# Proof of a self-adjoint linear transformation

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

• Hint: Think in terms of the eigenvalues. Commented Nov 16, 2019 at 23:56
• $V$ is two dimensional so how can your 3 vectors be a basis? Commented Nov 17, 2019 at 0:50
• Your matrix is $3x3$ so how can it represent $W$ wrt to your (now) basis of $2$ vectors? I think you may be confusing the fact that the vectors are three tuples but when you consider the matrix of $W$ it is with respect to the $coefficients$ of your arbitrary vector in $V$ written as a linear combination of your basis vectors (in $V$). Commented Nov 17, 2019 at 0:55

$$1).\$$The determinant, trace, eigenvectors and eigenvalues of a square matrix are invariant under change of basis.
$$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 $$a,b,\ ab\le \frac{1}{2}(a+b)^2$$