A question about the symmetric positive definite matrix A and $D^{-1/2}AD^{-1/2}$
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Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.
What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?
linear-algebra matrices positive-definite
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add a comment |
$begingroup$
Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.
What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?
linear-algebra matrices positive-definite
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add a comment |
$begingroup$
Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.
What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?
linear-algebra matrices positive-definite
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Assume that $Ainmathbb{R}^{ntimes n}$ is a symmetric positive definite matrix and $D=diag(d_1,ldots,d_n)$ is a diagonal matrix constructing by using the diagonal entries of $A$, indeed $d_i=a_{ii}$.
What can we say about the relationship between the eigenvalues and eigenvectors of $A$ and the eigenvalues and eigenvectors of $D^{-1/2}AD^{-1/2}$?
linear-algebra matrices positive-definite
linear-algebra matrices positive-definite
edited Dec 26 '18 at 15:16
gt6989b
35.8k22557
35.8k22557
asked Dec 26 '18 at 15:15
like_mathlike_math
29917
29917
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1 Answer
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There is not much to say.
$D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.
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I think you are right. Thank you
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– like_math
Dec 26 '18 at 22:00
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1 Answer
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1 Answer
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votes
$begingroup$
There is not much to say.
$D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.
$endgroup$
$begingroup$
I think you are right. Thank you
$endgroup$
– like_math
Dec 26 '18 at 22:00
add a comment |
$begingroup$
There is not much to say.
$D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.
$endgroup$
$begingroup$
I think you are right. Thank you
$endgroup$
– like_math
Dec 26 '18 at 22:00
add a comment |
$begingroup$
There is not much to say.
$D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.
$endgroup$
There is not much to say.
$D^{-1/2} A D^{-1/2}$ is positive definite. It has diagonal entries $1$, so the sum of its eigenvalues is its trace, which is $n$. The trace of $A$ could be any positive number.
answered Dec 26 '18 at 15:26
Robert IsraelRobert Israel
331k23221477
331k23221477
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I think you are right. Thank you
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– like_math
Dec 26 '18 at 22:00
add a comment |
$begingroup$
I think you are right. Thank you
$endgroup$
– like_math
Dec 26 '18 at 22:00
$begingroup$
I think you are right. Thank you
$endgroup$
– like_math
Dec 26 '18 at 22:00
$begingroup$
I think you are right. Thank you
$endgroup$
– like_math
Dec 26 '18 at 22:00
add a comment |
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