Let $A$ be a real symmetric non-negative definite $n\times n$ matrix . Set $B=I+A$ . Show that $B$ is non-singular and positive definite .
My work : Take any nonzero column $X\in \mathbb{R}^{n\times 1}$ . Then $X^tBX=X^tIX+ X^tAX= \sum (x_i^2) + X^tAX >0$ Since A is positive definite and by definition of it $X^tAX>0$ for non-zero $X$ .
Suppose $B$ is singular . Then there is a non-zero $X\in \mathbb{R}^{n\times 1}$ such that $BX=AX+X=0 $ i.e $AX=-X$ . Now $-||X||=-(X^tX)=(X^tAX)>0 $ .This shows that $X$ must be non-singular. Contradiction to the fact that such the fact $B$ is non-zero .
is my solution correct ? If not please provide your solution . Thank you a lot .