Exercise 6.5: Matrix Monotone Functions
(a)
Let $A, B \in \S^{d\times d}_+$. Note that \begin{equation} (A + t B)^2 - A^2 = t^2 B^2 + t(AB + BA). \end{equation} Therefore, if $AB + BA < 0$, then can make $t$ small to find the result. Consider \begin{equation} A = \begin{bmatrix} 1 & 0 \newline 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 1 \newline 1 & 1 \end{bmatrix}. \end{equation} Then \begin{equation} AB + BA = \begin{bmatrix} 2 & 1 \newline 1 & 0 \end{bmatrix}, \end{equation} and a quick calculation shows that $AB + BA$ has eigenvalues $1 \pm \sqrt{2}$.
(b)
Set $X = A$ and $H = tB$ and rearrange \begin{align} e^{X + H} &= 1 + (X + H) + \tfrac12(X + H)^2 + \tfrac1{3!}(X + H)^3 + \ldots \newline &= 1 + (X + H) + \tfrac12(X^2 + XH + HX + H^2) \nonumber \newline &\qquad + \tfrac1{3!}(X^3 + X^2 H + XHX + H X^2) + \ldots + o(t) \newline &= e^X + H + \tfrac12(XH + HX) + \tfrac1{3!}(X^2 H + XHX + H X^2) + \ldots + o(t). \end{align} Suppose that $X$ is a projection: $X^n = X$ for all $n \in \N$. Then \begin{align} e^{X + H} &= e^X + H + (\tfrac12 + \tfrac1{3!} + \ldots)(XH + HX) + \parens{\sum_{i=3}^\infty\frac{i - 2}{i!}}XHX + o(t) \newline &= e^X + H + (e - 2)(XH + HX) + (3 - e)XHX + o(t) \newline &= e^X + t \begin{bmatrix} e & e - 1 \newline e - 1 & 1 \end{bmatrix} + o(t), \end{align} where the latter matrix is negative definite.
(c)
Use that $A \mapsto A^p$ is monotone for $p \in [0, 1]$ and that \begin{equation} \log(x) = \lim_{h \to \infty} h\,(x^{1/h} - 1). \end{equation}