# Let A be a $n\times n$ non-singular symmetric matrix with entries in $(0,\infty)$.

Let $A$ be a $n\times n$ non-singular symmetric matrix with entries in $(0,\infty)$. Then we can conclude that

(a) $|A| > 0 (|A|$ denotes the determinant of A).

(b) $A$ is a positive definite matrix.

(c) $B = A^2$ is a positive definite matrix.

(d) $C = A^{-1}$ is a matrix with entries in $(0,\infty)$.

I took an example matrix $A= \begin{bmatrix} 1 & 2\\ 2 & 1 \\ \end{bmatrix}$, and saw that options (a), (b) and (d) are wrong.

But what is the proper approach for this problem ?

You are right about a,b,d. Here is another approach to c, which does not rely on eigenvalues.

Let $A$ be non-singular and symmetric. Then $B=A^2 = A^TA$ is positive definite. Take a vector $x\ne0$, then it holds $$x^TBx = x^TA^TAx=(Ax)^T(Ax) = \|Ax\|^2 >0,$$ where we used invertibility of $A$: since $x\ne 0$, $Ax$ cannot be zero. This proves positive definitness of $B$.

Edit: this result is true for symmetric, non-singular, real matrix $A$, the entries do not need to be positive. It is also true if $A\in \mathbb C^{n,n}$ is Hermitian, non-singular. However, the proof would be more complex.

You are right about (a), (b), (d).

Eigen values of $A^2$ can not be negative.. Again , eigen values of $A$ may not be $0$ , as $A$ is non-singular. So , eigen values of $A^2$ can never zero. Thus , all eigen values of $A^2$ are positive.

So, $A^2$ is positive definite. So, (c) is TRUE.

• Sorry about confusing you. You were right about $A^2$ being positive definite since it is assumed that $A$ is non-singular :) – GenericNickname May 12 '15 at 8:25
• Thanks for the help .. Is there a better way to start solving this problem .. what if I couldn't think of a proper example matrix .. ? – Stuck in a JAM May 12 '15 at 8:33
• Is it not a better way??? When a statement is FALSE then cite an example..and when a statement is TRUE then try to prove it – Empty May 12 '15 at 8:35

For a, b and d your approach is right you have a counterexample. Let's take a look at c. $A$ symmetric non singular and therefore is diagonalisable with real nonzero eigenvalues. $A^2$ will then have positive eigenvalues (the squares of those of $A$) on top of being symmetric so it is positive definite. Non singular is an essential assumption as $\begin{bmatrix} 1&1\\1&1\end{bmatrix}$ shows.