Mixing
Is there any relation between SVDs of two matrices with same range?
$begingroup$
Let $A$ and $B$ be two symmetric and positive semidefinite matrices with the same size.
Further, assume that $A$ and $B$ share the same column space (i.e., $mathcal R (A) = mathcal R (B)$ ).
Is there any relation (even in terms of strong inequalities) between the SVD of $A$ and $B$?
linear-algebra matrices linear-transformations numerical-linear-algebra
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be two symmetric and positive semidefinite matrices with the same size.
Further, assume that $A$ and $B$ share the same column space (i.e., $mathcal R (A) = mathcal R (B)$ ).
Is there any relation (even in terms of strong inequalities) between the SVD of $A$ and $B$?
linear-algebra matrices linear-transformations numerical-linear-algebra
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be two symmetric and positive semidefinite matrices with the same size.
Further, assume that $A$ and $B$ share the same column space (i.e., $mathcal R (A) = mathcal R (B)$ ).
Is there any relation (even in terms of strong inequalities) between the SVD of $A$ and $B$?
linear-algebra matrices linear-transformations numerical-linear-algebra
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
$endgroup$
Let $A$ and $B$ be two symmetric and positive semidefinite matrices with the same size.
Further, assume that $A$ and $B$ share the same column space (i.e., $mathcal R (A) = mathcal R (B)$ ).
Is there any relation (even in terms of strong inequalities) between the SVD of $A$ and $B$?
linear-algebra matrices linear-transformations numerical-linear-algebra
linear-algebra matrices linear-transformations numerical-linear-algebra
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
asked Jan 19 at 21:25
Bashir SadeghiBashir Sadeghi
424
424
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Yes. The number of singular values will be the same and the left singular vectors corresponding to those singular values will match. This is clear since if $A = USigma V^T$ then $$Ax = sum_{i} u_i (lambda_i v_i^T x), $$ which shows that the column space of $A$ is the same as the span of the columns of $U$. Since the column space of $A$ and $B$ are the same, the span of their left singular vectors will be the same (and since they have the same range, the rank will be the same). They may be different up to reordering however.
While the result above is true in general, note that since $A$ and $B$ are symmetric positive-semidefinite, this implies that they have the exact same singular value decomposition (except for singular values/eigenvalues). That is $A = USigma U^T$ and the columns of $U$ are the same as the corresponding columns of the eigendecomposition for $B$. Indeed, since they are both psd, their svd's will be the same as their eigendecomposition.
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
1
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes. The number of singular values will be the same and the left singular vectors corresponding to those singular values will match. This is clear since if $A = USigma V^T$ then $$Ax = sum_{i} u_i (lambda_i v_i^T x), $$ which shows that the column space of $A$ is the same as the span of the columns of $U$. Since the column space of $A$ and $B$ are the same, the span of their left singular vectors will be the same (and since they have the same range, the rank will be the same). They may be different up to reordering however.
While the result above is true in general, note that since $A$ and $B$ are symmetric positive-semidefinite, this implies that they have the exact same singular value decomposition (except for singular values/eigenvalues). That is $A = USigma U^T$ and the columns of $U$ are the same as the corresponding columns of the eigendecomposition for $B$. Indeed, since they are both psd, their svd's will be the same as their eigendecomposition.
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
1
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
add a comment |
$begingroup$
Yes. The number of singular values will be the same and the left singular vectors corresponding to those singular values will match. This is clear since if $A = USigma V^T$ then $$Ax = sum_{i} u_i (lambda_i v_i^T x), $$ which shows that the column space of $A$ is the same as the span of the columns of $U$. Since the column space of $A$ and $B$ are the same, the span of their left singular vectors will be the same (and since they have the same range, the rank will be the same). They may be different up to reordering however.
While the result above is true in general, note that since $A$ and $B$ are symmetric positive-semidefinite, this implies that they have the exact same singular value decomposition (except for singular values/eigenvalues). That is $A = USigma U^T$ and the columns of $U$ are the same as the corresponding columns of the eigendecomposition for $B$. Indeed, since they are both psd, their svd's will be the same as their eigendecomposition.
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
$endgroup$
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
1
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
add a comment |
$begingroup$
Yes. The number of singular values will be the same and the left singular vectors corresponding to those singular values will match. This is clear since if $A = USigma V^T$ then $$Ax = sum_{i} u_i (lambda_i v_i^T x), $$ which shows that the column space of $A$ is the same as the span of the columns of $U$. Since the column space of $A$ and $B$ are the same, the span of their left singular vectors will be the same (and since they have the same range, the rank will be the same). They may be different up to reordering however.
While the result above is true in general, note that since $A$ and $B$ are symmetric positive-semidefinite, this implies that they have the exact same singular value decomposition (except for singular values/eigenvalues). That is $A = USigma U^T$ and the columns of $U$ are the same as the corresponding columns of the eigendecomposition for $B$. Indeed, since they are both psd, their svd's will be the same as their eigendecomposition.
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
$endgroup$
Yes. The number of singular values will be the same and the left singular vectors corresponding to those singular values will match. This is clear since if $A = USigma V^T$ then $$Ax = sum_{i} u_i (lambda_i v_i^T x), $$ which shows that the column space of $A$ is the same as the span of the columns of $U$. Since the column space of $A$ and $B$ are the same, the span of their left singular vectors will be the same (and since they have the same range, the rank will be the same). They may be different up to reordering however.
While the result above is true in general, note that since $A$ and $B$ are symmetric positive-semidefinite, this implies that they have the exact same singular value decomposition (except for singular values/eigenvalues). That is $A = USigma U^T$ and the columns of $U$ are the same as the corresponding columns of the eigendecomposition for $B$. Indeed, since they are both psd, their svd's will be the same as their eigendecomposition.
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
answered Jan 19 at 21:48
OldGodzillaOldGodzilla
58227
58227
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
1
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
add a comment |
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
1
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
$begingroup$
Let $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$. I understand that the column space of $U_1$ and $U_2$ is the same. But, generally $U_1not = U_2$.
$endgroup$
– Bashir Sadeghi
Jan 19 at 22:04
1
1
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
$begingroup$
Sorry, I didn't mean exactly what I wrote. If $A = U_1 Sigma_1 U_1^T$ and $B = U_2 Sigma_2 U_2^T$, since $R(A) = R(B)$, the columns of $U_1$ and $U_2$ corresponding to the nonzero eigenvalues must span the same space. Yes, $U_1 neq U_2$ in general as you noted. I'm just pointing out that their eigendecompositions and svd's are (essentially) the same so that their eigenvectors must span the same space. You also will know that their ranks will be the same.
$endgroup$
– OldGodzilla
Jan 20 at 18:07
add a comment |
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