MHB Problem of the Week #125 - August 18th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Let $p$ be a quadratic form. Show that if $A$ is a nonzero $n\times n$ symmetric matrix such that $\mathrm{tr}(A)=0$, then $p(\mathbf{x}) = \mathbf{x}^TA\mathbf{x}$ is indefinite.

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[Chris L T521]

This week's delays are brought to you by yours truly starting work again now that Fall semester is underway.

This week's problem was correctly answered by laura123. You can find her solution below.

[sp]Every symmetric matrix is diagonalizable i.e. it is similar to a diagonal matrix, so $A$ is similar to a diagonal matrix $D$ .
The trace of a matrix is similarity-invariant therefore $\mathrm{tr}(A)=\mathrm{tr}(D)$, but $\mathrm{tr}(A)=0$ therefore $\mathrm{tr}(D)=0$.
The eigenvalues of $A$ are the diagonal elements of $D$ therefore $\mathrm{tr}(D)$ is the sum of the eigenvalues of $A$, it follows that $\displaystyle \sum_{i}\lambda_i=0$ (where $\lambda_i$ are the eigenvalues of $A$).
$A$ is a nonzero matrix so there are $\lambda_i\neq 0$ and $\lambda_j\neq 0$, eigenvalues of $A$, such that $\lambda_i\cdot\lambda_j< 0$ therefore $A$ is indefinite. So $p(\mathbf{x}) = \mathbf{x}^TA\mathbf{x}$ is indefinite.[/sp]