Mixing
Largest eigenvalue of matrix product $A^T B A$
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With $A in mathbb{S}^{d times d}_+$ (symmetric positive semi definite) and $B in mathbb{S}^{d times d}_{++}$ (symmetric positive definite), can we rewrite or upper bound $lambda_{max}(A^T B A)$ in terms of eigenvalues of $A$ and $B$?
linear-algebra eigenvalues-eigenvectors positive-definite symmetric-matrices positive-semidefinite
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
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add a comment |
$begingroup$
With $A in mathbb{S}^{d times d}_+$ (symmetric positive semi definite) and $B in mathbb{S}^{d times d}_{++}$ (symmetric positive definite), can we rewrite or upper bound $lambda_{max}(A^T B A)$ in terms of eigenvalues of $A$ and $B$?
linear-algebra eigenvalues-eigenvectors positive-definite symmetric-matrices positive-semidefinite
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
$endgroup$
add a comment |
$begingroup$
With $A in mathbb{S}^{d times d}_+$ (symmetric positive semi definite) and $B in mathbb{S}^{d times d}_{++}$ (symmetric positive definite), can we rewrite or upper bound $lambda_{max}(A^T B A)$ in terms of eigenvalues of $A$ and $B$?
linear-algebra eigenvalues-eigenvectors positive-definite symmetric-matrices positive-semidefinite
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
$endgroup$
With $A in mathbb{S}^{d times d}_+$ (symmetric positive semi definite) and $B in mathbb{S}^{d times d}_{++}$ (symmetric positive definite), can we rewrite or upper bound $lambda_{max}(A^T B A)$ in terms of eigenvalues of $A$ and $B$?
linear-algebra eigenvalues-eigenvectors positive-definite symmetric-matrices positive-semidefinite
linear-algebra eigenvalues-eigenvectors positive-definite symmetric-matrices positive-semidefinite
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
asked Jan 23 at 20:31
Chuanhao LiChuanhao Li
132
132
add a comment |
add a comment |
1 Answer
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For a quadratic positive semi-definite matrix $M$ we have $lambda_rm{max}(M) = | M |_2$, where $|cdot |_2$ is the operator norm corresponding to the euclidean norm. Since operator norms are submultiplicative:
$$lambda_rm{max}(A^T B A) = |A^T B A|_2 leq |A^T|_2 |B|_2 |A|_2 = lambda_rm{max}(A)^2 cdot lambda_rm{max}(B)$$
Equality can be easily achieved using diagonal matrices.
answered Jan 23 at 20:54
0x5390x539
1,445518
$endgroup$
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
For a quadratic positive semi-definite matrix $M$ we have $lambda_rm{max}(M) = | M |_2$, where $|cdot |_2$ is the operator norm corresponding to the euclidean norm. Since operator norms are submultiplicative:
$$lambda_rm{max}(A^T B A) = |A^T B A|_2 leq |A^T|_2 |B|_2 |A|_2 = lambda_rm{max}(A)^2 cdot lambda_rm{max}(B)$$
Equality can be easily achieved using diagonal matrices.
answered Jan 23 at 20:54
0x5390x539
1,445518
$endgroup$
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
add a comment |
$begingroup$
For a quadratic positive semi-definite matrix $M$ we have $lambda_rm{max}(M) = | M |_2$, where $|cdot |_2$ is the operator norm corresponding to the euclidean norm. Since operator norms are submultiplicative:
$$lambda_rm{max}(A^T B A) = |A^T B A|_2 leq |A^T|_2 |B|_2 |A|_2 = lambda_rm{max}(A)^2 cdot lambda_rm{max}(B)$$
Equality can be easily achieved using diagonal matrices.
answered Jan 23 at 20:54
0x5390x539
1,445518
$endgroup$
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
add a comment |
$begingroup$
For a quadratic positive semi-definite matrix $M$ we have $lambda_rm{max}(M) = | M |_2$, where $|cdot |_2$ is the operator norm corresponding to the euclidean norm. Since operator norms are submultiplicative:
$$lambda_rm{max}(A^T B A) = |A^T B A|_2 leq |A^T|_2 |B|_2 |A|_2 = lambda_rm{max}(A)^2 cdot lambda_rm{max}(B)$$
Equality can be easily achieved using diagonal matrices.
answered Jan 23 at 20:54
0x5390x539
1,445518
$endgroup$
For a quadratic positive semi-definite matrix $M$ we have $lambda_rm{max}(M) = | M |_2$, where $|cdot |_2$ is the operator norm corresponding to the euclidean norm. Since operator norms are submultiplicative:
$$lambda_rm{max}(A^T B A) = |A^T B A|_2 leq |A^T|_2 |B|_2 |A|_2 = lambda_rm{max}(A)^2 cdot lambda_rm{max}(B)$$
Equality can be easily achieved using diagonal matrices.
answered Jan 23 at 20:54
0x5390x539
1,445518
answered Jan 23 at 20:54
0x5390x539
1,445518
answered Jan 23 at 20:54
0x5390x539
1,445518
answered Jan 23 at 20:54
0x5390x539
1,445518
1,445518
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
add a comment |
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
$begingroup$
Thank you for the help! I didn't realize operator norm has sub-multiplicativity. For people who want to see proof for sub-multiplicativity. math.stackexchange.com/a/580838/637666
$endgroup$
– Chuanhao Li
Jan 23 at 22:45
add a comment |
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