{bmatrix} u_1 & cdots & u_nend{bmatrix}$ where $u_j$ was an eigenvector of $A$ then it would also be a basis for $mathbb{R}^n$, since we'd have $n$ linearly independent vectors. If I had this then I believe you can do the tedious matrix multiplication and get $PDP^T$ given the $D$ defined above and receive $A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$. Then I'd be done.
But to me the question implies that any orthonormal basis for $mathbb{R}^n$ would satisfy this. Perhaps I need to show that if $A$ can be expressed with those basis vectors then those basis vectors must be the eigenvectors of $A$. I'm kind of stuck on this part though!
Edit: To be clear: I have outlined here my approach to the proof and what I know to be true. I'm ultimately stuck on how to prove the quoted question. I am asking how one can prove this.
This is exercise 7.2.26 of Anton and Rorres' Elementary Linear Algebra, 11th ed.
linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices
linear-algebra eigenvalues-eigenvectors diagonalization orthogonal-matrices
edited Nov 21 at 2:30
JimmyK4542
39.9k245105
39.9k245105
asked Nov 20 at 23:57
Emily Horsman
111110
111110
1
I assume that the last term should be $c_nu_nu_n^T$ not $c_nu_n+u_n^T$?
– jgon
Nov 21 at 0:21
@jgon Correct, fixed.
– Emily Horsman
Nov 21 at 0:40
add a comment |
1
I assume that the last term should be $c_nu_nu_n^T$ not $c_nu_n+u_n^T$?
– jgon
Nov 21 at 0:21
@jgon Correct, fixed.
– Emily Horsman
Nov 21 at 0:40
1
1
I assume that the last term should be $c_nu_nu_n^T$ not $c_nu_n+u_n^T$?
– jgon
Nov 21 at 0:21
I assume that the last term should be $c_nu_nu_n^T$ not $c_nu_n+u_n^T$?
– jgon
Nov 21 at 0:21
@jgon Correct, fixed.
– Emily Horsman
Nov 21 at 0:40
@jgon Correct, fixed.
– Emily Horsman
Nov 21 at 0:40
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
I'm not quite sure what you're asking in your question, but if its helpful, here's how I would write this proof.
1) If $$A=sum_{i=1}^n c_iu_iu_i^T,$$then observe that
$$A^T=left(sum_{i=1}^nc_iu_iu_i^Tright)^T=sum_{i=1}^n c_i(u_i^T)^Tu_i^T=sum_{i=1}^n c_iu_iu_i^T=A,$$
where the second equality follows since taking transposes reverses the order of multiplication for matrices, and we can always pull constants out front.
2) If $A$ has the form above, then to show $c_j$ is an eigenvalue, consider the following product:
$$Au_j= sum_{i=1}^nc_iu_iu_i^Tu_j=sum_{i=1}^nc_iu_idelta_{ij}=c_ju_j.$$
The second equality follows from the fact that the $u_i$ form an orthonormal basis so $u_i^Tu_j=delta_{ij}$ (by definition of orthonormal).
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
add a comment |
up vote
3
down vote
By inspection from the hypotesis we have that
$$A^T= (c_1u_1u_1^T + cdots + c_nu_nu_n^T)^T=A$$
and
$$Acdot u_i=c_iu_i$$
therefore the thesis follows.
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
add a comment |
up vote
1
down vote
Note that
$(u_i u_i^T)^T = (u_i^T)^Tu_i^T = u_i u_i^T; tag 1$
thus each matrix $u_i u_i^T$ is symmetric; hence every $c_i u_i u_i^T$ and hence their sum. This shows that
$A^T = A. tag 2$
We further note that, since the $u_i$ are orthnormal,
$u_i^T u_j = delta_{ij}, tag 3$
whence
$A u_j = displaystyle left ( sum_{i = 1}^n c_i u_i u_i^T right ) u_j = sum_{i = 1}^n c_i u_i u_i^Tu_j = sum_{i = 1}^n c_iu_i delta_{ij} = c_j u_j, tag 4$
which shows that $c_j$ is an eigenvalue of $A$ with associated eigenvector $u_j$, $1 le j le n$.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
I'm not quite sure what you're asking in your question, but if its helpful, here's how I would write this proof.
1) If $$A=sum_{i=1}^n c_iu_iu_i^T,$$then observe that
$$A^T=left(sum_{i=1}^nc_iu_iu_i^Tright)^T=sum_{i=1}^n c_i(u_i^T)^Tu_i^T=sum_{i=1}^n c_iu_iu_i^T=A,$$
where the second equality follows since taking transposes reverses the order of multiplication for matrices, and we can always pull constants out front.
2) If $A$ has the form above, then to show $c_j$ is an eigenvalue, consider the following product:
$$Au_j= sum_{i=1}^nc_iu_iu_i^Tu_j=sum_{i=1}^nc_iu_idelta_{ij}=c_ju_j.$$
The second equality follows from the fact that the $u_i$ form an orthonormal basis so $u_i^Tu_j=delta_{ij}$ (by definition of orthonormal).
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
add a comment |
up vote
4
down vote
accepted
I'm not quite sure what you're asking in your question, but if its helpful, here's how I would write this proof.
1) If $$A=sum_{i=1}^n c_iu_iu_i^T,$$then observe that
$$A^T=left(sum_{i=1}^nc_iu_iu_i^Tright)^T=sum_{i=1}^n c_i(u_i^T)^Tu_i^T=sum_{i=1}^n c_iu_iu_i^T=A,$$
where the second equality follows since taking transposes reverses the order of multiplication for matrices, and we can always pull constants out front.
2) If $A$ has the form above, then to show $c_j$ is an eigenvalue, consider the following product:
$$Au_j= sum_{i=1}^nc_iu_iu_i^Tu_j=sum_{i=1}^nc_iu_idelta_{ij}=c_ju_j.$$
The second equality follows from the fact that the $u_i$ form an orthonormal basis so $u_i^Tu_j=delta_{ij}$ (by definition of orthonormal).
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
I'm not quite sure what you're asking in your question, but if its helpful, here's how I would write this proof.
1) If $$A=sum_{i=1}^n c_iu_iu_i^T,$$then observe that
$$A^T=left(sum_{i=1}^nc_iu_iu_i^Tright)^T=sum_{i=1}^n c_i(u_i^T)^Tu_i^T=sum_{i=1}^n c_iu_iu_i^T=A,$$
where the second equality follows since taking transposes reverses the order of multiplication for matrices, and we can always pull constants out front.
2) If $A$ has the form above, then to show $c_j$ is an eigenvalue, consider the following product:
$$Au_j= sum_{i=1}^nc_iu_iu_i^Tu_j=sum_{i=1}^nc_iu_idelta_{ij}=c_ju_j.$$
The second equality follows from the fact that the $u_i$ form an orthonormal basis so $u_i^Tu_j=delta_{ij}$ (by definition of orthonormal).
I'm not quite sure what you're asking in your question, but if its helpful, here's how I would write this proof.
1) If $$A=sum_{i=1}^n c_iu_iu_i^T,$$then observe that
$$A^T=left(sum_{i=1}^nc_iu_iu_i^Tright)^T=sum_{i=1}^n c_i(u_i^T)^Tu_i^T=sum_{i=1}^n c_iu_iu_i^T=A,$$
where the second equality follows since taking transposes reverses the order of multiplication for matrices, and we can always pull constants out front.
2) If $A$ has the form above, then to show $c_j$ is an eigenvalue, consider the following product:
$$Au_j= sum_{i=1}^nc_iu_iu_i^Tu_j=sum_{i=1}^nc_iu_idelta_{ij}=c_ju_j.$$
The second equality follows from the fact that the $u_i$ form an orthonormal basis so $u_i^Tu_j=delta_{ij}$ (by definition of orthonormal).
answered Nov 21 at 0:29
jgon
10.1k11638
10.1k11638
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
add a comment |
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
$u_i^Tu_j = 0$ if $i neq j$. So $Au_j = c_ju_ju_j^Tu_j$ right? Ah and $u_j^Tu_j = 1$ since its orthonormal.
– Emily Horsman
Nov 21 at 1:02
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
@EmilyHorsman exactly.
– jgon
Nov 21 at 1:13
add a comment |
up vote
3
down vote
By inspection from the hypotesis we have that
$$A^T= (c_1u_1u_1^T + cdots + c_nu_nu_n^T)^T=A$$
and
$$Acdot u_i=c_iu_i$$
therefore the thesis follows.
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
add a comment |
up vote
3
down vote
By inspection from the hypotesis we have that
$$A^T= (c_1u_1u_1^T + cdots + c_nu_nu_n^T)^T=A$$
and
$$Acdot u_i=c_iu_i$$
therefore the thesis follows.
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
add a comment |
up vote
3
down vote
up vote
3
down vote
By inspection from the hypotesis we have that
$$A^T= (c_1u_1u_1^T + cdots + c_nu_nu_n^T)^T=A$$
and
$$Acdot u_i=c_iu_i$$
therefore the thesis follows.
By inspection from the hypotesis we have that
$$A^T= (c_1u_1u_1^T + cdots + c_nu_nu_n^T)^T=A$$
and
$$Acdot u_i=c_iu_i$$
therefore the thesis follows.
edited Nov 21 at 0:26
answered Nov 21 at 0:04
gimusi
88.7k74394
88.7k74394
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
add a comment |
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon Opsssss yes I think you are right!
– gimusi
Nov 21 at 0:23
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
@jgon in that way I think it reduces to a simple check by the definition.
– gimusi
Nov 21 at 0:28
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
I agree, this whole question confuses me to be honest
– jgon
Nov 21 at 0:30
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
It seems I was vastly overcomplicating it
– Emily Horsman
Nov 21 at 0:47
add a comment |
up vote
1
down vote
Note that
$(u_i u_i^T)^T = (u_i^T)^Tu_i^T = u_i u_i^T; tag 1$
thus each matrix $u_i u_i^T$ is symmetric; hence every $c_i u_i u_i^T$ and hence their sum. This shows that
$A^T = A. tag 2$
We further note that, since the $u_i$ are orthnormal,
$u_i^T u_j = delta_{ij}, tag 3$
whence
$A u_j = displaystyle left ( sum_{i = 1}^n c_i u_i u_i^T right ) u_j = sum_{i = 1}^n c_i u_i u_i^Tu_j = sum_{i = 1}^n c_iu_i delta_{ij} = c_j u_j, tag 4$
which shows that $c_j$ is an eigenvalue of $A$ with associated eigenvector $u_j$, $1 le j le n$.
add a comment |
up vote
1
down vote
Note that
$(u_i u_i^T)^T = (u_i^T)^Tu_i^T = u_i u_i^T; tag 1$
thus each matrix $u_i u_i^T$ is symmetric; hence every $c_i u_i u_i^T$ and hence their sum. This shows that
$A^T = A. tag 2$
We further note that, since the $u_i$ are orthnormal,
$u_i^T u_j = delta_{ij}, tag 3$
whence
$A u_j = displaystyle left ( sum_{i = 1}^n c_i u_i u_i^T right ) u_j = sum_{i = 1}^n c_i u_i u_i^Tu_j = sum_{i = 1}^n c_iu_i delta_{ij} = c_j u_j, tag 4$
which shows that $c_j$ is an eigenvalue of $A$ with associated eigenvector $u_j$, $1 le j le n$.
add a comment |
up vote
1
down vote
up vote
1
down vote
Note that
$(u_i u_i^T)^T = (u_i^T)^Tu_i^T = u_i u_i^T; tag 1$
thus each matrix $u_i u_i^T$ is symmetric; hence every $c_i u_i u_i^T$ and hence their sum. This shows that
$A^T = A. tag 2$
We further note that, since the $u_i$ are orthnormal,
$u_i^T u_j = delta_{ij}, tag 3$
whence
$A u_j = displaystyle left ( sum_{i = 1}^n c_i u_i u_i^T right ) u_j = sum_{i = 1}^n c_i u_i u_i^Tu_j = sum_{i = 1}^n c_iu_i delta_{ij} = c_j u_j, tag 4$
which shows that $c_j$ is an eigenvalue of $A$ with associated eigenvector $u_j$, $1 le j le n$.
Note that
$(u_i u_i^T)^T = (u_i^T)^Tu_i^T = u_i u_i^T; tag 1$
thus each matrix $u_i u_i^T$ is symmetric; hence every $c_i u_i u_i^T$ and hence their sum. This shows that
$A^T = A. tag 2$
We further note that, since the $u_i$ are orthnormal,
$u_i^T u_j = delta_{ij}, tag 3$
whence
$A u_j = displaystyle left ( sum_{i = 1}^n c_i u_i u_i^T right ) u_j = sum_{i = 1}^n c_i u_i u_i^Tu_j = sum_{i = 1}^n c_iu_i delta_{ij} = c_j u_j, tag 4$
which shows that $c_j$ is an eigenvalue of $A$ with associated eigenvector $u_j$, $1 le j le n$.
answered Nov 21 at 0:49
Robert Lewis
42k22760
42k22760
add a comment |
add a comment |
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1
I assume that the last term should be $c_nu_nu_n^T$ not $c_nu_n+u_n^T$?
– jgon
Nov 21 at 0:21
@jgon Correct, fixed.
– Emily Horsman
Nov 21 at 0:40