Why is $||Sigma - I||_2 =max|lambda_i|$ where $lambda_i$ is an eigenvalue of $Sigma - I$
$begingroup$
$||Sigma - I||_2 = max|lambda_i|$ 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
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
$endgroup$
add a comment |
$begingroup$
$||Sigma - I||_2 = max|lambda_i|$ 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
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$||Sigma - I||_2 = max|lambda_i|$ 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
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
$endgroup$
$||Sigma - I||_2 = max|lambda_i|$ 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
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
edited Dec 22 '18 at 21:40
pablo_mathscobar
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
asked Dec 22 '18 at 21:26
pablo_mathscobarpablo_mathscobar
1277
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
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.
answered Dec 23 '18 at 16:27
tchtch
833310
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Dec 23 '18 at 16:27
tchtch
833310
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 23 '18 at 16:27
tchtch
833310
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 23 '18 at 16:27
tchtch
833310
$endgroup$
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.
answered Dec 23 '18 at 16:27
tchtch
833310
edited Dec 23 '18 at 16:34
answered Dec 23 '18 at 16:27
tchtch
833310
answered Dec 23 '18 at 16:27
tchtch
833310
answered Dec 23 '18 at 16:27
tchtch
833310
833310
add a comment |
add a comment |
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$begingroup$
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