Finding a non-singular matrix $C$ such that $C^{-1}AC$ is diagonal

Question

$A = \left(\begin{array}{cc}1 & 0 \\1 & 3\end{array}\right)$.

I find the eigenvalues = 1,3.

The eigenvector corresponding to 1 = $t(1,-2)^t$. The eigenvector corresponding to 3 = $r(1,0)$.

Then I form $C=\left(\begin{array}{cc}1 & 1 \\-2 & 0\end{array}\right)$. But $C^{-1}AC$ is not a diagonal matrix. Rather it is:$\left(\begin{array}{cc}2.5 & -.5 \\-1.5 & 1.5\end{array}\right)$.

What am I doing wrong?

Answer 1

Your eigenvectors are wrong. They should be $(-2,1)$ and $(0,1)$. The $C$ formed by this works.

http://www.wolframalpha.com/input/?i=%28%7B%7B-2%2C0%7D%2C%7B1%2C1%7D%7D%5E%28-1%29%29%7B%7B1%2C0%7D%2C%7B1%2C3%7D%7D%7B%7B-2%2C0%7D%2C%7B1%2C1%7D%7D

Answer 2

Your eigenvectors $\lambda_1=1$ and $\lambda_2=3$ are correct. To get the eigenvecors, we need to solve the equation

$$\left(\begin{array}{cc}1 & 0 \\1 & 3\end{array}\right) \left(\begin{array}{cc}a \\b \end{array}\right) = \lambda \left(\begin{array}{cc}a \\b \end{array}\right) $$

For $\lambda=1$, we get the following equations:

$$a = a$$ $$a + 3b = b$$

By the first equation, $a$ can be whatever we like, so let's set $a=1$. From the second equation we get $b=-\frac{1}{2}$. So our first eigenvector is $v_1 = \left(\begin{array}{cc}1 \\-\frac{1}{2} \end{array}\right) = \left(\begin{array}{cc}-2 \\1 \end{array}\right)$.

Try a similar procedure for the second eigenvector.