I am lost in relation to a way to solve these eigenvalues and eigenvector questions. Please, someone could guide me on how can I take a solution? I will appreciate your help.
Let A be a real n x n matrix such that A^T = A. Show that if all n eigenvalues λj, j=1,..., n are distinct (λi different of λj for j different i), then the eigenvectors are mutually orthogonaç so that Vi^T.Vj =0 if i different of j.
Let A be a symmetric positive definite nxn matrix and define the Rayleigh quotient for any x different of 0 by
R(x)=(x^T.Ax)/(x^T.x)
Let λ be an eigenvalue of A with corresponding eigenvalue v. Show that
∇R(v)=0,
and that λ=R(v)
i.e. the eigenvectors are the stationary points of the Rayleigh quotient and the eigenvalues are the local maxima, minima or saddle points. Hint: note that
∇R(x) = 2(1/x^T x)Ax - 2(x^T Ax/x^T x)x
Let $AX=aX$, take transoose and use $A^T=A$ then $$A X_1=a_1 X_1 \implies X^T_1 A=a_1 X^T_1~~~~(1)$$ and $$A X_2=a_2 X_1 \implies X^T_2 A=a_2 X^T_2~~~~(2)$$ Multiply (1) by $X_2$ from right on both sides, we get $$X_1^T A X_2=a_1 X_1^T X_2~~~~(3)$$ Next, multiply (2) by $X_1$ and take transpose to gry $$X_2^T A X_1=a_2 X_2^T \implies X_1^T A X_2=a_2 X_1^T X_2~~~~(4)$$ Subtract (4) from (3) to get $$(a_1-a_2)X_1^TX_2=0$$ This implies that if $a_1 \ne a_2$, then $X_!^TX_2=0$, implyimg that eigenvectors $X_1, X_2$ are orthogonal.