If $P,Q\ge 0;P+Q=I$, how to prove $||P-Q||_2\le1$

Question

We define $$\|A\|_2=\max_{x\ne0}\frac{\|Ax\|_2}{\|x\|_2}$$ as the matrix norm (Spectral Norm, http://mathworld.wolfram.com/SpectralNorm.html).

If $P,Q$ are definite symmetry matrices and $P+Q=I$. How to prove that $$\|P-Q\|_2\le1$$

My thinking:

In fact, from $P+Q=I$, we can conclude that each pair eigenvalue of P and Q are sum up to 1. As a result, their eigenvalue are all less than 1. But what's the next step to say that $\|P-Q\|_2\le1$?

Answer 1

In general the statement is not true. Take $P = 2I$ and $Q = -I$. Then $P+Q = I$ but $P-Q = 3I$ and $\|3I\|_2 = 3 > 1$

However if you assume that $P$ and $Q$ are both either or negative definite, it could work:

(EDIT it does:) Let $P,Q$ be symmetric, positive definite (spd). Let $v$ be an eigenvector of $Q$. Then

$(P+Q)v = Pv + \lambda v = v \Rightarrow Pv = (1-\lambda)v$

Therefore $v$ is also an eigenvector of $P$ with eigenvalue $(1-\lambda)$. A similar argument shows that every eigenvector of $Q$ is an eigenvector of $P$. Full rank of $P$ and $Q$ implies that every eigenvector of $P-Q$ must be an eigenvector of $P$ and $Q$ also.

Now let $v$ be an eigenvector of both $P - Q$. Then $(P-Q)v = (1-\lambda)v - \lambda v = (1-2\lambda)v$. However both $\lambda$ and $1-\lambda$ are bigger than $0$, since $P$ and $Q$ are spd. That means $\lambda \in [0,1]$ and that implies $(1-2\lambda) \in [-1,1]$. That is true for every eigenvalue of $(P-Q)$, therefore $\|(P-Q)\|_2 \leq 1$.

If $P$ and $Q$ are both negative definite replace $P := -P$ and $Q:= -Q$ and the same argument as above holds.

Answer 2

Here's an alternative solution using the trace. As shown in comments, the claim does not hold if $P, Q$ are not nonnegative definite. So assume that they both are. In this case the eigenvalues $\lambda_j$ of $P$ and the eigenvalues $\mu_j$ of $Q$ satisfy (EDIT: Warning, there is an issue here see comments) $$\tag{1} \begin{array}{ccc} 0\le \lambda_j, & 0\le \mu_j,& \lambda_j+\mu_j \le 1. \end{array} $$ The first two inequalities come from nonnegative definiteness, the last one is obtained by taking the trace of $P+Q=I$.

Since $P-Q$ is symmetric its norm equals its spectral radius, that is, the maximum absolute value of its eigenvalues.

Here comes the key step: as strongly hinted by don-joe's answer, the fact that $P+Q=I$ implies that $P$ and $Q$ commute, because $$ PQ=(P+Q)Q-Q^2 = Q-Q^2= Q(P+Q)-Q^2=QP.$$ And so the eigenvalues of $P-Q$ are $\lambda_j-\mu_j$. Because of (1) we have $$|\lambda_j-\mu_j|\le 1, $$ so in particular $$ \|P-Q\|_2=\max_j |\lambda_j-\mu_j|\le 1, $$ as claimed.