2
$\begingroup$

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 ?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Sorry about confusing you. You were right about $A^2$ being positive definite since it is assumed that $A$ is non-singular :) $\endgroup$ – GenericNickname May 12 '15 at 8:25
  • $\begingroup$ 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 .. ? $\endgroup$ – Stuck in a JAM May 12 '15 at 8:33
  • $\begingroup$ 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 $\endgroup$ – Empty May 12 '15 at 8:35
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.