# Trace Inequality Tr(ABAB)

1. Apr 22, 2014

### jostpuur

Here's the claim: Assume that $A$ and $B$ are both symmetric matrices of the same size. Also assume that at least other one of them does not have negative eigenvalues. Then

$$\textrm{Tr}(ABAB)\geq 0$$

I don't know how to prove this!

2. Apr 22, 2014

### jbunniii

If $\lambda$ is an eigenvalue of $AB$ then $\lambda^2$ is an eigenvalue of $ABAB$. Since the trace is the sum of the eigenvalues, it suffices to show that the eigenvalues of $AB$ are all real. This would be true if $AB$ were symmetric, but the product of two symmetric matrices need not be symmetric. But can you show that $AB$ is similar to a symmetric matrix?

3. Apr 22, 2014

### jostpuur

Never mind, I succeeded in proving this now. After months of wondering the proof appears in few minutes after posting to PF as usual

My hint for this is that you should first prove that you can assume $A$ to be diagonal with non-negative entries.

4. Apr 22, 2014

### AlephZero

5. Apr 22, 2014

### jbunniii

Only positive semidefinite in this case, so that proof needs to be modified ($K$ may not be invertible). I scratched something together this morning but upon reviewing what I wrote, I see that it contains a mistake. Will take another look this evening if I get a chance.

6. Apr 22, 2014

### jostpuur

According to my formulas this is true too: If $A$ is positive definite, and $B$ is non-zero (and the other assumptions), then

$$\textrm{Tr}(ABAB) > 0$$

It is not clear which would be the most desired form for this problem.

7. Apr 22, 2014

### AlephZero

My bad. I tripped over the meaning of
Too many negatives!

8. Apr 22, 2014

### AlephZero

Clearly if $A$ can be semi-definite, it can be zero, and $ABAB = 0$.

The basic idea of my thinking is to show that $AB$ has a full set of real eigenvalues, by simultaneously diagonalizing $A$ and $B$. The eigenvalues of $(AB)(AB)$ are then all positive.

If the diagonalization fails because $A$ is singular, replace $A$ with the nonsingular matrix $A + eI$ where $e > 0$, and take the limit as $e \to 0$.

9. Apr 22, 2014

### jbunniii

I was thinking of doing something similar to how one can "compress" the singular value decomposition when some of the singular values are zero. Here's a sketch:

Since $A$ is symmetric, it can be orthogonally diagonalized: $A = QDQ^T$ where $D$ is diagonal. Since $A$ is positive semidefinite, $D$ has nonnegative diagonal elements: namely, the eigenvalues of $A$. We may remove any zero eigenvalues as follows. For each $i$ for which $[D]_{i,i} = 0$, delete the $i$'th row and column from $D$. Call the result $D_0$. Also for each such $i$, delete the $i$'th column from $Q$. Call the result $Q_0$.

Thus if there is a zero eigenvalue with algebraic multiplicity $k$ and the dimension of $A$ is $n\times n$, then $D_0$ is $(n-k)\times(n-k)$ and $Q_0$ is $n \times (n-k)$.

We have $A = Q_0 D_0 Q_0^T$ but now $D_0$ is invertible whereas $D$ was not.

Caveat for any subsequent calculations: $Q_0$ is now rectangular, and we still have $Q_0^T Q_0 = I$ but $Q_0 Q_0^T$ will not be $I$.

I haven't checked whether this delivers the goods.