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Let $A$ be a non-scalar $2 \times 2$ complex matrix with a single eigenvalue $\lambda$. I want to show that $A$ is non-diagonalisable.

For $A$ to be diagonalisable, it must be that the geometric multiplicity of $\lambda$ is 2. This means that the eigenspace associated to $\lambda$ has dimension 2, and that there are two linearly independent eigenvectors associated to $\lambda$.

The problem is that I have no good idea about how to apply the fact that $A$ is non-scalar, since this is necessary to prove the statement.

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"non-scalar" - so, not of the form $\lambda\mathbf I$, then? –  J. M. Aug 27 '12 at 13:15
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Just go from the other side: Assume that $A$ is diagonalisable, and show that the condition of only one eigenvalue leads to a scalar matrix. Hint: Write $A$ as function of its eigenvectors and eigenvalues. –  celtschk Aug 27 '12 at 13:24
    
@celtschk If you write that as an answer, I will mark it as correct. –  utdiscant Aug 27 '12 at 13:38
    
In "For $A$ to be non-diagonalisable, it must be that the geometric multiplicity of $\lambda$ is $2$", you mean to say instead "For $A$ to be diagonalisable...", don't you? Now "$A$ is non-scalar" precisely means that this (the geometric multiplicity being $2$) does not happen. –  Marc van Leeuwen Aug 27 '12 at 15:01
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2 Answers

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Just go from the other side: Assume that A is diagonalisable, and show that the condition of only one eigenvalue leads to a scalar matrix. Hint: Write A as function of its eigenvectors and eigenvalues.

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Assume $A$ is a $2\times2$ matrix with a single eigenvalue $\lambda$, assume by negation that $A$ is diagonalize then there is a matrix $P$ s.t. $$P^{-1}\begin{pmatrix}\lambda\\ & \lambda \end{pmatrix}P=A$$ thus $$A=P^{-1}\lambda IP=\lambda P^{-1}IP=\lambda P^{-1}P=\lambda I=\begin{pmatrix}\lambda\\ & \lambda \end{pmatrix}$$

That is: if $A$ is a $2\times2$ matrix with a single eigenvalue $\lambda$that is diagonalize then $A=\begin{pmatrix}\lambda\\ & \lambda \end{pmatrix}$.

Now conclude that if $A$ is not a scalar matrix it must be non-diagonalisable.

Another solution uses Jordan form: any $2\times2$ matrix is similar to a matrix of the form $$\begin{pmatrix}\lambda_{1} & x\\ & \lambda_{2} \end{pmatrix}$$ where $\lambda_{1},\lambda_{2}$ are the eigenvalues of $A$ and $x\in\{0,1\}$.

In your case $\lambda_{1}=\lambda_{2}=\lambda$ and if $x=0$ then the matrix is a diagonalize scalar matrix, and if $x=1$ you can check that the matrix is non-diagonalisable.

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