If $A,B$ symmetric positive semidefinite, show tr$(AB) \geq 0$ Supposing $V$ is a finite dimensional vector space (over $\mathbb{R}$) of dimension $n$, and $A,B$ are symmetric positive definite linear mappings from $V$ to $V$, how can I show that in any orthonormal basis $\mathrm{tr}(AB) \geq 0$?
I noticed that since they are symmetric we have that 
$$\mathrm{tr}(AB) = \sum_{i=1}^n\sum_{j=1}^nA_{ij}B_{ji} = \sum_{i=1}^n\sum_{j=1}^nA_{ij}B_{ij}$$ which is the sum of the elements of the element-wise product of $A,B$. I don't know if this is helpful.
 A: Here's another derivation (7 years later):
Let $A,B\succeq0$. Then the eigendecomposition of symmetric $B$ gives $B=\sum_{i=1}^n \lambda_i v_i v_i^T$. Therefore,
$$\begin{align}
\operatorname{Tr}[AB]&=\operatorname{Tr}[A\sum_{i=1}^n \lambda_i v_i v_i^T]\\
&=\sum_{i=1}^n \lambda_i \operatorname{Tr}[Av_i v_i^T]\\
&=\sum_{i=1}^n \underbrace{\lambda_i}_{\geq0} \underbrace{v_i^TAv_i}_{\geq0} \\
&\geq 0
\end{align}$$
where the last equality is from the cyclic property of the trace.
Feel free to ask for any clarifications neeeded.
Edit: Here's the explanation of the eigendecomposition. 
In matrix form, the eigen-equation is: $BV=V\Lambda$, where $V$ is a matrix whose columns are the eigenvectors $\{v_i\}$ of $B$, and where $\Lambda=\operatorname{diag}(\lambda_1,...,\lambda_n)$ is a diagonal matrix with the eigenvalues of $B$ along the diagonal. Because $B$ is symmetric, these $V$ matrices are orthogonal, meaning their columns are orthonormal, so $VV^T=V^TV=I_n$. We can then re-write the matrix equation: $BVV^T=B=V\Lambda V^T$.
$$\begin{align}
\rightarrow B &=
\begin{bmatrix}
v_1 & \cdots & v_n
\end{bmatrix}
\begin{bmatrix}
\lambda_1 & \cdots & 0\\
\vdots & \ddots & \vdots\\
0 & \cdots & \lambda_n
\end{bmatrix}
\begin{bmatrix}
v_1^T\\
\vdots\\
v_n^T
\end{bmatrix}\\
&=
\begin{bmatrix}
v_1 & \cdots & v_n
\end{bmatrix}
\begin{bmatrix}
\lambda_1v_1^T\\
\vdots\\
\lambda_nv_n^T
\end{bmatrix}\\
&=
\sum_{i=1}^n \lambda_i v_i v_i^T
\end{align}$$
where this is a sum of outer products of the eigenvectors. Note that I wrote my matrices above as vector with elements that are vectors. This shorthand is valid and very convenient, but feel free to write it out and check me!
A: As others have remarked, you might as well suppose that $A$ and $B$ are positive semidefinite matrices. We may write $A = X^{t}X$ and $B = Y^{t}Y$ where $X$ and $Y$ are $n \times n$ real matrices. Then ${\rm tr}(AB) = {\rm tr}(X^{t}X Y^{t}Y)$ = ${\rm tr}((YX^{t})(XY^{t})).$ The latter matrix has the form ${\rm tr}(UU^{t})$ for a real $n \times n$ matrix $U$, and such a trace is always non-negative.
A: Since this may be homework, I will only give hints. 


*

*Without loss of generality you may assume that $V=R^n$. 

*Trace is independent of the basis you use. Thus it suffices to 
show this in the basis where $A$ is diagonal. 

*A positive semi-definite matrix has nonnegative diagonal. Why?

*Putting 1-3 together, one needs to show that the $tr(AB)\geq 0$ where $A$ is a nonnegative diagonal matrix and $B$ has nonnegative diagonal. 
