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Suppose A and B are $m\times K$ and $n\times K$ matrices respectively. Some people state that $$\sigma_K^2(AB^T)=\lambda_{\text{min}}(B^TBA^TA)\geq\lambda_{\text{min}}(B^TB)\lambda_{\text{min}}(A^TA).$$ But I don't understand the proof of the statement. The following is my try so far. Thanks in advance for any help.

The second inequality is proved in this link.

Ѕᴀᴀᴅ
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Daaaaa
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    You need $K\le\min{m,n}$ to make sense of $\sigma_K(AB^T)$. The inequality can be rewritten as $\sigma_K(AB^T)\ge\sigma_\min(A)\sigma_\min(B)$, which is quite obvious if you consider the restriction of $AB^T$ on $\ker(B^T)^\perp$. Alternatively, since singular values are unitarily invariant, by SVD you may assume that $B^T=\pmatrix{S&0}$ where $S$ is a $K\times K$ nonnegative diagonal matrix. Then $AB^T=\pmatrix{AS&0}$ and hence $\sigma_K(AB^T)=\sigma_\min(AS)\ge\sigma_\min(A)\sigma_\min(S)=\sigma_\min(A)\sigma_\min(B)$. – user1551 Sep 26 '22 at 12:28
  • @user1551 Hi, I am actually very puzzeled with $\sigma_K^2(AB^T)=\lambda_{K}(AB^TBA^T)=\lambda_{min}(B^TBA^TA)$. Could you please help me understand why the interchange is legitimate. Thanks! – Daaaaa Sep 27 '22 at 02:07
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    By Sylvester's eigenvalue theorem, if both $XY$ and $YX$ are square matrices and the size of $XY$ is larger than or equal to the size of $YX$, then the spectrum of $XY$ is the spectrum of $YX$ together with an appropriate number of zero eigenvalues (see a proof here). Now put $X=A$ and $Y=B^TBA^T$. Then $YX=B^TBA^TA$ shares its $K$ eigenvalues with $XY=AB^TBA^T$. Since $XY$ is positive semidefinite, these $K$ eigenvalues must be nonnegative... – user1551 Sep 27 '22 at 03:53
  • ...As the other $m-K$ eigenvalues of $XY$ are zero, the the $K$-th largest eigenvalue of $XY$ is precisely the $K$-th largest eigenvalue of $YX$. Yet, $YX$ is $K\times K$. Its $K$-th largest eigenvalue is precisely its minimum eigenvalue. Thus $\sigma_K(XY)=\sigma_K(YX)=\sigma_\min(YX)$. – user1551 Sep 27 '22 at 03:53

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