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I realize that this is because when the eigenvalues are either 0 or 1 they will have the same square root. But why does this happen?

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    $\begingroup$ Eigenvalues are not either 0 or 1. N times any unit matrice has N as eigenvalue. $\endgroup$ – Lord_Gestalter Jul 15 '14 at 6:09
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    $\begingroup$ The statement "eigenvalues are either $0$ or $1$" is false. For example, $2I$ is a positive definite matrix with an eigenvalue $2$. $\endgroup$ – Oria Gruber Jul 15 '14 at 6:13
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Let $\lambda, v$ be an eigenpair for a positive definite matrix $P$, i.e. $Pv = \lambda v$. Multiply both sides of this equation by $P^\top$ to get $P^{\top}Pv = P^{\top}\lambda v$. We have $P^\top = P$ and hence $P^{\top}Pv = \lambda^2 v$. Therefore $\lambda^2$ is an eigenvalue for $P^{\top}P$, which is the square of a singular value for the matrix $P$. Since $P$ is positive definite, $\lambda > 0$ and hence $\sqrt{\lambda^2} = \lambda$. Therefore, the singular value is equal to the eigenvalue.

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  • $\begingroup$ Doesn't "Therefore λ2 is an eigenvalue for P⊤P, which is the square of a singular value for the matrix P." use the assumption you want to prove? $\endgroup$ – Lord_Gestalter Jul 15 '14 at 6:11
  • $\begingroup$ Nope. When you have $P^{\top}Pv = \lambda^2 v$, $\lambda^2$ is by construction an eigenvalue for $P^{\top}P$. $\endgroup$ – Calculon Jul 15 '14 at 6:14
  • $\begingroup$ I forgot to add that singular values of any real matrix $A$ ($A$ does not even have to be square) are the square roots of the eigenvalues of $A^{\top}A$. $\endgroup$ – Calculon Jul 15 '14 at 6:29
  • $\begingroup$ My comment was more about the singular value side of the sentence, but your're right. I'll have to repeat the basics once in a while ... $\endgroup$ – Lord_Gestalter Jul 15 '14 at 6:35
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Another approach:

if $A$ is positive definite, then $A$ can be diagonlized by an orthogonal matrix $P$:

$PAP^T=D$ where $D$ is diagonal matrix with the eigenvalues on the diagonal.

Since $A=A^T$:

$AA^T=P^TDPP^TDP=P^TD^2P$

Or in other words:

$PAA^TP^T=D^2$, so the diagonal form of $AA^T$ is $D^2$ and it has $AA^T$'s eigenvalues on the diagonal.

This shows us that if $\alpha$ is an eigenvalue of $A$, then $\alpha^2$ is an eigenvalue of $AA^T$. Since $A$ is positive definite, we know $\alpha >0$, and so $\alpha=\sqrt{\alpha ^2}$

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If a matrix $A$ is positive definite, it will be unitarily similar to a diagonal matrix $D$. That is, $A = PDP^T$ for some unitary matrix $P$. Notice that this is the eigendecomposition of $A$ and, therefore, $D$ will have $A$'s eigenvalues on the diagonal. Additionally, since $A$ is positive definite, its eigenvalues, and therefore all the diagonal elements of $D$, will be positive numbers. Finally, notice that $PDP^T$ is also the singular value decomposition of $A$.

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