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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

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    $\begingroup$ Hint: Think in terms of the eigenvalues. $\endgroup$
    – Mohit
    Commented Nov 16, 2019 at 23:56
  • $\begingroup$ $V$ is two dimensional so how can your 3 vectors be a basis? $\endgroup$ Commented Nov 17, 2019 at 0:50
  • $\begingroup$ 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 $V$). $\endgroup$ Commented Nov 17, 2019 at 0:55

1 Answer 1

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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 $a,b,\ ab\le \frac{1}{2}(a+b)^2$

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  • $\begingroup$ Is my calculation for H^2 correct where I get 1/4*(traceW)^2. The 1/4 doesn't fit $\endgroup$ Commented Nov 17, 2019 at 1:25
  • $\begingroup$ I can't really follow your proof. Have you tried using the hints in my answer? $\endgroup$ Commented Nov 18, 2019 at 14:39

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