PS0solution

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CS229 2018Autumn 的PS0的解答

  1. \[ \nabla f(x) = Ax+b^T \]

  2. \[ \nabla f(x) = g^{'}(h(x))\nabla h(x) \]important

  3. \[ \nabla^2 f(x) = A \]

  4. \[ \nabla f(x) = g^{'}(a^Tx)a \\ \nabla^2 f(x) = g^{''}(a^Tx)aa^T \]

  1. $ A^T = (zzT)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

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

  1. (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\(A = U\Lambda U^T \geq 0\) \(U\Lambda U^T = [u^{(i)}]\Lambda [u^{(i)}]^T\) \(= \lambda^{(i)}(t^{(i)})^2\) \(\therefore \lambda^{(i)} \geq 0\)