Proof of a self-adjoint linear transformation

Question

Let V be a subset of ${R^3}$ and be a two-dimensional subspace. Let $W: V\rightarrow V$ be a self-adjoint linear transformation. Define K=det(W) and H=$(1/2)$trace(W).

Prove that $K \leq H^2$

What I have so far: Since W is self-adjoint, W is represented by a matrix:

$\begin{bmatrix} \lambda_1 &0 &0 \\ 0&\lambda_2 &0 \\ 0& 0 & \lambda_3 \end{bmatrix}$

With eigenvalues along the diagonal.

$K=det(W)=\lambda_1\lambda_2\lambda_3$ $\ $$H^2=1/4(\lambda_1+\lambda_2+\lambda_3)^2$

Any direction on how to go about solving this proof is appreciated

Answer 1

Hints:

$1).\ $The determinant, trace, eigenvectors and eigenvalues of a square matrix are invariant under change of basis.

$2).\ $ Every self-adjoint linear transformation on a finite- dimensional inner product space has an orthonormal basis of eigenvectors.

$3).\ $ the eigenvalues of a self-adjoint operator are real.

$4).\ $ for any two real numbers $a,b,\ ab\le \frac{1}{2}(a+b)^2$