Finding eigenvectors to eigenvalues, and diagonalization

I just finished solving a problem on finding eigenvectors corresponding to eigenvalues, however, I'm not sure if it is correct. I was wondering if someone could check my work:

For the matrix $W = \begin{bmatrix} 1 & 2 \\ 3 & 2\\ \end{bmatrix}$, I must find the eigenvectors corresponding to the eigenvalues, as well as a diagonal matrix similar to W.

I was able to find that the eigenvalues were equal to $\lambda = 4, -1$. Then, I used the equation $(A - \lambda I)v = 0$ to solve for the vector.

When $\lambda = 4$, I set up the equation $\begin{bmatrix} 1 & 2 \\ 3 & 2\\ \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4\\ \end{bmatrix}$ = $\begin{bmatrix} -3 & 2 \\ 3 & -2\\ \end{bmatrix}$, which gave me the eigenvector $\begin{bmatrix} 2\\ 3\\ \end{bmatrix}$.

For $\lambda = -1$, I did the exact same procedure and received the eigenvector which gave me the eigenvector $\begin{bmatrix} 1\\ -1\\ \end{bmatrix}$.

Did I do this part correctly? How do I find a diagonal matrix similar to $W$?

• Change the rest of the A's to W's as well! – NickD Apr 4 '17 at 2:37
• You have two distinct eigenvalues for a $2\times2$ matrix, so you can write down the similar diagonal matrix without further ado: it’s just a matrix with the eigenvalues along its diagonal. – amd Apr 4 '17 at 3:00
• thank you so much for your help. could please help me a last question I have here? math.stackexchange.com/questions/2217044/… – user400359 Apr 4 '17 at 3:26

we can use Row operations to obtain a diagonal matrix similar to W

W = \begin{bmatrix} 1 & 2 \\ 3 & 2\\ \end{bmatrix} $r_1-r_2=R_1$ gives $$W = \begin{bmatrix} -2 & 0 \\ 3 & 2\\ \end{bmatrix}$$ then $R_2=2r_2$ gives W = \begin{bmatrix} -2 & 0 \\ 6 & 4\\ \end{bmatrix} now $R_2=r_2+3r_1$ gives $W = \begin{bmatrix} -2 & 0 \\ 0 & 4\\ \end{bmatrix}$ and $R_1=\frac{1}{2}r_1$ gives $W = \begin{bmatrix} -1 & 0 \\ 0 & 4\\ \end{bmatrix}$ which is in diagonal form, as required, as you can see the diagonal entries are the eigenvalues you calculated

• How do you know that the matrix you have arrived upon is similar to W? – Doug M Apr 4 '17 at 3:34

I think it is worth the exercise to verify that

$W\mathbf v = \lambda \mathbf v$

$W \begin {bmatrix} 2\\3 \end{bmatrix} = 4\begin {bmatrix} 2\\3 \end{bmatrix}$ and $W \begin {bmatrix} 1\\-1 \end{bmatrix} = -\begin {bmatrix} 1\\-1 \end{bmatrix}$

which it does...in both cases.

In which case:

$W\begin{bmatrix} \mathbf v_1&\mathbf v_2 \end{bmatrix} = \begin{bmatrix} \mathbf v_1&\mathbf v_2 \end{bmatrix}\begin{bmatrix} \lambda_1\\&\lambda_2\end{bmatrix}$

Let $P = \begin{bmatrix} \mathbf v_1&\mathbf v_2 \end{bmatrix}$ and $\Lambda = \begin{bmatrix} \lambda_1\\&\lambda_2\end{bmatrix}$

$WP = P\Lambda\\ P^{-1}WP = \Lambda$

$\lambda$ is a diagonal matrix similar to W

• Okay, so the part I have done is correct. How do I find a diagonal matrix similar to W ? – user400359 Apr 4 '17 at 2:35
• @stackofhay42 subtract the second row away from the first then take 3 lots of the first row away from the second? – user395952 Apr 4 '17 at 2:38
• I have outlined the theory and the process. $\begin{bmatrix}0.2&0.2\\0.6&-0.4\end{bmatrix}W \begin{bmatrix}2&1\\3&-1\end{bmatrix}= \begin{bmatrix}4\\&-1\end{bmatrix}$ – Doug M Apr 4 '17 at 2:39