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What I want:

If $A$ is SPD then there is a matrix $X$ such that $A = X^2$

I am able to prove this using the Schur Decomposition, but I was asked to prove it using SVD decomposition.

I was trying to prove that if $A = UDV^T = VDU^T = A^T$ then $U = V$, but I'm failing to do that.

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    $\begingroup$ hint: also use the positive defineted-ness of the matrix $\endgroup$
    – learner
    Sep 9 '20 at 19:13
  • $\begingroup$ @learner I don't see why it's useful to prove $ U = V $ since $D$ for the SVD has positive values anyway $\endgroup$
    – Figurinha
    Sep 9 '20 at 19:19
  • $\begingroup$ @learner tried a lot, but all I can get is that $A^2 = A^TA = VD^2V^T = U D^2 U^T = AA^T$... But I don't know how to define $X$ :( $\endgroup$
    – Figurinha
    Sep 9 '20 at 19:49
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    $\begingroup$ there is no merit in finding $A^2$. If at all you should be squaring anything it should be $X$ ;) $\endgroup$
    – learner
    Sep 9 '20 at 20:12
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$A$ is symmetric. $A$ possesses complete set of orthogonal eigenvectors. So we have $Ax = \lambda x$, where $x$ a right eigenvector corresponding to eigenvalue $\lambda$. Now, it is also true that $x^TA = \lambda x^T$, and so $x$ is also a left eigenvector. Therefore we have in the SVD, the orthonormal matrices $U$ and $V$ are the same. Now \begin{align} A &= UDV^T \\ &\stackrel{a}= UD^{\frac{1}{2}}D^{\frac{1}{2}}V^T \\ &\stackrel{b}= UD^{\frac{1}{2}}D^{\frac{1}{2}}U^T \\ &\stackrel{c}= UD^{\frac{1}{2}}V^TVD^{\frac{1}{2}}U^T \\ &\stackrel{d}= X.X \\ &= X^2 \end{align}

  • a: Since A is PD, the eigenvalues are positive so a halving operation is allowed.
  • b: $U=V$
  • c: $V^TV$ = I
  • d: Defining $X$ to be a matrix with eigenvalues half power of the corresponding eigenvalues of $A$, eigenvectors same as that of $A$, and the fact that $U=V$

For every symmetric matrix $\in \mathbb{R}^{n \times n}$ there exist $n$-linearly independent eigenvectors. And for any matrix, for any eigenvalue, we have algebraic multiplicity $\geq$ geometric multiplicity of the. Using the above two statements, one can conclude that a symmetric matrix has $n$ eigenvalues.

One can represent any symmetric PD matrix in its eigenvalue decomposition form. The way one would derive this is take the idea $Ax = \lambda x$ to a matrix level to get $AX = XD$, where $D$ is a diagonal matrix with eigenvalues in its diagonal. These eigenvalues are arranged so that they get multiplied with their corresponding eigenvector in $X$ (columns of $X$). Since $X$ is orthonormal, $X^T = X^{-1}$ and one can write $A =XDX^T$, which also happens to be the SVD of $A$ with $U=V=X$.

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  • $\begingroup$ I still can't understand why $U = V$. I have never used this argument of right and left eigenvectors for solving SVD problems. Could you detail more this part, please? The conclusion after $U = V$ I can understand. :) $\endgroup$
    – Figurinha
    Sep 10 '20 at 13:59
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    $\begingroup$ @Figurinha I have added more explanation. Let me know if it is clear. Like we have a definition for right eigenvector (or "eigenvector") which is $Ax = \lambda x$, we have a similar definition for left eigenvector which is, $x$ is left eigenvector if $x \neq 0, x^TA = \lambda x$. In the definition of SVD, the columns of $U$ are left eigenvalues. $\endgroup$
    – learner
    Sep 10 '20 at 15:29

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