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 .

  • $\begingroup$ looks good to me! $\endgroup$ – Owen Sizemore May 16 '17 at 20:27
  • $\begingroup$ You might want to be a little more careful with the difference between positive definite and non-negative definite. $\endgroup$ – Michael Biro May 16 '17 at 20:34
  • $\begingroup$ I'm confused. Isn't any (strictly) positive-definite matrix already non-signular? $\endgroup$ – tomasz May 16 '17 at 20:46

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.


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.


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