I was reading a proof that uses the following matrix relation;
$$
A \leq \text{Tr}(A)I
$$
Where $\text{Tr}(\cdot)$ denotes the trace operator and $I$ is the identity matrix of appropriate dimension.
In the proof, $A = A^{\text{T}} \geq 0$ is a covariance matrix. However, the proof does not discuss restrictions on $A$. In particular, it is not indicated whether this relation applies to any $A$, or if only to a real, symmetric, positive semi-definite $A$.
Is this relation valid? How do we show it? Are there any restrictions on $A$?
Best Answer
It is a valid inequality whenever $A$ is positive semi-definite.
The standard partial order on $M_n(\mathbb{C})$ is valid for self-adjoint matrices. For self-ajoint matrices $A$ and $B$ we write $$B\leq A\text{ if $A-B$ is positive semi-definite}$$ (i.e $A-B\geq 0$).
So to show that $A\leq tr(A)I$, for $A$ positive, we only need to show that $tr(A)I-A$ is positive semidefinite.
To prove this, let $U$ be the unitary that diagonalizes $A$, and let $D$ be the resulting diagonal matrix. In this basis, we need only to show that $tr(A)I-D$ is positive semi-definite. Note it is sufficient to see that the eigenvalues are all non-negative.
Since $tr(A)=tr(U^*DU)=\sum_{i=1}^n d_{ii}$, and $d_{ii}$ for $1\leq i \leq n$ are the non-negative eigenvalues of $A$ (since $A$ is positive), we see that each entry $(tr(A)I-D)_{jj}$ for $1\leq j\leq n$ is non-negative. This is because $$\sum_{i=1}^n d_{ii}-d_{jj}=\sum_{i\neq j}d_{ii}\geq 0$$ for all $1\leq j\leq n$, and we are done.
Note that this could fail if ever $\sum_{i\neq j}d_{ii}<0$ for some $j$, so it won't hold in for all $A\in M_n(\mathbb{C})$$.