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I'm having some difficulty proving the following:

Let $P \in M_n(\mathbb C)$ be idempotent. Prove that all nonzero singular values of $P$ satisfy $\sigma_i \ge 1$.

By definition I know that $P$ being idempotent means $P^2 = P$. Likely I have to invoke the Singular Value Decomposition Theorem to prove the problem. So, let the singular value decomposition of $P$ be given by $$P = U\Sigma V^* = P^2 = U\Sigma V^*U\Sigma V^*.$$ By definition I know that the singular values of $P$ are the square roots of the eigenvalues of $P^*P.$ And unfortunately I am not sure where to go from here.

I was inclined to say that $$\sigma_1 = \|P\|_2 = \|P^2\|_2 \le \|P\|_2\|P\|_2 = \sigma_1^2 \implies 1 \le \sigma_1,$$ for nonzero $\sigma_1$, but this doesn't tell me enough. What about $\sigma_2$? and further?

Can anyone provide a hint?

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  • $\begingroup$ Two nice options: one, note that $P$ can be unitarily upper-triangularized. Two: use the Rayleigh-Ritz characterization of singular values. $\endgroup$ – Omnomnomnom Dec 13 '16 at 15:06
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Hint. Let $P=USV$ be a singular value decomposition and let $W=VU$. Then $P^2=P$ implies that $SWS=S$.

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Thoughts so far: Without loss of generality, suppose that $P$ is upper triangular. Noting that $P^2 = P$, we may take $P$ to have the form $$ P = \pmatrix{I&Q\\0&0} $$ Where $I$ is the identity matrix of size $r$ ($r$ is the rank of $P$). We have $$ P^*P = \pmatrix{I&Q\\Q^*&Q^*Q} $$ Now, if $x$ is an eigenvector of $P^*P$ associated with a non-zero eigenvalue, then $x$ is in $\ker(P^*P)^\perp = im(P^*P) = im(P^*)$. Thus, $x = P^*y$ for some $y$. So, $$ (P^*P)x = \pmatrix{I&Q\\Q^*&Q^*Q} \pmatrix{y\\Q^*y} = ? $$

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  • $\begingroup$ Is $Q$ meant to be unitary? $\endgroup$ – Decaf-Math Dec 13 '16 at 16:23
  • $\begingroup$ Nope. $Q$ isn't even necessarily square $\endgroup$ – Omnomnomnom Dec 13 '16 at 16:30
  • $\begingroup$ I am not sure with $I$ in the block. Since it is triangular, it can have some nonzero entries above the diagonal of $I$. Similarly with $0$. $\endgroup$ – i707107 Dec 13 '16 at 16:44
  • $\begingroup$ @i707107 but $P^2 = P$, which means that the upper-left block must be idempotent as well. $\endgroup$ – Omnomnomnom Dec 13 '16 at 16:45
  • $\begingroup$ @Omnomnomnom I see. Thank you. $\endgroup$ – i707107 Dec 13 '16 at 17:08

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