Let $A \in \mathbb{R}^{N \times N}$ be symmetric.
a) The respective Eigenvalue $\lambda$ to an approximately defined Eigenvector $0 \neq x \mathbb \in {R}^n$ from $A$ has to be calculated, so that $||Ax-\lambda x||^2_2$ is minimal. Specify a formula to calculate $\lambda$.
This is an introductory class to numerical methods. Can I use here the same formula that I have always used? Considering $Ax=\lambda x$, solving the linear system $(A-\lambda I)x=0$ should give us the (approximated) Eigenvalue $\lambda$, right? Why wouldn't $||Ax-\lambda x||^2_2$ be $0$?
$A$ has furthermore the Eigenvalues $\lambda_1 \leq \lambda_2 \leq ...\leq \lambda_n$. Show that:
b) $\lambda_1 x^T x \leq x^T Ax \leq \lambda_n x^T x$
I think I'm missing something. Can I simply use $Ax=\lambda x$? Thus $x^T Ax$ is the same as $\lambda x^T x$, as $\lambda$ can be any arbitrary Eigenvalue between the smallest ($\lambda_1$) and the largest ($\lambda_n$), it follows that $\lambda_1 x^T x \leq \lambda x^T x \leq \lambda_n x^T x$ is valid.
c) $\lambda_1 = \min_{x \neq 0} \frac{x^T Ax}{x^T x}= \min_{x^T x = 1} x^T Ax$
I'm a little lost here. I would attempt to use the same trick again, thus $\lambda_1 = \min_{x^T x = 1} x^T \lambda x = \min_{x^T x = 1} \lambda x^T x = \min_{x^T x = 1} \lambda$. However I don't think is proving what they want?
d) $\lambda_n = \max_{x \neq 0} \frac{x^T Ax}{x^T x}= \max_{x^T x = 1} x^T Ax$
Analogue to c)
Many thanks in advance!
Edit: Thanks to Martin I know understand better what they want.
For b) and following the tip that I can diagonalize the Matrix, I now have
Let $D=Q^T AQ$. The eigenvalues of A are in the diagonal from $D$. With $y=Q^T x$ it follows that:
$x^T Ax=x^T Q D Q^T x = y^T D y = \sum \lambda_i y_i^2$
and therefore
$\lambda_1 x^T x=\lambda_1 y^T y \leq \sum \lambda_i y_i^2 \leq \lambda_n y^T y = \lambda_n x^T x$
I think I'm very near to being able to answer c) and d) but I'm still missing something. Any help? Thanks!