Given that $\|x\|_A=\sqrt{x^tAx}$ . Suppose $A $ is symmetric positive definite matrix and $\lambda_i$ are the eigenvalues such that $0\le \lambda_1\le ...\le \lambda_n$ then prove that
$$\sqrt{\lambda_1} \|x\|_2 \le \|x\|_A \le \sqrt{\lambda_n}\|x\|_2$$
I consider for $2\times 2$ matrix . Then $\|x\|_A^2 = x^tAx = a_{11}x_1^2 +a_{22}x_2^2 +2 a_{12}x_1x_2 $ , then we have to prove that $\lambda_1(x_1^2 +x_2^2) \le a_{11}x_1^2 +a_{22}x_2^2 +2 a_{12}x_1x_2 \le\lambda_n (x_1^2+x_2^2) $. But how prove these two inequalities ???
Thankyou for helping :)
Since $A$ is symmetric, it can be diagonalized, i.e. it can always written in the form
$$ A = R D R^T $$
where $D$ is the diagonal matrix of eigenvalues, and $R$ is an orthogonal matrix whose columns are the corresponding normalized eigenvectors.
Now
$$x^T A x = x^T R D R^T x $$
If you define vector $ y = R^T x $ , then $y^T y = x^T R R^T x = x^T x $
We can now write
$$ x^T A x = y^T D y = \displaystyle \sum_i \lambda_i y_i^2 $$
But,
$$ \displaystyle \lambda_{Min} \sum_i y_i^2 \le \sum_i \lambda_i y_i^2 \le \lambda_{Max} \sum_i y_i^2 $$
And as stated above, $\sum_i y_i^2 = \sum_i x_i^2 $
Therefore,
$$ \lambda_{Min} \sum_i x_i^2 \le x^T A x \le \lambda_{Max} \sum_i x_i^2 $$