# Eigenvalue of self-adjoint

adjoint-operatorseigenvalues-eigenvectorsinner-productslinear algebra

I am solving an exercise:

Let $$T: V \rightarrow W$$ be a linear transformation.
$$V$$ and $$W$$ are finite-dimensional inner product spaces. Prove $$T^*T$$ and $$TT^*$$ are semidefinite.

This is a solution that I don't understand:
$$T^*T$$ and $$TT^*$$ are self-adjoint, then we have $$T^*T(x) = \lambda x$$. Hence:
$$\lambda = ⟨T^*T(x),x⟩ = ⟨T(x),T(x)⟩ ≥ 0.$$
$$\lambda$$ is $$≥ 0$$, hence $$T^*T$$ is semidefinite.

I don't understand why the eigenvalue is equal to $$⟨T^*T(x),x⟩$$.
Thank you for any kind of help!

#### Best Answer

The proposed solution is not well written, but it appears that $$\lambda$$ is an arbitrary eigenvalue of $$T^*T$$ and $$x$$ is a normalized eigenvector associated with $$\lambda$$. In that case, $$\lambda = \lambda \langle x,x\rangle = \langle \lambda x,x\rangle = \langle T^*T(x),x\rangle$$ The point of the argument was to demonstrate that all eigenvalues of $$T^*T$$ are nonnegative.

I would, however, suggest a more direct approach. Note $$T^*T$$ is self-adjoint, and given $$x\in V$$, $$\langle T^*Tx,x\rangle = \langle Tx,Tx\rangle \ge 0$$. Hence, $$T^*T$$ is positive semidefinite. By a similar argument $$TT^*$$ is positive semidefinite.