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Let $A$ be an $m \times n$ matrix with rank$ (A)=r$. Use the spectral decomposition of $A'A$ to show that there exists an $n \times(n-r)$ matrix $X$ such that $AX=0$ and $X'X=I$, where $I$ is identity matrix $(n-r)\times (n-r)$.

I know that for a symmetric matrix $A=VDV'$, where $V$ are orthogonal matrices and $D$ is diagonal of corresponding eigenvalues.

I found the spectral decomposition of $A'A$ to be :$VDDV$, so $V(D^2)V$.

I am stuck at this point. I know rank $(A)$=rank $(A'A)=r$, but cannot how to apply SVD to this problem- my other idea was that $X=V$ since I think $V'V=I$ since they are orthogonal and if $V$ is an orthogonal then $AX=0$. But I don't think this would have the right rank...

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  • $\begingroup$ Yes, that helps. I haven't fully written out the proof yet- but if you're curious you rewrite the above equation as V'A'AV=D, and you can partition the columns of V into a matrix X that correspond to V'A'AV=0, and this X is now a n*(n-r) matrix $\endgroup$ – user3731561 Oct 7 '14 at 11:57

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