linear algebra - Finding the diagonalizing matrix.

Find a nonsingular matrix $C$ such that $C^{-1}AC$ is a diagonal matrix.

$$A=\begin{pmatrix} 1 & 0 \\ 1 & 3 \\ \end{pmatrix}$$

I have found the eigenvalues to be 1 and 3. I also tried solving it by taking $C=\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$
and I got nowhere. What is the correct method for finding such $C$ in general?

Thank you in advance

Answer

To diagonalize a matrix, we find a matrix consisting of the eigenvectors of the matrix we wish to diagonalize.

To find the eigenvectors, we must find vectors in the kernel of $A-\lambda I$ where $\lambda$ is an eigenvalue.

In this case it means we need vectors in the kernels of

$$\begin{bmatrix} 0&0\\ 1&2 \end{bmatrix}\text{ and } \begin{bmatrix} -2 & 0\\ 1 & 0 \end{bmatrix}.$$

For instance, we may take the vectors $\langle 2,-1\rangle$ and $\langle 0,1\rangle$ which we see to be in each kernel. We then put the eigenvectors into a matrix $\left( \begin{array}{cc} 2 & 0 \\ -1 & 1 \\ \end{array} \right)$, which, along with its inverse will serve to diagonalize the matrix.

$$\left( \begin{array}{cc} 1/2 & 0 \\ 1/2 & 1 \\ \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ 1 & 3 \\ \end{array} \right)\left( \begin{array}{cc} 2 & 0 \\ -1 & 1 \\ \end{array} \right)=\left( \begin{array}{cc} 1 & 0 \\ 0 & 3 \\ \end{array} \right)$$

This all works because we are simply putting our original matrix into the basis consisting of its eigenvectors and we know exactly how the matrix behaves with respect to this basis.