Eigenvalue and algebraic multiplicity of some matrix Determine the eigenvalue and algebraic multiplicity of $$P=\pmatrix{0 & 1 & 0 \\ 0 & 0 & 1 \\1 & -3 & 3  \\ }$$ My calculations led me to the eigenvalues $0,\pm 1,3$ each of algebraic multiplicity $ 1$. While the book's answer is: the only eigenvalue is $ 1$,of algebraic multiplicity $ 3$. Am I right, if not may you explain why? $$$$Thanks in advance
 A: Hint: if you solve 
$$|P- \lambda I| = 0$$
The characteristic polynomial is:
$$-\lambda^3+3 \lambda^2-3 \lambda + 1 = 0$$
This factors as:
$$-(\lambda-1)^3 = 0$$
You get a single eigenvalue $\lambda = 1$
It has algebraic multiplicity $3$.
A: The matrix is in companion form, so the characteristic polynomial can be read off from the bottom row (the polynomial coefficients are the negatives of the bottom row) as $\lambda^3 + (-3) \lambda^2 + (3) \lambda + (-1) = (\lambda-1)^3$.
A: I get the same as the book. Let's have a look at where you might have gone wrong. If $M$ is the matrix above and $I$ is the three-by-three identity matrix then the eigenvalues are found by solving $\det(M-\lambda I) = 0$.
$$M-\lambda I = \left[ \begin{array}{ccc} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 1 & -3 & 3-\lambda \end{array}\right]$$
I'm going to expand the first row to get the determinant. Let $|N|$ be the determinant:
$$\begin{array}{ccc}
\left| \begin{array}{ccc} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 1 & -3 & 3-\lambda \end{array}\right| &=& -\lambda\left|\begin{array}{cc} -\lambda & 1 \\ -3 & 3-\lambda \end{array}\right|-1\left|\begin{array}{cc} 0 & 1 \\ 1 & 3-\lambda \end{array}\right|+0\left|\begin{array}{cc} 0 & -\lambda \\ 1 & -3 \end{array}\right| \\
&=& -\lambda(-\lambda(3-\lambda)+3)-1(0-1)+0(0+\lambda) \\
&=& -\lambda^3+3\lambda^2-3\lambda+1 \\
&=& (1-\lambda)^3
\end{array}$$
Hence $\lambda = 1$ is the only eigenvalue of the matrix. Putting $\lambda = 1$ gives
$$M- I = \left[ \begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 1 & -3 & 2 \end{array}\right]$$
The kernel vectors, $[u,v,w]^{\top}$, are given by solving
$$\left[ \begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 1 & -3 & 2 \end{array}\right]\left[\begin{array}{c} u \\ v \\ w \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]$$
In other words: $v-u=0$, $w-v=0$ and $u - 3v + 2w = 0$. The first equation tells us that $u=v$ and the second tells us that $w=v$. Putting this into the third gives $v - 3v + 2v = 0$. This holds for all $v$, and so the eigenspace is $[v,v,v]^{\top}$, where you are free to choose $v$, i.e. the line spanned by $[1,1,1]^{\top}$.
