Why is $||Sigma - I||_2 =max|lambda_i|$ where $lambda_i$ is an eigenvalue of $Sigma - I$












1












$begingroup$


$||Sigma - I||_2 = max|lambda_i|$ where $lambda_i$ is an eigenvalue of $Sigma - I$



$Sigma $ is a diagonal matrix and $I$ is the identity matrix



I know that since $Sigma - I$ is diagonal, its eigenvalues are the values along the diagonal



I also know that the singular values of $Sigma - I$ are the square roots of the eigenvalues of $(Sigma - I)^T(Sigma - I) = (Sigma - I)^2$



I can't seem to see why this is true?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your question is a special case of the more general fact.
    $endgroup$
    – A.Γ.
    Dec 22 '18 at 21:38






  • 3




    $begingroup$
    I don't think this is true. Take $Sigma=begin{bmatrix} 2 & 0\ 0 & 2end{bmatrix}$ for example. Then $Sigma-I$ is the identity matrix which has $2$-norm $sqrt 2$ but the maximum eigenvalue is $1$. I think you should consider another norm on the matrix space (for example the operator norm).
    $endgroup$
    – Levent
    Dec 22 '18 at 21:39










  • $begingroup$
    @A.Γ. You should give an official answer, even it is essentially a reference to another question plus Levent's comment..
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 22:46






  • 2




    $begingroup$
    @Levent the matrix 2-norm is generally defined as $sup_{xneq 0} langle Ax,xrangle/langle x,xrangle$, while the Frobenius norm is the square root of sum of square of entries.
    $endgroup$
    – tch
    Dec 23 '18 at 16:05












  • $begingroup$
    @TylerChen Oh okay, thanks for pointing it out. I thought it meant the Euclidean norm.
    $endgroup$
    – Levent
    Dec 23 '18 at 18:09
















1












$begingroup$


$||Sigma - I||_2 = max|lambda_i|$ where $lambda_i$ is an eigenvalue of $Sigma - I$



$Sigma $ is a diagonal matrix and $I$ is the identity matrix



I know that since $Sigma - I$ is diagonal, its eigenvalues are the values along the diagonal



I also know that the singular values of $Sigma - I$ are the square roots of the eigenvalues of $(Sigma - I)^T(Sigma - I) = (Sigma - I)^2$



I can't seem to see why this is true?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your question is a special case of the more general fact.
    $endgroup$
    – A.Γ.
    Dec 22 '18 at 21:38






  • 3




    $begingroup$
    I don't think this is true. Take $Sigma=begin{bmatrix} 2 & 0\ 0 & 2end{bmatrix}$ for example. Then $Sigma-I$ is the identity matrix which has $2$-norm $sqrt 2$ but the maximum eigenvalue is $1$. I think you should consider another norm on the matrix space (for example the operator norm).
    $endgroup$
    – Levent
    Dec 22 '18 at 21:39










  • $begingroup$
    @A.Γ. You should give an official answer, even it is essentially a reference to another question plus Levent's comment..
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 22:46






  • 2




    $begingroup$
    @Levent the matrix 2-norm is generally defined as $sup_{xneq 0} langle Ax,xrangle/langle x,xrangle$, while the Frobenius norm is the square root of sum of square of entries.
    $endgroup$
    – tch
    Dec 23 '18 at 16:05












  • $begingroup$
    @TylerChen Oh okay, thanks for pointing it out. I thought it meant the Euclidean norm.
    $endgroup$
    – Levent
    Dec 23 '18 at 18:09














1












1








1





$begingroup$


$||Sigma - I||_2 = max|lambda_i|$ where $lambda_i$ is an eigenvalue of $Sigma - I$



$Sigma $ is a diagonal matrix and $I$ is the identity matrix



I know that since $Sigma - I$ is diagonal, its eigenvalues are the values along the diagonal



I also know that the singular values of $Sigma - I$ are the square roots of the eigenvalues of $(Sigma - I)^T(Sigma - I) = (Sigma - I)^2$



I can't seem to see why this is true?






linear-algebra







share|cite|improve this question











$endgroup$




$||Sigma - I||_2 = max|lambda_i|$ where $lambda_i$ is an eigenvalue of $Sigma - I$



$Sigma $ is a diagonal matrix and $I$ is the identity matrix



I know that since $Sigma - I$ is diagonal, its eigenvalues are the values along the diagonal



I also know that the singular values of $Sigma - I$ are the square roots of the eigenvalues of $(Sigma - I)^T(Sigma - I) = (Sigma - I)^2$



I can't seem to see why this is true?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 21:40







pablo_mathscobar

















asked Dec 22 '18 at 21:26









pablo_mathscobarpablo_mathscobar

1277




1277












  • $begingroup$
    Your question is a special case of the more general fact.
    $endgroup$
    – A.Γ.
    Dec 22 '18 at 21:38






  • 3




    $begingroup$
    I don't think this is true. Take $Sigma=begin{bmatrix} 2 & 0\ 0 & 2end{bmatrix}$ for example. Then $Sigma-I$ is the identity matrix which has $2$-norm $sqrt 2$ but the maximum eigenvalue is $1$. I think you should consider another norm on the matrix space (for example the operator norm).
    $endgroup$
    – Levent
    Dec 22 '18 at 21:39










  • $begingroup$
    @A.Γ. You should give an official answer, even it is essentially a reference to another question plus Levent's comment..
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 22:46






  • 2




    $begingroup$
    @Levent the matrix 2-norm is generally defined as $sup_{xneq 0} langle Ax,xrangle/langle x,xrangle$, while the Frobenius norm is the square root of sum of square of entries.
    $endgroup$
    – tch
    Dec 23 '18 at 16:05












  • $begingroup$
    @TylerChen Oh okay, thanks for pointing it out. I thought it meant the Euclidean norm.
    $endgroup$
    – Levent
    Dec 23 '18 at 18:09


















  • $begingroup$
    Your question is a special case of the more general fact.
    $endgroup$
    – A.Γ.
    Dec 22 '18 at 21:38






  • 3




    $begingroup$
    I don't think this is true. Take $Sigma=begin{bmatrix} 2 & 0\ 0 & 2end{bmatrix}$ for example. Then $Sigma-I$ is the identity matrix which has $2$-norm $sqrt 2$ but the maximum eigenvalue is $1$. I think you should consider another norm on the matrix space (for example the operator norm).
    $endgroup$
    – Levent
    Dec 22 '18 at 21:39










  • $begingroup$
    @A.Γ. You should give an official answer, even it is essentially a reference to another question plus Levent's comment..
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 22:46






  • 2




    $begingroup$
    @Levent the matrix 2-norm is generally defined as $sup_{xneq 0} langle Ax,xrangle/langle x,xrangle$, while the Frobenius norm is the square root of sum of square of entries.
    $endgroup$
    – tch
    Dec 23 '18 at 16:05












  • $begingroup$
    @TylerChen Oh okay, thanks for pointing it out. I thought it meant the Euclidean norm.
    $endgroup$
    – Levent
    Dec 23 '18 at 18:09
















$begingroup$
Your question is a special case of the more general fact.
$endgroup$
– A.Γ.
Dec 22 '18 at 21:38




$begingroup$
Your question is a special case of the more general fact.
$endgroup$
– A.Γ.
Dec 22 '18 at 21:38




3




3




$begingroup$
I don't think this is true. Take $Sigma=begin{bmatrix} 2 & 0\ 0 & 2end{bmatrix}$ for example. Then $Sigma-I$ is the identity matrix which has $2$-norm $sqrt 2$ but the maximum eigenvalue is $1$. I think you should consider another norm on the matrix space (for example the operator norm).
$endgroup$
– Levent
Dec 22 '18 at 21:39




$begingroup$
I don't think this is true. Take $Sigma=begin{bmatrix} 2 & 0\ 0 & 2end{bmatrix}$ for example. Then $Sigma-I$ is the identity matrix which has $2$-norm $sqrt 2$ but the maximum eigenvalue is $1$. I think you should consider another norm on the matrix space (for example the operator norm).
$endgroup$
– Levent
Dec 22 '18 at 21:39












$begingroup$
@A.Γ. You should give an official answer, even it is essentially a reference to another question plus Levent's comment..
$endgroup$
– Paul Frost
Dec 22 '18 at 22:46




$begingroup$
@A.Γ. You should give an official answer, even it is essentially a reference to another question plus Levent's comment..
$endgroup$
– Paul Frost
Dec 22 '18 at 22:46




2




2




$begingroup$
@Levent the matrix 2-norm is generally defined as $sup_{xneq 0} langle Ax,xrangle/langle x,xrangle$, while the Frobenius norm is the square root of sum of square of entries.
$endgroup$
– tch
Dec 23 '18 at 16:05






$begingroup$
@Levent the matrix 2-norm is generally defined as $sup_{xneq 0} langle Ax,xrangle/langle x,xrangle$, while the Frobenius norm is the square root of sum of square of entries.
$endgroup$
– tch
Dec 23 '18 at 16:05














$begingroup$
@TylerChen Oh okay, thanks for pointing it out. I thought it meant the Euclidean norm.
$endgroup$
– Levent
Dec 23 '18 at 18:09




$begingroup$
@TylerChen Oh okay, thanks for pointing it out. I thought it meant the Euclidean norm.
$endgroup$
– Levent
Dec 23 '18 at 18:09










1 Answer
1






active

oldest

votes


















1












$begingroup$

As @A.Γ noted, this is a special case of a more general fact about normal matrices, but we can prove it without that fact.



Let $Lambda$ be a diagonal matrix with entries $lambda_i$. We take $lVert{A}rVert_2 = sup_{xneq 0} lVert{Ax}rVert_2 / lVert{x}rVert_2 = sup_{lVert x rVert_2 = 1} lVert{Ax}rVert_2$.



Taking $x$ to be the eigenvector corresponding to the largest magntiude eigenvalue gives,
$$
lVert{Ax}rVert_2 = lVert{(max|lambda_i|)xrVert}_2 = (max|lambda_i|) lVert{x}rVert_2 = max|lambda_i|
$$



So, for any matrix, the matrix 2-norm is always at least the size of the largest magnitude eigenvalue.



Now, let $x=[x_1,x_2,ldots, x_n]^T$. Then $Ax = [lambda_1 x_1, lambda_2x_2, ldots, lambda_nx_n]^T$. Therefore,
$$
lVert{Ax}rVert_2 = sum_{i=1}^{n} lambda_i^2x_i^2
$$



Since $lVert x rVert_2 = 1$ we have that $sum_{i=1}^{n} x_i^2 = 1$ , which implies that $ 0leq x_i^2 leq 1$. Therefore,
$$
lVert{Ax}rVert_2^2 = sum_{i=1}^{n} lambda_i^2 x_i^2
leq sum_{i=1}^{n} (max|lambda_i|)^2 x_i^2
= (max|lambda_i|)^2 sum_{i=1}^{n} x_i^2
= (max|lambda_i|)^2
$$



Therefore, for diagonal matrices, the matrix 2-norm is bounded above by the size of the largest magnitude eigenvalue, and so the two quantities must be equal.






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    1 Answer
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    1 Answer
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    active

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    active

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    1












    $begingroup$

    As @A.Γ noted, this is a special case of a more general fact about normal matrices, but we can prove it without that fact.



    Let $Lambda$ be a diagonal matrix with entries $lambda_i$. We take $lVert{A}rVert_2 = sup_{xneq 0} lVert{Ax}rVert_2 / lVert{x}rVert_2 = sup_{lVert x rVert_2 = 1} lVert{Ax}rVert_2$.



    Taking $x$ to be the eigenvector corresponding to the largest magntiude eigenvalue gives,
    $$
    lVert{Ax}rVert_2 = lVert{(max|lambda_i|)xrVert}_2 = (max|lambda_i|) lVert{x}rVert_2 = max|lambda_i|
    $$



    So, for any matrix, the matrix 2-norm is always at least the size of the largest magnitude eigenvalue.



    Now, let $x=[x_1,x_2,ldots, x_n]^T$. Then $Ax = [lambda_1 x_1, lambda_2x_2, ldots, lambda_nx_n]^T$. Therefore,
    $$
    lVert{Ax}rVert_2 = sum_{i=1}^{n} lambda_i^2x_i^2
    $$



    Since $lVert x rVert_2 = 1$ we have that $sum_{i=1}^{n} x_i^2 = 1$ , which implies that $ 0leq x_i^2 leq 1$. Therefore,
    $$
    lVert{Ax}rVert_2^2 = sum_{i=1}^{n} lambda_i^2 x_i^2
    leq sum_{i=1}^{n} (max|lambda_i|)^2 x_i^2
    = (max|lambda_i|)^2 sum_{i=1}^{n} x_i^2
    = (max|lambda_i|)^2
    $$



    Therefore, for diagonal matrices, the matrix 2-norm is bounded above by the size of the largest magnitude eigenvalue, and so the two quantities must be equal.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      As @A.Γ noted, this is a special case of a more general fact about normal matrices, but we can prove it without that fact.



      Let $Lambda$ be a diagonal matrix with entries $lambda_i$. We take $lVert{A}rVert_2 = sup_{xneq 0} lVert{Ax}rVert_2 / lVert{x}rVert_2 = sup_{lVert x rVert_2 = 1} lVert{Ax}rVert_2$.



      Taking $x$ to be the eigenvector corresponding to the largest magntiude eigenvalue gives,
      $$
      lVert{Ax}rVert_2 = lVert{(max|lambda_i|)xrVert}_2 = (max|lambda_i|) lVert{x}rVert_2 = max|lambda_i|
      $$



      So, for any matrix, the matrix 2-norm is always at least the size of the largest magnitude eigenvalue.



      Now, let $x=[x_1,x_2,ldots, x_n]^T$. Then $Ax = [lambda_1 x_1, lambda_2x_2, ldots, lambda_nx_n]^T$. Therefore,
      $$
      lVert{Ax}rVert_2 = sum_{i=1}^{n} lambda_i^2x_i^2
      $$



      Since $lVert x rVert_2 = 1$ we have that $sum_{i=1}^{n} x_i^2 = 1$ , which implies that $ 0leq x_i^2 leq 1$. Therefore,
      $$
      lVert{Ax}rVert_2^2 = sum_{i=1}^{n} lambda_i^2 x_i^2
      leq sum_{i=1}^{n} (max|lambda_i|)^2 x_i^2
      = (max|lambda_i|)^2 sum_{i=1}^{n} x_i^2
      = (max|lambda_i|)^2
      $$



      Therefore, for diagonal matrices, the matrix 2-norm is bounded above by the size of the largest magnitude eigenvalue, and so the two quantities must be equal.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        As @A.Γ noted, this is a special case of a more general fact about normal matrices, but we can prove it without that fact.



        Let $Lambda$ be a diagonal matrix with entries $lambda_i$. We take $lVert{A}rVert_2 = sup_{xneq 0} lVert{Ax}rVert_2 / lVert{x}rVert_2 = sup_{lVert x rVert_2 = 1} lVert{Ax}rVert_2$.



        Taking $x$ to be the eigenvector corresponding to the largest magntiude eigenvalue gives,
        $$
        lVert{Ax}rVert_2 = lVert{(max|lambda_i|)xrVert}_2 = (max|lambda_i|) lVert{x}rVert_2 = max|lambda_i|
        $$



        So, for any matrix, the matrix 2-norm is always at least the size of the largest magnitude eigenvalue.



        Now, let $x=[x_1,x_2,ldots, x_n]^T$. Then $Ax = [lambda_1 x_1, lambda_2x_2, ldots, lambda_nx_n]^T$. Therefore,
        $$
        lVert{Ax}rVert_2 = sum_{i=1}^{n} lambda_i^2x_i^2
        $$



        Since $lVert x rVert_2 = 1$ we have that $sum_{i=1}^{n} x_i^2 = 1$ , which implies that $ 0leq x_i^2 leq 1$. Therefore,
        $$
        lVert{Ax}rVert_2^2 = sum_{i=1}^{n} lambda_i^2 x_i^2
        leq sum_{i=1}^{n} (max|lambda_i|)^2 x_i^2
        = (max|lambda_i|)^2 sum_{i=1}^{n} x_i^2
        = (max|lambda_i|)^2
        $$



        Therefore, for diagonal matrices, the matrix 2-norm is bounded above by the size of the largest magnitude eigenvalue, and so the two quantities must be equal.






        share|cite|improve this answer











        $endgroup$



        As @A.Γ noted, this is a special case of a more general fact about normal matrices, but we can prove it without that fact.



        Let $Lambda$ be a diagonal matrix with entries $lambda_i$. We take $lVert{A}rVert_2 = sup_{xneq 0} lVert{Ax}rVert_2 / lVert{x}rVert_2 = sup_{lVert x rVert_2 = 1} lVert{Ax}rVert_2$.



        Taking $x$ to be the eigenvector corresponding to the largest magntiude eigenvalue gives,
        $$
        lVert{Ax}rVert_2 = lVert{(max|lambda_i|)xrVert}_2 = (max|lambda_i|) lVert{x}rVert_2 = max|lambda_i|
        $$



        So, for any matrix, the matrix 2-norm is always at least the size of the largest magnitude eigenvalue.



        Now, let $x=[x_1,x_2,ldots, x_n]^T$. Then $Ax = [lambda_1 x_1, lambda_2x_2, ldots, lambda_nx_n]^T$. Therefore,
        $$
        lVert{Ax}rVert_2 = sum_{i=1}^{n} lambda_i^2x_i^2
        $$



        Since $lVert x rVert_2 = 1$ we have that $sum_{i=1}^{n} x_i^2 = 1$ , which implies that $ 0leq x_i^2 leq 1$. Therefore,
        $$
        lVert{Ax}rVert_2^2 = sum_{i=1}^{n} lambda_i^2 x_i^2
        leq sum_{i=1}^{n} (max|lambda_i|)^2 x_i^2
        = (max|lambda_i|)^2 sum_{i=1}^{n} x_i^2
        = (max|lambda_i|)^2
        $$



        Therefore, for diagonal matrices, the matrix 2-norm is bounded above by the size of the largest magnitude eigenvalue, and so the two quantities must be equal.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 23 '18 at 16:34

























        answered Dec 23 '18 at 16:27









        tchtch

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