# Diagonalization eigenvalues

by hpayandah
Tags: algebra, diagonalization, eigenvalue, linear
 P: 23 I get the same result as your teacher for $$\lambda = 2$$ $$A= \begin{bmatrix} 3 & 1 & 0\\ 0 & 2 & 1\\ 1 & 1 & 3 \end{bmatrix}$$ so for lamda = 2, $$(A-2I)\vec{v}=\vec{0}$$ $$\begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$$ I've personally always found it easier to not do row operations here and just jump straight in. From that you get $$v_3 = 0$$ and $$v_1 = -v_2$$ so therefore $$\vec{v} = \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}$$ Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable. Hope this helps.
 P: 23 You seem to have made a mistake in the step $$(A-2I)\vec{v}=\vec{0}$$ Why is your second row 1 0 1 instead of 0 0 1 ? Have you said $$v_2= \delta$$ ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e. $$\begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix}$$