yourMATHsolver: How to diagonalize a matrix

Consider a matrix

$A=\begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{bmatrix}.$

This matrix has eigenvalues

$\lambda_1 = 3, \quad \lambda_2 = 2, \quad \lambda_3= 1.$

So A is a 3-by-3 matrix with 3 different eigenvalues, therefore it is diagonalizable. Note that if there are exactly n distinct eigenvalues in an n×n matrix then this matrix is diagonalizable.

These eigenvalues are the values that will appear in the diagonalized form of matrix $A$ , so by finding the eigenvalues of $A$ we have diagonalized it. We could stop here, but it is a good check to use the eigenvectors to diagonalize $A$ .

The eigenvectors of A are

$v_1 = \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad v_3 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix}.$

One can easily check that $A v_k = \lambda_k v_k.$

Now, let P be the matrix with these eigenvectors as its columns:

$P= \begin{bmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{bmatrix}.$

Note there is no preferred order of the eigenvectors in P; changing the order of the eigenvectors in P just changes the order of the eigenvalues in the diagonalized form of A.

Then P diagonalizes A, as a simple computation confirms:

$P^{-1}AP = \begin{bmatrix} 0 & -1 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 0 & 0 \\ 2 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{bmatrix}.$