Verify that if P is an orthogonal matrix and x = Py then $y^Ty = x^Tx$. Let A be a real symmetric n × n matrix. Then we know that there exists a real orthogonal matrix P such that $P^TAP$ is diagonal. By using the transformation x = Py, or otherwise, prove that for every x ∈$R_n$,
$$mx^Tx ≤ x^TAx ≤ Mx^Tx$$ where m and M are the smallest and greatest eigenvalues of A respectively. For which x is it true that $x^TAx = Mx^Tx$?
$x = Py$ If $P$ is orthogonal
$x^Tx = $$(Py)^T(Py)\\ y^TP^TPy\\ y^Ty$
If $A$ is a symmetric matrix then there exists an orthogonal $P$ such that: $A = P^TDP$
$x^TAx = (Px)^TD(Px)\\ y = Px\\ x^TAx = y^TDy$
$y^TDy = \sum \lambda_iy_i^2$
Since each $y_i^2 >0$
$\lambda_{min}\sum y_i^2\le \sum \lambda_iy_i^2\le \lambda_{max}\sum y_i^2$
$\lambda_{min}y^Ty \le y^TDy \le \lambda_{max}y^Ty\\ \lambda_{min}y^Ty \le x^TAx \le \lambda_{max}y^Ty$