Eigenvalues of PCA transformed data are the same as original data, why?
$begingroup$
We have an arbitrary Matrix $A^{n×m}$ with $m$ measurements and $n$ samples
To compute the PCA transform of $A$ we compute the eigenvectors for the covariance matrix of $A$
$COV(A) x_A = lambda x_A$
we get Matrix $B^{ntimes m}$ as PCA transformation with
$B = A x_A$
If we compute a PCA transformation for $B$ we get
$COV(B) x_B = lambda_B x_B$
and in this case
$COV(A x_A) = COV(B) = lambda_A = lambda_B$
also ${x_A}^T x_A = I = COV(x_A)$
I can intuitively understand that another PCA won't work because the new dimensions are already maximized for variance. But I would like to know if there is a proof for that.
can somebody explain to me why this is?
Thanks
eigenvalues-eigenvectors covariance
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
$endgroup$
add a comment |
$begingroup$
We have an arbitrary Matrix $A^{n×m}$ with $m$ measurements and $n$ samples
To compute the PCA transform of $A$ we compute the eigenvectors for the covariance matrix of $A$
$COV(A) x_A = lambda x_A$
we get Matrix $B^{ntimes m}$ as PCA transformation with
$B = A x_A$
If we compute a PCA transformation for $B$ we get
$COV(B) x_B = lambda_B x_B$
and in this case
$COV(A x_A) = COV(B) = lambda_A = lambda_B$
also ${x_A}^T x_A = I = COV(x_A)$
I can intuitively understand that another PCA won't work because the new dimensions are already maximized for variance. But I would like to know if there is a proof for that.
can somebody explain to me why this is?
Thanks
eigenvalues-eigenvectors covariance
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
$endgroup$
add a comment |
$begingroup$
We have an arbitrary Matrix $A^{n×m}$ with $m$ measurements and $n$ samples
To compute the PCA transform of $A$ we compute the eigenvectors for the covariance matrix of $A$
$COV(A) x_A = lambda x_A$
we get Matrix $B^{ntimes m}$ as PCA transformation with
$B = A x_A$
If we compute a PCA transformation for $B$ we get
$COV(B) x_B = lambda_B x_B$
and in this case
$COV(A x_A) = COV(B) = lambda_A = lambda_B$
also ${x_A}^T x_A = I = COV(x_A)$
I can intuitively understand that another PCA won't work because the new dimensions are already maximized for variance. But I would like to know if there is a proof for that.
can somebody explain to me why this is?
Thanks
eigenvalues-eigenvectors covariance
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
$endgroup$
We have an arbitrary Matrix $A^{n×m}$ with $m$ measurements and $n$ samples
To compute the PCA transform of $A$ we compute the eigenvectors for the covariance matrix of $A$
$COV(A) x_A = lambda x_A$
we get Matrix $B^{ntimes m}$ as PCA transformation with
$B = A x_A$
If we compute a PCA transformation for $B$ we get
$COV(B) x_B = lambda_B x_B$
and in this case
$COV(A x_A) = COV(B) = lambda_A = lambda_B$
also ${x_A}^T x_A = I = COV(x_A)$
I can intuitively understand that another PCA won't work because the new dimensions are already maximized for variance. But I would like to know if there is a proof for that.
can somebody explain to me why this is?
Thanks
eigenvalues-eigenvectors covariance
eigenvalues-eigenvectors covariance
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
edited Dec 13 '18 at 23:42
Jan Scholz
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
asked Dec 13 '18 at 21:32
Jan ScholzJan Scholz
12
12
add a comment |
add a comment |
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