Consider the form $ \frac{x^T B x}{x^T x} $ I am asked to find a constant $C$ such that $\frac{x^T B x}{x^T x} \le C \lambda (B)$ where $\lambda(B)$ is the largest eigenvalue of $B$, or prove no such constant exists. I am also given the following assumptions about $B$, but I think 1 and 2 are not necessary
- $B$ is a 2x2 matrix
- $B$ is a real matrix
- $B$ has real eigenvalues
- $x\in \mathbb{R}^2$
We can decompose $B$ into symmetric $S$ and antisymmetric parts giving $\frac{x^T B x}{x^T x}=\frac{x^T S x}{x^Tx}$ and now diagonalise $S$, the following result then comes easily $$\frac{x^T S x}{x^Tx}\le \lambda(S)$$
Then all I need is $\lambda(S)\le\lambda(B)$ but I have no idea if this is true or not. It is mentioned in the comments of the downvoted answer on this post that it is true, but when the author was asked for a proof, he disappeared. I would like to know
- Am I overcomplicating this problem and theres just another way to do it?
Or 2. Is this inequality true and I've done it right?
