Finding the diagonalizing matrix.

Question

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 1

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.

Answer 2

The method you tried is (the) a correct one. You find the eigenvalues and you put them in a diagonal matrix $D$. You write $C$ as you did, you find $C^{-1}$ via Cramer's rule (for example) and you solve the system $CAC^{-1} = D$.

You may have a calculation wrong. Here you have a link to the solution of your problem. I'll try to give a more developed answer in a moment.

In more detail, you have:

$C = \begin{pmatrix} a&b\\ c&d \end{pmatrix}$, thus $C^{-1}= \frac{1}{ad-cb} \begin{pmatrix} d&-b\\ -c&a\end{pmatrix}$ (in virtue of this).

As the eigenvalues are $1$ and $3$, we may write:

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

Thus, solving the system (recall that this would not be possible if the matrix didn't diagonalize):

$\begin{pmatrix} a&b\\ c&d \end{pmatrix} \begin{pmatrix} 1&0\\ 1&3 \end{pmatrix}\begin{pmatrix} d&-b\\ -c&a\end{pmatrix} = (ad-cb) \begin{pmatrix} 1&0\\ 0&3 \end{pmatrix}$

which is a system (I think it's even linear) of four equations and four variables and should be solvable. You shall find $a=-2$, $b=0$, $c=d=1$, that is:

$C = \begin{pmatrix} -2&0\\ 1&1 \end{pmatrix}$.

Answer 3

Check that $\;\binom{-2}1\;,\;\;\binom01\;$ are eigenvectors belonging to the eigenvalues $\;1,3\;$ resp., and form the matrix

$$C:=\begin{pmatrix}\!\!\!-2&0\\1&1\end{pmatrix}$$

Check now that $\;C^{-1}AC\;$ is diagonal

Answer 4

Now that you have the eigenvalues, find the eigenvectors.

$A\xi_1=1\xi_1 \implies \xi_1 = (-2,1)^T$

$A\xi_3=3\xi_3 \implies \xi_3 = (0,1)^T$

Since you have distinct eigenvalues, these eigenvectors will be linearly independent. Consider the matrix $C$ where the columns are the eigenvectors. $C$ is non-singular since the columns are linearly independent. Then $C^{-1} A C$ is diagonal.