$B=I+A$ is non-singular and positive definite 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 .
 A: Your solution is fine. An alternative solution is to prove that the eigenvalues of $A$ are non-negative. In fact if $\lambda$ is an eigenvalue of $A$ and $X$ is an eigenvector associated then
$$X^TAX=\lambda X^TX\ge0\implies \lambda\ge0$$
hence $$\det(A+I)=\chi_A(-1)\ne0$$
because $-1$ isn't an eigenvalue of $A$. Hence $A+I$ is invertible.
A: Yet another way:
Lemma:  Let $\lambda$ be any eigenvalue of the matrix $A$, with eigenvector $v$:
$Av = \lambda v; \tag{1}$
then for any scalar $\mu$, $\lambda + \mu$ is an eigenvalue of $A + \mu I$, also with eigenvector $v$:
$(A + \mu I)v = (\lambda + \mu)v. \tag{2}$
Proof of Lemma:  We have
$(A + \mu I)v = Av + \mu I v = \lambda v + \mu v = (\lambda + \mu)v. \tag{3}$
End:  Proof of Lemma.
Applying this lemma to the case at hand we immediately see that the eigenvalues of $B = I + A$ must all be of the form $\lambda + 1$, where $\lambda$ is an eigenvalue of $A$.  It follows that every eigenvalue of $B$ is positive.  Thus $B$ is non-singular and fact positive definite.
