Proving that a matrix is symmetric if it can be expressed as a spectral decomposition












3















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:





  1. $A$ 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 $D^T = D$.


  2. $c_1, cdots, c_n$ 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.










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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
















3















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:





  1. $A$ 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 $D^T = D$.


  2. $c_1, cdots, c_n$ 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.










share|cite|improve this question




















  • 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














3












3








3


1






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:





  1. $A$ 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 $D^T = D$.


  2. $c_1, cdots, c_n$ 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.










share|cite|improve this question
















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:





  1. $A$ 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 $D^T = D$.


  2. $c_1, cdots, c_n$ 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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edited Nov 21 at 2:30









JimmyK4542

40.4k245105




40.4k245105










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














  • 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










3 Answers
3






active

oldest

votes


















4














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).






share|cite|improve this answer





















  • $u_i^Tu_j = 0$ 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



















3














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.






share|cite|improve this answer























  • @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



















1














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 $u_j$, $1 le j le n$.






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    3 Answers
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    3 Answers
    3






    active

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    active

    oldest

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    active

    oldest

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    4














    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).






    share|cite|improve this answer





















    • $u_i^Tu_j = 0$ 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
















    4














    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).






    share|cite|improve this answer





















    • $u_i^Tu_j = 0$ 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














    4












    4








    4






    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).






    share|cite|improve this answer












    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).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 21 at 0:29









    jgon

    12.4k21940




    12.4k21940












    • $u_i^Tu_j = 0$ 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












    • @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











    3














    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.






    share|cite|improve this answer























    • @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
















    3














    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.






    share|cite|improve this answer























    • @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














    3












    3








    3






    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 21 at 0:26

























    answered Nov 21 at 0:04









    gimusi

    1




    1












    • @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 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











    1














    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 $u_j$, $1 le j le n$.






    share|cite|improve this answer


























      1














      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 $u_j$, $1 le j le n$.






      share|cite|improve this answer
























        1












        1








        1






        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 $u_j$, $1 le j le n$.






        share|cite|improve this answer












        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 $u_j$, $1 le j le n$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 0:49









        Robert Lewis

        43.5k22863




        43.5k22863






























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