Show that this matrix is not diagonalizable

Say I have a matrix:

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

1. Is this matrix diagonalizable?

2. Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?

Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:

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

$$eigenvector = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ t \end{bmatrix} = t * \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns: • You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity). – hardmath Sep 28 '18 at 16:06

1. No, it's not diagonalizable. If the two eigenvalues of a $$2 \times 2$$ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).

2. The eigenvalues of an $$n \times n$$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$$n$$ polynomial. Since every degree-$$n$$ polynomial has $$n$$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $$n \times n$$ matrix has $$n$$ eigenvalues (when counted with algebraic multiplicities).

By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.

• "when counted with multiplicities" algebraic multiplicities. – Acccumulation Sep 28 '18 at 18:33
• Thanks; I added that. – John Hughes Sep 28 '18 at 21:11

HINT

Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $$2$$ with arithmetic multiplicity $$2$$ and geometric multiplicity $$1$$, that is an eigenspace with dimension $$1$$.

1. No. $$A - 2I$$ has only one linearly independent column. (The second column of $$A-2I$$ is zero.)
2. To give a quick example, let's consider the 2D rotation matrix. $$\begin{pmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{pmatrix}$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.

The geometric multiplicity of an eigenvalue (that is, the number of independent eigenvectors corresponding to that eigenvalue), is the dimension of the null space of the matrix minus the identity matrix times that eigenvalue (that is, you take the matrix and subtract the eigenvalue from each main diagonal entry). The dimension of the null space of a $$n \times n$$ matrix is $$n$$ minus the rank of the matrix. So if $$n=2$$, then to have a geometric multiplicity of 2, you have to have the rank be zero. But a rank zero matrix is simply he zero matrix. So once you subtract 2 from the main diagonal and see that you don't get the zero matrix, you know that there's only one linearly independent eigenvector for that eigenvalue.