TITLE: Telling when an eigenvalue has a multiplicity of one or more
QUESTION [0 upvotes]: Let $A=\begin{bmatrix}1&2&3\\1&2&3\\1&2&3\end{bmatrix}$  Find all eigenvalues of A. 
Solution: $A=\begin{vmatrix}1-\lambda&2&3\\1&2-\lambda&3\\1&2&3-\lambda\end{vmatrix}$
$A=\begin{vmatrix}1-\lambda&2&3\\1&2-\lambda&3\\0&\lambda&\lambda\end{vmatrix}$
$A=\begin{vmatrix}1-\lambda&2&3\\ \lambda&-\lambda&0 \\0&\lambda&\lambda\end{vmatrix}$
$A=\lambda^2\begin{vmatrix}1-\lambda&2&3\\ 1&-1 &0 \\0&1&-1\end{vmatrix}$
$0 = \lambda^2(6−\lambda)$
The eigenvalues are $0$, $0$ and $6$.

However, in the matrix: $\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\begin{vmatrix}1-\lambda&1\\0&1-\lambda\end{vmatrix}$
$(1-\lambda)^2=0$
There is only one eigenvalue and it's $1$. How would you be able to tell when an eigenvalue is supposed to have a multiplicity of $1$ or more?

REPLY [0 votes]: It's convenient to write the matrix as an outer product, namely $$A=\begin{bmatrix}1&2&3\\1&2&3\\1&2&3\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}\begin{bmatrix}1&2&3\end{bmatrix}=ab^T.$$ 
For the first eigenvalue, let $v$ be a vector orthogonal to $b$ i.e. in the orthogonal complement of $b$. Then $Av=ab^Tv=a(0)=0v$ so $v$ is a right-eigenvector with eigenvalue $\lambda=0$, and since the orthogonal complement is 2D for $b\in\mathbb{R}^3$ we conclude that the (geometric!) multiplicity is 2. For the other eigenvalue, note that $Aa=ab^Ta=a(6)$ so $a$ is a right-eigenvector with eigenvalue $\lambda=6$.
PS: It is important that $a$ is not orthogonal to $b$. If it were, then $a$ would already lie in the orthogonal complement and would not yield a distinct eigenvector. So the geometric multiplicity of the zero eigenvalue would still be 2 and we wouldn't have another eigenvalue of multiplicity one.
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