PS0solution

1. (a)
   $$
   \nabla f(x) = Ax+b^T
   $$
   (b)
   $$
   \nabla f(x) = g^{‘}(h(x))\nabla h(x)
   $$important

(c)
$$
\nabla^2 f(x) = A
$$
(d)
$$
\nabla f(x) = g^{‘}(a^Tx)a \
\nabla^2 f(x) = g^{‘’}(a^Tx)aa^T
$$

2. (a)
   $ A^T = (zz^T)^T = zz^T = A $
   对于$\forall x \in R^n$
   有$x^TAx = x^Tzz^Tx \
   = (x^Tz)(x^Tz)^T$

因为x^Tz 正定
所以A正定
(b)
$\because$ z为非零向量
$\therefore R(A) = 1$
$N(A = {x|z^Tx=0})$ N 是NullSpace

(c)
对于$\forall x \in R^m$
有$x^TBAB^Tx = (x^TB)A(x^TB)^T$
因为A正定，所以 $ (B^Tx)^TA(B^Tx) \geq 0$
所以$BAB^T$PSD

3. (a).
   $$
   \begin{align*}
   \because A &= T\Lambda T^{-1} \
   \therefore AT &= T\Lambda \
   A[t^{(i)}] &= [t^{(i)}]\Lambda \
   [At^{(0)},At^{(1)},\cdots,At^{(n-1)}] &= \begin{bmatrix} t^{(0)} & t^{(1)}& \cdots & t^{(n-1)}\end{bmatrix} \cdot \begin{bmatrix}
   \lambda^{0} & 0 & 0 & \cdots & 0 \
   0 & \lambda^{1} & 0 & \cdots & 0 \
   0 & 0 & \lambda^{2} & \cdots & 0 \
   \vdots & \vdots & \vdots & \ddots & \vdots \
   0 & 0 & 0 & \cdots & \lambda^{n-1}
   \end{bmatrix} \
   \therefore At^{(i)}&=\lambda t{(i)}
   \end{align*}
   $$
   (b).
   $$
   A=U\Lambda U^T \
   AU = U\Lambda \
   $$
   然后和a相同
   (c).
   $\because$ A is PSD
   $\therefore$ 对于A的特征矩阵 有$A = U\Lambda U^T \geq 0$
   $U\Lambda U^T = [u^{(i)}]\Lambda [u^{(i)}]^T$
   $= \lambda^{(i)}(t^{(i)})^2$
   $\therefore \lambda^{(i)} \geq 0$