problem about symmetric positive semi-definite matrix

Question

Let $A,B$ be symmetric positive semi-definite matrix with real entries I have to show that

1. $ Im(A) \subset Im(A+B)$

2. if $tr(AB)=0$ then $ AB=O $

I know that a symmetric matrix A is positive semi-definite iff any principle minors of A $\geq 0$ (There are also other properties such as all eigenvalue is non negative ,for all $x\neq 0 ,x^t(A)x \geq 0$ etc.) also A+B is still symmetric positive semi-definite but I don't know how to apply this property to find $Im(A)$ and $Im(A+B)$ and for 2. If I can show that A and B can be simultaneously diagonalized (I found that this is true? but cannot show it) then $X^{-1}AX=M ,X^{-1}BX=N$ for some diagonal matrix M,N with all entries nonnegative.This mean $X^{-1}ABX=MN$ therefore $0=tr(AB)=tr(MN) $but MN is diagonal matrix with all entries nonnegative so $MN=O$ so $AB=XMNX^{-1} =O$ Do I miss anything? Any ideas to complete this proof? and how to show 1.?

Thanks for your help.

Answer 1

Here is an outline for the first, with some gaps to fill in.

Show instead that ker($A+B) \subset$ ker $A$ (by the dimension formula, this is the same). So we need to show that an $x$ in the kernel of the sum is in the kernel of $A$.

But $0 = (A+ B)x$ is the same as $Ax = -Bx$, and so $x^tAx = - x^tBx$, and so (why?) $$x^tAx=0$$

Now, $A$ has a Cholesky decomposition. If you replace $A$ by it, then you get that the norm of a vector (not $x$ !) is zero. But this vector is (...), and if you left-multiply this vector by part of the Cholesky decomposition again, it stays zero. This leads you to $Ax=0$, which is what we wanted to show.

Answer 2

If $A$ is a real symmetric matrix, then $\text{Im}(A)=\text{ker}(A)^\perp$. Now if $v\in\text{ker}(A+B)$, then $\langle (A+B)v, v\rangle=0$. But as $A$ and $B$ are both positive semi-definite, $\langle Av, v\rangle\geq 0$ and $\langle Bv, v\rangle\geq 0$. This forces $\langle Av, v\rangle=0$. But if we let $\sqrt{A}$ be the positive semi-definite square root of $A$, then $\langle\sqrt{A}v, \sqrt{A}v\rangle=0\Longrightarrow \sqrt{A}v=0\Longrightarrow Av=0\Longrightarrow v\in\text{ker}(A)$. So $\text{ker}(A+B)\subseteq\text{ker}(A)\Longrightarrow \text{Im}(A)\subseteq\text{Im}(A+B)$.

For the second problem: Let $\{v_i\}_{i=1}^n$ be the orthonormal basis of eigenvectors of $B$. Then \begin{align*} \text{tr}(AB)=0\Longrightarrow &\sum_{i=1}^n\langle ABv_i, v_i\rangle=0\\ &\sum_{i=1}^n\lambda_i\langle Av_i, v_i\rangle=0\\ \Longrightarrow &\lambda_i\langle Av_i, v_i\rangle=0\ \ \text{for all }i \end{align*} If $\lambda_i=0$, then $ABv_i=\lambda_iAv_i=0$. If $\langle Av_i, v_i\rangle=0$, then by the previous argument, $Av_i=0$. So $ABv_i=\lambda_i Av_i=0$. All in all, $AB=0$ on the span of $v_1, \cdots, v_n$ and hence $AB=0$.