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A $n\times n$ projective matrix $A$ is a Hermitian and idempotent matrix, that is, $$A^*=A=A^2.$$ Suppose that $A,B$ are both projective, then $\|A-B\|\leq 1$, here $\|\cdot\|$ is the spectral norm, $$\|A\|=\max_{\|x\|_2=1}\|Ax\|_2, \|x\|_2^2=\sum_{i=1}^n |x_i|^2,\ x=(x_1,\cdots,x_n).$$

The way I do is to first diagonalize $A,B$ as $$A=U^*\operatorname{diag}(I_r,0)U,\quad B=V^*\operatorname{diag}(I_s,0)V,$$ where $U,V$ are unitary matrices, $U^*U=I$, and $I_n$ is the identity matrix. Then $$\|A-B\|=\|U^*\operatorname{diag}(I_r,0)U-V^*\operatorname{diag}(I_s,0)V\|$$ $$=\|\operatorname{diag}(I_r,0)W-W\operatorname{diag}(I_s,0)\|,\quad W=UV^*$$ Direct computations show that the entries of the matrix $\operatorname{diag}(I_r,0)W-W\operatorname{diag}(I_s,0)$ is $0$ or that of the unitary matrix $W$. Do the spectral norm of such matrices $\leq 1$?

Michael Hardy
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xldd
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1 Answers1

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So you want to prove that $\|A - B \| \leq 1$. Basically, you can get this from elementary geometry. The details are as follows.

We pick an $x \in \mathbb C^n$ and set $$a = Ax, \quad b = x - Ax, \quad c = Bx, \quad d = x - Bx.$$Then we must show that $\|a - c\| \leq \|x\|$. We use the usual inner product on $\mathbb C^n$, i.e. $\left<u,v\right> = u^T\overline v$ for (column vectors) $u, v \in \mathbb C^n$. With that, we get from the projection properties of $A$ $$\left<a,b\right> = \left<Ax,x - Ax\right> = \left<x, A^*x - A^*Ax\right> = \left<x,Ax - A^2x\right> = \left<x,Ax - Ax\right> = 0.$$ And similarly $$\left<b,d\right> = 0.$$

By construction, we have the decompositions $$x = a + b = c + d,$$ from which we get $$\left<x,x\right> = \left<a,a\right> + \left<b,b\right> = \left<c,c\right> + \left<d,d\right>.$$

Next, from $x = a + b$ we get $$a - \frac{1}{2}x = \frac{1}{2}\left(a-b\right).$$ This in turn implies $$\left<a - \frac{1}{2}x,a - \frac{1}{2}x\right> = \frac{1}{4}\left<a-b,a-b\right> = \frac{1}{4}\left(\left<a,a\right> + \left<b,b\right>\right) = \frac{1}{4}\left<x,x\right>.$$

By a completely analogous calculation we find $$\left<c - \frac{1}{2}x,c - \frac{1}{2}x\right> = \frac{1}{4}\left<x,x\right>.$$

Recall that $\|u\|^2 = \left<u,u\right>$ for any $u \in \mathbb C^n$. Thus, we have just shown that $$\|a - \frac{1}{2}x\| = \frac{1}{2}\|x\| = \|c - \frac{1}{2}x\|.$$

Now, recall the triangle inequality for the norm $\|\cdot\|$. This gives us $$\|a - c\| = \|a - \frac{1}{2}x + \frac{1}{2}x - c\| \leq \|a - \frac{1}{2}x\| + \|\frac{1}{2}x - c\| = \frac{1}{2}\|x\| + \frac{1}{2}\|x\| = \|x\|.$$ This is exactly what we want.

And believe it or not, behind all these calculations is a simple geometric intuition, namely that the distance between two points on a sphere is at most the diameter of that sphere.

In the calculations above, the sphere has center $\frac{1}{2}x$ and diameter $\|x\|$, and the two points on the sphere are $a$ and $c$.

jflipp
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