Mixing
A good formula for singular value matrix of SVD?
$begingroup$
For the SVD, $A_{mtimes n}=U_mSigma_{mtimes n}V^T_n$ where $U$ & $V$ are orthogonal matrices & $Sigma$ is diagonal, I am trying to obtain a formula for $Sigma$...
If $SigmaSigma^T$ was found using $AA^T=USigmaSigma^TU^T$, then:
$$Sigma = left{begin{array}{ll}
(SigmaSigma^T)^{1/2}begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}, & m > n\
(SigmaSigma^T)^{1/2}, & m = n\
(SigmaSigma^T)^{1/2}begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}, & m < n
end{array}
right.$$
Likewise, if $Sigma^TSigma$ was found using $A^TA=VSigma^TSigma V^T$, then:
$$Sigma = left{begin{array}{ll}
begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m > n\
(Sigma^TSigma)^{1/2}, & m = n\
begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m < n
end{array}
right.$$
Did I make any mistakes? If not, is there a more elegant/simpler formula?
Notes:
- I knew that the square roots of $SigmaSigma^T$ & $Sigma^TSigma$ would have only real entries bc they are just the eigenvalue matrices of the positive semi-definite $AA^T$ & $A^TA$, resp. whose eigenvalues are always positive.
$SigmaSigma^T$ & $Sigma^TSigma$ are square diagonal matrices so finding the square roots of their diagonal elements is all that is required to find $(SigmaSigma^T)^{1/2}$ & $(Sigma^TSigma)^{1/2}$.- From $AV=USigma$ I found that $Sigma=U^TAV$
linear-algebra eigenvalues-eigenvectors matrix-decomposition singularvalues
asked Jan 31 at 2:03
LandonLandon
1139
$endgroup$
add a comment |
$begingroup$
For the SVD, $A_{mtimes n}=U_mSigma_{mtimes n}V^T_n$ where $U$ & $V$ are orthogonal matrices & $Sigma$ is diagonal, I am trying to obtain a formula for $Sigma$...
If $SigmaSigma^T$ was found using $AA^T=USigmaSigma^TU^T$, then:
$$Sigma = left{begin{array}{ll}
(SigmaSigma^T)^{1/2}begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}, & m > n\
(SigmaSigma^T)^{1/2}, & m = n\
(SigmaSigma^T)^{1/2}begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}, & m < n
end{array}
right.$$
Likewise, if $Sigma^TSigma$ was found using $A^TA=VSigma^TSigma V^T$, then:
$$Sigma = left{begin{array}{ll}
begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m > n\
(Sigma^TSigma)^{1/2}, & m = n\
begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m < n
end{array}
right.$$
Did I make any mistakes? If not, is there a more elegant/simpler formula?
Notes:
- I knew that the square roots of $SigmaSigma^T$ & $Sigma^TSigma$ would have only real entries bc they are just the eigenvalue matrices of the positive semi-definite $AA^T$ & $A^TA$, resp. whose eigenvalues are always positive.
$SigmaSigma^T$ & $Sigma^TSigma$ are square diagonal matrices so finding the square roots of their diagonal elements is all that is required to find $(SigmaSigma^T)^{1/2}$ & $(Sigma^TSigma)^{1/2}$.- From $AV=USigma$ I found that $Sigma=U^TAV$
linear-algebra eigenvalues-eigenvectors matrix-decomposition singularvalues
asked Jan 31 at 2:03
LandonLandon
1139
$endgroup$
add a comment |
$begingroup$
For the SVD, $A_{mtimes n}=U_mSigma_{mtimes n}V^T_n$ where $U$ & $V$ are orthogonal matrices & $Sigma$ is diagonal, I am trying to obtain a formula for $Sigma$...
If $SigmaSigma^T$ was found using $AA^T=USigmaSigma^TU^T$, then:
$$Sigma = left{begin{array}{ll}
(SigmaSigma^T)^{1/2}begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}, & m > n\
(SigmaSigma^T)^{1/2}, & m = n\
(SigmaSigma^T)^{1/2}begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}, & m < n
end{array}
right.$$
Likewise, if $Sigma^TSigma$ was found using $A^TA=VSigma^TSigma V^T$, then:
$$Sigma = left{begin{array}{ll}
begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m > n\
(Sigma^TSigma)^{1/2}, & m = n\
begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m < n
end{array}
right.$$
Did I make any mistakes? If not, is there a more elegant/simpler formula?
Notes:
- I knew that the square roots of $SigmaSigma^T$ & $Sigma^TSigma$ would have only real entries bc they are just the eigenvalue matrices of the positive semi-definite $AA^T$ & $A^TA$, resp. whose eigenvalues are always positive.
$SigmaSigma^T$ & $Sigma^TSigma$ are square diagonal matrices so finding the square roots of their diagonal elements is all that is required to find $(SigmaSigma^T)^{1/2}$ & $(Sigma^TSigma)^{1/2}$.- From $AV=USigma$ I found that $Sigma=U^TAV$
linear-algebra eigenvalues-eigenvectors matrix-decomposition singularvalues
asked Jan 31 at 2:03
LandonLandon
1139
$endgroup$
For the SVD, $A_{mtimes n}=U_mSigma_{mtimes n}V^T_n$ where $U$ & $V$ are orthogonal matrices & $Sigma$ is diagonal, I am trying to obtain a formula for $Sigma$...
If $SigmaSigma^T$ was found using $AA^T=USigmaSigma^TU^T$, then:
$$Sigma = left{begin{array}{ll}
(SigmaSigma^T)^{1/2}begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}, & m > n\
(SigmaSigma^T)^{1/2}, & m = n\
(SigmaSigma^T)^{1/2}begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}, & m < n
end{array}
right.$$
Likewise, if $Sigma^TSigma$ was found using $A^TA=VSigma^TSigma V^T$, then:
$$Sigma = left{begin{array}{ll}
begin{bmatrix}I_n\0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m > n\
(Sigma^TSigma)^{1/2}, & m = n\
begin{bmatrix}I_m & 0end{bmatrix}_{mtimes n}(Sigma^TSigma)^{1/2}, & m < n
end{array}
right.$$
Did I make any mistakes? If not, is there a more elegant/simpler formula?
Notes:
- I knew that the square roots of $SigmaSigma^T$ & $Sigma^TSigma$ would have only real entries bc they are just the eigenvalue matrices of the positive semi-definite $AA^T$ & $A^TA$, resp. whose eigenvalues are always positive.
$SigmaSigma^T$ & $Sigma^TSigma$ are square diagonal matrices so finding the square roots of their diagonal elements is all that is required to find $(SigmaSigma^T)^{1/2}$ & $(Sigma^TSigma)^{1/2}$.- From $AV=USigma$ I found that $Sigma=U^TAV$
linear-algebra eigenvalues-eigenvectors matrix-decomposition singularvalues
linear-algebra eigenvalues-eigenvectors matrix-decomposition singularvalues
asked Jan 31 at 2:03
LandonLandon
1139
asked Jan 31 at 2:03
LandonLandon
1139
edited Jan 31 at 2:19
Landon
asked Jan 31 at 2:03
LandonLandon
1139
asked Jan 31 at 2:03
LandonLandon
1139
asked Jan 31 at 2:03
LandonLandon
1139
1139
add a comment |
add a comment |
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