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Let $A$ be $n\times n$ non-singular matrix with real entries. Let $B=A^T$ denote the transpose of $A$. Which of the following matrices are positive definite?

$1. A+B \\2.A^{-1}+B^{-1}\\3.AB\\4.ABA$

Actually, by finding some counter-examples we can cross out the options but I want to have some argument to check their positive definiteness via eigenvalues. Kindly help if it can be done in this way too. Thank You!

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    3 is the only one which is necessarily positive definite. – 2017-01-16

2 Answers 2

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Take $A = -I$ as an counterexample for $1,2,4$.

$AB$, however, is necessarily positive definite. That is, for any non-singular $A$, $AA^T$ is positive definite. The ideal argument here is not by eigenvalues, but the definition of positive definite.

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  1. The matrix $A + A^T$ is symmetric, but not necessarily positive definite. For example, we can take a symmetric $A$ $$ A = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix} $$ and then $$ A + A^T = 2A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ has eigenvalues $x^2 - \operatorname{tr}(A + A^T)x + \det(A + A^T) = x^2 -1 = (x - 1)(x+1)$.
  2. We have $(A^T)^{-1} = (A^{-1})^T$. Hence, $A^{-1} + (A^T)^{-1} = A^{-1} + (A^{-1})^T$. By taking $A$ to be the inverse of the matrix in the previous item, we obtain a counterexample.
  3. This matrix is positive semi-definite because it is symmetric $(AA^T)^T = (A^T)^T A^T = AA^T$) and we have $\left< AA^Tx, x \right> = \left< A^Tx, A^T x \right> \geq 0$. Since $A$ is invertible, so is $A^T$ and hence we see that $AA^T$ is positive definite (as $\det(AA^T) \neq 0$).
  4. We have $(AA^TA)^T = A^TAA^T$ which leads us to suspect that the product is not even symmetric. And indeed, if we take $$ A = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} $$ we have $$ AA^TA = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}= \begin{pmatrix} 2 & 2 \\ 0 & 0 \end{pmatrix}. $$