Linear algebra - eigenvectors of $B$ not diagonalizable.

Question

Explain why the eigenvectors of $B$ does not constitute a basis for $\Bbb R^3$ and $B$ are thus not diagonalizable.

$B= \left ( \begin{matrix}2 & 0 & 0\\ 1 & 2 & -1 \\ 0 & 0 & 1 \end{matrix} \right ) $

How i solved it:

First I got the eigenvalues ($x_1=2, x_2 = 2, x_3 =1$)and eigenvectors $[(0,1,0),(0,0,0),(0,1,1)]$.

My answer:

So, because we have a Null vector and the set of eigenvectors is linearly dependent, it not constitute a basis for $\Bbb R^3$ and $B$ is thus not diagonalizable, .

Is my answer correct?

Answer 1

let an eigenvector corresponding to the repeating eigenvalue $2$ be $(x, y, z.)$ then $(A - 2I)(x,y,z)^T = (0,0,0)^T$ writing this out in matrix form you have $$\pmatrix{0&0&0\\1&0&-1\\0&0&-1}\pmatrix{x\\y\\z} = \pmatrix{0\\0\\0}.$$ this means $$x - z = 0, z = 0 \to x = 0, z = 0, y \text{ arbitrary}. $$ that is the eigenspace , $\ker(A - 2I)$ corresponding to the eigenvalue has dimension $1$ and basis is $(0,1,0)^T$

therefore the repeating eigenvalue $2$ is missing an eigenvector. this implies that $A$ cannot be diagonalized. you will have to find a generalized eigenvector and find jordan form for $A.$

Answer 2

You can note that if $B$ were diagonalizable, then there would be two linearly independent eigenvectors with eigenvalue $2$ (just inspecting the diagonalization) but you only have one linearly independent eigenvector with eigenvalue $2$. (The zero vector is not an eigenvector, and even if it were, it wouldn't be linearly independent of the other eigenvector). So yes, $B$ is not diagonalizable.

Answer 3

I would reformulate your answer like this:

So, because we have only two independent eigenvectors this does not constitute a basis for $\mathbb{R}^3$ (which would need three linear independent vectors) and $B$ is thus not diagonalizable.