# 勉強日記

## (x'Ax)^2<=(x'AA'x)(x'x)

Let $$A$$ be any real square matrix (not necessarily symmetric).  Prove that: $$(x'A x)^2 \leq (x'A A'x)(x'x)$$

The key point in proving this inequality is to recognize that $$x'A A'x$$ can be expressed as vector norm of $$A'x$$.

Proof:

If $$x=0$$, then the inequality is trival.

Suppose $$x \neq 0$$.

$$\frac{x'A x}{x'x} = \frac{(A'x)'x}{\| x \|^2} = (A'\frac{x}{\| x \|})'\frac{x}{\| x \|}$$

Because $$\frac{x}{\| x \|}$$ is a unit vector, $$A'\frac{x}{\| x \|}$$ can be considered as scale and rotation of $$\frac{x}{\| x \|}$$ by $$A'$$.  Thus, the resulting vector norm of $$A'\frac{x}{\| x \|}$$ is $$\alpha$$ for some $$\alpha > 0$$.  And $$(A'\frac{x}{\| x \|})'\frac{x}{\| x \|}=\alpha \, cos(\beta)$$ for some $$-\pi \leq \beta \leq \pi$$, which is the angle between before and after premulitplying $$A'$$.

Now:

$$( \frac{x'A x}{x'x} )^2$$

$$= ( (A'\frac{x}{\| x \|})'\frac{x}{\| x \|} )^2$$

$$=\alpha^2 \, cos(\beta)^2$$

$$\leq \alpha^2$$

$$= (A'\frac{x}{\| x \|})'A'\frac{x}{\| x \|}$$

$$= \frac{(A'x)'A'x}{\| x \|^2}$$

$$= \frac{x'A A'x}{x'x}$$

Finally, multiplying both sides by $$(x'x)^2$$ completes the proof.

## Ab=0 iff A^+b=0, when A is symmetric.

Let $$A'$$ be transpose of $$A$$, $$A^{+}$$ be MP-inverse of $$A$$.

Prove that if $$A=A'$$, then $$A b=0$$ if and only if $$A^+b=0$$.

Proof:

$$A b=0$$ implies that $$b=(I-A^+A)q$$ for some $$q$$.  Then,

$$A^+ b \\ =A^+(I-A^+A)q \\ =A^+(I-A'^+A')q \\ =A^+(I-(AA^+)')q \\ =A^+(I-AA^+)q \\ =A^+q-A^+AA^+q \\ =A^+q-A^+q \\ =0$$

Because $$A^{++}=A$$, the proof is completed.

## 連続冪等行列関数の階数は定数である

Fは冪等であるため、$$tr(F)=rank(F)$$。Fは連続ですから、$$tr(F)$$も連続です。しかし、$$rank(F)$$の変化は常に1の倍数で飛びます。両方合わせば $$tr(F)=rank(F)$$ は定数でなければならないことが分かります。