## Section5.6Degeneracy

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

\begin{equation} 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} \end{equation}

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

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

we of course have

\begin{equation} Bv = 3v, \qquad Bw = 3w\text{.}\tag{5.6.3} \end{equation}

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

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

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.