Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities

Question

I have the following matrix:

$A = \begin{bmatrix} 1 && 7 && -2 \\ 0 && 3 && -1 \\ 0 && 0 && 2 \end{bmatrix}$

and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity.

What I have so far:

$\det(A - \lambda I) = \det\begin{bmatrix} 1-\lambda && 7 && -2 \\ 0 && 3-\lambda && -1 \\ 0 && 0 && 2-\lambda \end{bmatrix}$

I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me

$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).

Following that I move on to the geometric multiplicity:

$ A - 3I = \begin{bmatrix} -2 && 7 && -2 \\ 0 && 0 && -1 \\ 0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 1 && -\frac{7}{2} && 0 \\ 0 && 0 && 1 \\ 0 && 0 && 0 \end{bmatrix}$ which yields

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \frac{7}{2}\\ 1 \\ 0\end{bmatrix} s$ which has a geometric multiplicity of 1.

$ A - 2I = \begin{bmatrix} -1 && 7 && -2 \\ 0 && 1 && -1 \\ 0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 1 && 0 && -5 \\ 0 && 1 && -1 \\ 0 && 0 && 0 \end{bmatrix}$ which yields

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5\\ 1 \\ 1\end{bmatrix} s$ which has a geometric multiplicity of 1.

Finally,

$ A - I = \begin{bmatrix} 0 && 7 && -2 \\ 0 && 2 && -1 \\ 0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 0 && 1 && 0 \\ 0 && 0 && 1 \\ 0 && 0 && 0 \end{bmatrix}$

This is where I am stuck, I'm not sure what is the resulting $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ that column of 0's is confusing me.

Any help would be appreciated.

Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ ? or is that wrong?

Answer 1

That RRE $\begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}$ means $y=0$ and $z=0$, so $$\begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} x\\0\\0 \end{bmatrix} =x\begin{bmatrix} 1\\0\\0 \end{bmatrix}.$$

Answer 2

$$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\iff \begin{bmatrix}y\\z\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}$$

meaning that $y=z=0$, while leaving $x$ to be whatever you want, results in a kernel vector.

Answer 3

Yes, it works.

$$\begin{bmatrix} 0 && 1 && 0 \\ 0 && 0 && 1 \\ 0 && 0 && 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

Thus you are correct in choosing the third eigenvector as well.