## Section5.6Degeneracy

It is not always the case that an $$n\times n$$ matrix has $$n$$ distinct eigenvectors. For example, consider the matrix

$$B = \begin{pmatrix} 3 \amp 0 \amp 0\\ 0 \amp 3 \amp 0\\ 0 \amp 0 \amp 5 \end{pmatrix}\text{,}\tag{5.6.1}$$

whose eigenvectors are again clearly the standard basis. But what are the eigenvalues of $$B\text{?}$$ Again, the answer is obvious: $$3$$ and $$5\text{.}$$ In such cases, the eigenvalue $$3$$ is a degenerate eigenvalue of $$B\text{,}$$ since there are two independent eigenvectors of $$B$$ with eigenvalue $$3\text{.}$$ Degenerate eigenvalues are also referred to as repeated eigenvalues. In this case, one also says that $$3$$ is a repeated eigenvalue of multiplicity $$2$$.

However, that's not the whole story. Setting

$$v = \begin{pmatrix}1\\0\\0 \end{pmatrix}, \qquad w = \begin{pmatrix}0\\1\\0 \end{pmatrix}\text{,}\tag{5.6.2}$$

we of course have

$$Bv = 3v, \qquad Bw = 3w\text{.}\tag{5.6.3}$$

But, by the linearity of matrix operations, we also have

$$B(av+bw) = a\,Bv+b\,Bw = 3(av+bw)\text{,}\tag{5.6.4}$$

so that any linear combination of $$v$$ and $$w$$ is also an eigenvector of $$B$$ with eigenvalue $$3\text{.}$$ The eigenspace of $$B$$ corresponding to eigenvalue $$3$$ is therefore a 2-dimensional vector space (a plane), rather than the 1-dimensional eigenspaces (lines) that occur when the eigenvalues are distinct.