Diagonalizing a 3x3 matrix

Question

Prove that matrix $A$ is diagonalizable, find the bases for the eigenspaces, the diagonalizing matrix $P$, and compute $P^{-1} A P$ where $A= \left(\begin{array}{ccc} 2 & 0 & 3 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right).$

Answer 1

A matrix is diagonalizable if it is similar to a diagonal matrix. So in order to prove this matrix is diagonalizable, why don't we make it similar to a diagonal matrix.

As it is upper triangular, we can just read the eigenvalues off of the diagonal (to convince yourself of this, try finding $det(A-\lambda I)$ for any diagonal A or remember that finding the determinant of a diagonal matrix is just summing the diagonal entries): $\lambda=2,3$.

Then finding the eigenvectors, we look at the null space of $A-I\lambda$ for the two eigenvalues.

For $\lambda=2$, we find the nullspace of $\begin{bmatrix}0&0&3\\0&1&0\\0&0&1 \end{bmatrix}$, which eyeballing we can see is the vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$

or $\lambda=3$, we find the nullspace of $\begin{bmatrix}-1&0&3\\0&0&0\\0&0&0 \end{bmatrix}$ which we can eyeball one easily as $\begin{bmatrix}0\\1\\0\end{bmatrix}$. Then we need one more for this matrix to be diagonalizable, and fortunately this one is pretty clear too we need the first input in row 1 to sum with the third input to 0, $\begin{bmatrix}3\\0\\1\end{bmatrix}$ fits the bill.

Now we have put the eigenvectors together to get our $P=\begin{bmatrix}1\\0\\0\end{bmatrix}\begin{bmatrix}0\\1\\0\end{bmatrix}\begin{bmatrix}3\\0\\1\end{bmatrix}$ which we can invert and plug into the formula $PDP^{-1}=A$, with D the matrix with eigenvalues on the diagonal,

$D=\begin{bmatrix}2&0&0\\0&3&0\\0&0&3 \end{bmatrix}$ I leave it to you to compute $P^{-1}$ and multiply by the appropriate values to go from $PDP^{-1}=A$ to $P^{-1}AP=D$

Answer 2

The characteristic polynomial is $ \text{det}(A−λI)=(2−λ)(3−λ)^2$ so the eigenvalues of your matrix are $2$ and $ 3$. Therefore $2$ is an eigenvalue with algebraic multiplicity $1,$ and $3$ is an eigenvalue with algebraic multiplicity $2$. Recall the geometric multiplicity can also be described as the dimension of the nullspace of $A-λI$.

So for $λ = 2$ we have $A-λI = \left(\begin{array}{ccc} 0 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)$ so the null space has dimension $1.$ Thus the geometric multiplicity for $λ = 2$ is $1$.

Now for $λ = 3$ we have $A-λI = \left(\begin{array}{ccc} -1 & 0 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$ so the null space has dimension $2.$ Thus the geometric multiplicity for $λ = 3$ is $2$.

Because the algebraic multiplicities equal the geometric multiplicities for all the eigenvalues, by definition the matrix is non-defective and hence diagonalizable.

For a good explanation of diagonalizability see Chapter 24 of "Numerical Linear Algebra" by Trefethen and Bau.