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I'm confused about how to find the possible dimension of an eigenspace given that a matrix has exactly one eigenvalue.

Suppose $A$ is a $3\times 3$ matrix, with exactly one eigenvalue $\lambda$. I assume I'm working over $\mathbb{R}$, so $\lambda$ is the only real root of the characteristic polynomial of $A$. Since this is a cubic, $\lambda$ has multiplicity $1$ or $3$.

Since the dimension of the eigenspace is at most the algebraic multiplicity of the eigenvalue, I think the dimension is either $0$ or $1$, or $0,1,2$ or $3$.

But the possible answers (it is a multiple choice question) are

  • $1$
  • $2$
  • $3$
  • $1$ or $2$
  • $1$, $2$, or $3$

How can I more precisely determine the dimension?

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    $\begingroup$ Do you know what Jordan normal form is ? $\endgroup$
    – Belgi
    Commented Sep 28, 2012 at 23:39
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    $\begingroup$ Assuming $\lambda$ is real, the dimension certainly can't be zero, or it wouldn't be an eigenvalue at all.On what grounds have you excluded $2$ as possibility? $\endgroup$ Commented Sep 28, 2012 at 23:41

2 Answers 2

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The case that there is a non real root is simple since then $A$ is similar to $diag(\alpha,\lambda,\overline{\lambda})$ where $\alpha$ is the real eigenvalue and $\lambda$ is a non-real eigenvalue.

In this case it is clear that the dimension equals to algebraic multiplicity in the characteristic polynomial hence it is $1$.

The other case is that the characteristic polynomial is $(x-\alpha)^{3}$ where $\alpha\in\mathbb{R}$.

In this case we can have some options, for example, we can have $A=diag(\alpha,\alpha,\alpha)$ hence the dimension of the eigenspace is $3$.

We can also have $$A=\begin{pmatrix}\alpha & 1\\ & \alpha\\ & & \alpha \end{pmatrix}$$ and it is straightforward to check that the dimension of the eigenspace is $2$

We can also have $$A=\begin{pmatrix}\alpha & 1\\ & \alpha & 1\\ & & \alpha \end{pmatrix}$$ and it is straightforward to check that the dimension of the eigenspace is $1$

Finally, we can never have it that the dimension is $0$ since if $\alpha$ is an eigenvalue than by definition there is $v\neq0$ that is a corresponding eigenvector, since the span of $v$ is of dimension $1$ (because $v\neq0$) and since the span of $v$ is contained by definition, in the eigenspace of $\alpha$ we have that the dimension of the eigenspace is at least $1$.

So the correct answer is the last one.

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If I get it right, do you wish to find out the dimension of the space of the eigen values? Then, you could probably try and find the dimension of the Null Space of $A-\lambda I$. That should give you the dimension of the space of eigen vectors.

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