How to determine if a 3x3 matrix is diagonalizable?

Question

The matrix is given as:

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

So the matrix has eigenvalues of $0$ ,$0$,and $3$.

The matrix has a free variable for $x_1$ so there are only $2$ linear independent eigenvectors. So this matrix is not diagonalizable.

What conditions would be necessary for $A$ to be diagonalizable?

Is it simply all $3$ eigenvectors must be linearly independent? Or perhaps the opposite?

Answer 1

As you remarked correctly, the eigenvalues, with multiplicity, are $0,0,3$.

A matrix is diagonalizable if and only of for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.

For the eigenvalue $3$ this is trivially true as its multiplicity is only one and you can certainly find one nonzero eigenvector associated to it.

For the eigenvector $0$ however you would need to find $2$ linearly indepedent eigenvectors Yet as you said, indirectly, the eigenspace associated to $0$ is the space generated by $(1,0,0)$. Its dimension is thus one and you cannot find two independent eigenvectors for the eigenvalue $0$.

The matrix is thus not diagonalizable.

Necessary conditions for diagonalizable include:

• There exists a basis of eigenvectors, which in you case would mean there exist $3$ linearly independent eigenvectors. (This is almost what you said, but note that I said "there exists" and not "the three eigenvectors" since there are infinitely many eigenvectors.)

• For each eigenvalue the dimension of the respective eigenspace is equal to the multiplicity of the eigenvalue. (As mentioned at the start.)

Answer 2

A matrix $A$ is diagonalisable with distinct eigenvalues $\lambda_1,\ldots,\lambda_k$ (or a subset thereof) if and only if the product $(A-\lambda_1I)\ldots(A-\lambda_kI)$ is zero. In your example clearly $0,3$ are the only eigenvalues, but $$ A(A-3I)=\begin{pmatrix}0&-3&4\\0&0&0\\0&0&0\end{pmatrix} $$ is nonzero, so $A$ is not diagonalisable.