Finite dimensional operator with zero trace is $0$

Question

I am working in the book "A Hilbert space Problem book", written by Halmos.

There is a claim on page 109 stating that:

"On a finite dimensional space, every hyponormal operator is normal".

In which, "hyponormal operator" implies some operator $A$ with $A^*A \ge AA^*$.

In order to prove it, one assume that $A$ is hyponormal, then $\operatorname{tr} \left( A^*A- AA^* \right) = 0$.

They also claimed that every positive operator on finite dimensional vector space with trace $0$ is $0$ operator.

Could you please prove the last claim? I have no idea to verify it.

Answer 1

Suppose $\{e_\alpha \}$ is an orthonormal basis of $H$, then $\newcommand{\tr}{\operatorname{tr}}$ $$ \tr(u)=\sum_\alpha\langle ue_\alpha, e_\alpha\rangle, u\in B(H).$$ If $u$ is positive, then $\langle ue_\alpha, e_\alpha\rangle=\|u^{\frac{1}{2}}e_\alpha\|^2\geq 0$, thus $\tr(u)=0$ implies $u^{\frac{1}{2}}e_\alpha=0\forall \alpha,$ hence $u^{\frac{1}{2}}=0$ and $u=0$.

Answer 2

Suppose $tr(A)=0$, this implies that $A$ is nilpotent since the eigenvalues are positive. If $A\neq 0$, there exists $x$ such that $A(x)\neq 0$ and $A^2(x)=0$, for any integer $n$, $<x-nA(x),A(x-nA(x))>=<x,A(x)>-n<A(x),A(x)>$, for $n$ enough big, $<x,A(x)><n<A(x),A(x)>$ contradiction.

Answer 3

The finite dimensional version of the spectral theorem says that for any positive operator there's an orthogonal basis for which its matrix is diagonal with nonnegative entries on the diagonal. So if the trace (the sum of those entries) is zero, then so is the matrix.

For a basis-free proof, consider that if $V$ is an inner product space then $V^*\otimes V$ acquires an inner product by $<a\otimes b,c\otimes d>=<a,c><b,d>$. When $V$ is finite dimensional we have $\mathrm{Hom}(V,V)=V^*\otimes V$, and so we get an inner product on $\mathrm{Hom}(V,V)$. This is the Frobenius inner product, $A\bullet B = \mathrm{tr}(A^*B)$. Since this is an inner product we know that if $A\bullet A = 0$ then $A=0$.

Answer 4

Perhaps this will be considered killing a fly with a sledge-hammer, but here is another way of thinking about it. Letting $A =[a_{ij}] \in M_n$ denote the matrix representation of the operator, since $A$ is positive semi-definite (PSD), it is also a Gram matrix, which means there exists an inner product space $(V,\langle \cdot \rangle)$ and vectors $v_1,...,v_n \in V$ such that $a_{ij} = \langle v_i,v_j\rangle$. Thus, if $tr(A) =0$, $||v_i||=0$ and therefore $v_i=0$ for each $i$, whereby we conclude $a_{ij} = 0$ for every $i,j$.

That all PSD matrices are Gram, and vice-versa, is a useful little factoid to keep in mind (at least I think so).