Linear Algebra – How to Show That a Matrix is Not Diagonalizable

diagonalizationeigenvalues-eigenvectorslinear algebra

Say I have a matrix:

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

  1. Is this matrix diagonalizable?

  2. Does a 2×2 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:

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Best Answer

  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.

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