Suppose we have a matrix V given by
The columns of V are orthogonal to each other. We have proved that $V^TV = I$. We also have proved that the columns of V form a basis for RN, and that $<v_i, b>$ = $a$ 1 if $b = a1v1+ ... +anVn$.
Now we are given Q, and its eigenvalues such that
We've proved that the eigenvectors of Q are given by the columns of V.
Now we have to show that if $Q_2 = Q - λ_1 v_1 v_1^T$, $v_1$ is in Nul($Q_2$) and $v_2 ... v_n$ are eigenvectors
My approach so far was substituting $Q_2=Q - λ_1 v_1 v_1^T$ in $Q_2 x =_? 0$ But when I distribute $v_1$ through inner products, I'm not sure if I can say that $<v_1, v_i v_i^T>$ are 0. I mean, I know that $<v_1, v_i>$ because they're orthonormal, but can I make the other claim? Do I have the right approach to this whole proof?