How does the matrix multiplication change the smallest singular value?
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Assume the matrix $X in R^{Mtimes N}$ has rank $r$. Suppose we have an arbitrary semi-unitary matrix $W in R^{Mtimes r}$ such that $W^T W = I$.
In general, the matrix $Y=W^TX in R^{rtimes N}$ will have rank $r$. Suppose that the $r$th largest, i.e., the smallest singular value, of $X$ is $sigma_r$, what is the smallest singular value of $Y$ in terms of $W$?
matrices matrix-decomposition
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
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add a comment |
$begingroup$
Assume the matrix $X in R^{Mtimes N}$ has rank $r$. Suppose we have an arbitrary semi-unitary matrix $W in R^{Mtimes r}$ such that $W^T W = I$.
In general, the matrix $Y=W^TX in R^{rtimes N}$ will have rank $r$. Suppose that the $r$th largest, i.e., the smallest singular value, of $X$ is $sigma_r$, what is the smallest singular value of $Y$ in terms of $W$?
matrices matrix-decomposition
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
$endgroup$
$begingroup$
The rank might not be $r$. For instance, if some of columns of $W$ are orthogonal to the range of $X$ then the rank will be less than $r$. This means the smallest singular value of $Y$ could be 0. I'm not sure there is a closed for formula for the smallest singular value.
$endgroup$
– tch
Dec 28 '18 at 17:56
add a comment |
$begingroup$
Assume the matrix $X in R^{Mtimes N}$ has rank $r$. Suppose we have an arbitrary semi-unitary matrix $W in R^{Mtimes r}$ such that $W^T W = I$.
In general, the matrix $Y=W^TX in R^{rtimes N}$ will have rank $r$. Suppose that the $r$th largest, i.e., the smallest singular value, of $X$ is $sigma_r$, what is the smallest singular value of $Y$ in terms of $W$?
matrices matrix-decomposition
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
$endgroup$
Assume the matrix $X in R^{Mtimes N}$ has rank $r$. Suppose we have an arbitrary semi-unitary matrix $W in R^{Mtimes r}$ such that $W^T W = I$.
In general, the matrix $Y=W^TX in R^{rtimes N}$ will have rank $r$. Suppose that the $r$th largest, i.e., the smallest singular value, of $X$ is $sigma_r$, what is the smallest singular value of $Y$ in terms of $W$?
matrices matrix-decomposition
matrices matrix-decomposition
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
asked Dec 28 '18 at 13:20
ZHANG WeiZHANG Wei
388
388
$begingroup$
The rank might not be $r$. For instance, if some of columns of $W$ are orthogonal to the range of $X$ then the rank will be less than $r$. This means the smallest singular value of $Y$ could be 0. I'm not sure there is a closed for formula for the smallest singular value.
$endgroup$
– tch
Dec 28 '18 at 17:56
add a comment |
$begingroup$
The rank might not be $r$. For instance, if some of columns of $W$ are orthogonal to the range of $X$ then the rank will be less than $r$. This means the smallest singular value of $Y$ could be 0. I'm not sure there is a closed for formula for the smallest singular value.
$endgroup$
– tch
Dec 28 '18 at 17:56
$begingroup$
The rank might not be $r$. For instance, if some of columns of $W$ are orthogonal to the range of $X$ then the rank will be less than $r$. This means the smallest singular value of $Y$ could be 0. I'm not sure there is a closed for formula for the smallest singular value.
$endgroup$
– tch
Dec 28 '18 at 17:56
$begingroup$
The rank might not be $r$. For instance, if some of columns of $W$ are orthogonal to the range of $X$ then the rank will be less than $r$. This means the smallest singular value of $Y$ could be 0. I'm not sure there is a closed for formula for the smallest singular value.
$endgroup$
– tch
Dec 28 '18 at 17:56
add a comment |
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$begingroup$
The rank might not be $r$. For instance, if some of columns of $W$ are orthogonal to the range of $X$ then the rank will be less than $r$. This means the smallest singular value of $Y$ could be 0. I'm not sure there is a closed for formula for the smallest singular value.
$endgroup$
– tch
Dec 28 '18 at 17:56