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We will only be looking at one specific eigenvalue, eigenspace (there are three eigenvalues in total, I know two of them are fine and last one seems not but as I'm not sure I need to ask you).

We have matrix $A=\begin{pmatrix} 3 & -1 & 0\\ 2 & 0 & 0\\ -2 & 2 & -1 \end{pmatrix}$

The eigenvalues are $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=-1$

We will only concentrate on $\lambda_{3}=-1$ because I have checked the others already and I'm very sure they are fine (they show that the matrix is diagonalisable).

We insert $\lambda_{3}=-1$ here $\begin{pmatrix} 3-\lambda & -1 & 0\\ 2 & -\lambda & 0\\ -2 & 2 & -1-\lambda \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

Then we get $\begin{pmatrix} 4 & -1 & 0\\ 2 & 1 & 0\\ -2 & 2 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$

$I: 4x-y=0 \Leftrightarrow y=4x$

$II: 2x+y=0$

$III: -2x+2y=0$

We see that there is no value for $z$, so we can choose an arbitrary $z$. We get the eigenspace $E_{A}(-1)=\left\{ \left. \begin{pmatrix} x\\ 4x\\ z \end{pmatrix} \right | x,z \in \mathbb{R} \right\}$

Two variables mean the basis is made up by $2$ vectors and this means the dimension of the eigenspace is $2$ and this means the matrix is not diagonalisable because $\lambda_{3}=-1$ is just a single eigenvalue?


Can you please tell me if this is correct?

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    $\begingroup$ $x$ is not free: if $y=4x$ then $2x+y=6x$ which must be zero by equation II. $\endgroup$ – Ian Mar 9 '17 at 20:56
  • $\begingroup$ @Ian Thank you for the comment and the edit too. Now I also know how to make this longer line^^ $\endgroup$ – cnmesr Mar 9 '17 at 21:00
  • $\begingroup$ Do you know how to do row-reduction/Gaussian elimination? That lets you read a basis for the null space of a matrix directly from the rref. I find doing that to be much less error-prone than trying to solve a system of linear equations “by hand.” In this case, the rref is $\operatorname{diag}(1,1,0)$, from which you can instantly see that the eigenspace is the span of $(0,0,1)^T$. $\endgroup$ – amd Mar 10 '17 at 0:13
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From I and II, we have $y = 4x$ and $y = -2x$. The only way these can simultaneously hold is if $x = y = 0$. However, $z$ is indeed a free variable. We find that $$ E_A(-1) = \{(0,0,z)^T : z \in \Bbb R\} $$

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  • $\begingroup$ Oh thanks I don't know how I could miss that..! :s But let's say what I did above was correct (the eigenspace), would my reasoning be correct too? $\endgroup$ – cnmesr Mar 9 '17 at 20:59
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    $\begingroup$ It would not. We would indeed conclude that the eigenspace of $-1$ is $2$-dimensional. If we found that $A$ had another eigenspace with dimension $1$, then the sum of the dimensions is $3$, which means that $A$ is diagonalizable (i.e. has a basis consisting of eigenvectors). $\endgroup$ – Omnomnomnom Mar 9 '17 at 21:01
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    $\begingroup$ As an explanatory example: consider the $2 \times 2$ identity matrix, which is diagonalizable (since it is diagonal!). However, every vector from the 2-dimensional $\Bbb R^2$ is an eigenvector associated with $1$, and $1$ is the only eigenvalue. $\endgroup$ – Omnomnomnom Mar 9 '17 at 21:02
  • $\begingroup$ Ouch, so there is something I misunderstood badly. The sum of the dimensions always has to be exactly $3$, in every other example where we are working with $3 \times 3$ matrix, and then the matrix is diagonalisable? $\endgroup$ – cnmesr Mar 9 '17 at 21:03
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    $\begingroup$ You said that a bit strangely. A $3 \times 3$ matrix is diagonalizable if and only if the sum of the dimensions of all eigenspaces is $3$. $\endgroup$ – Omnomnomnom Mar 9 '17 at 21:05
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Another way of saying that is- no matter how many distinct eigenvalues an by n matrix has, it is diagonalizable if and only if there are n independent Eigenvectors. Of course, eigenvectors corresponding to distinct eigenvalues are necessarily independent so if an n by n matrix has n distinct eigenvalues then it must be diagonalizable- but it is the eigenvectors that are important!

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