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I wish to prove that any $2\times 2$ matrix $T$ with only one eigenvalue $ \lambda$ of geometric multiplicity 1 is similar to one of the form

$\left[\begin{array}{cc} \lambda & 1 \\ 0 &\lambda \end{array}\right]$.

Now if I choose my basis to be $v$, where $v$ is some eigenvector of $\lambda$ and any vector $w$ not in null $(T - \lambda I)$ this should work. It is obvious in this basis how the first column of the matrix above comes about,

and for the second column, we note that

$T(w) = (T-\lambda I)w + \lambda w$ and by Cayley-Hamilton it is clear that $(T - \lambda I)w$ is an eigenvector of $T$. However this only means that in the matrix above I have a constant, not necessarily 1 in the slot where it is sitting.

How can I choose $w$ so that $(T-\lambda I)w$ is an eigenvector of $T$ of eigenvalue 1?


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up vote 2 down vote accepted

Once you have $\left[\begin{array}{cc} \lambda & c \\ 0 &\lambda \end{array}\right]$ with $c\neq0$, you can use


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You're right, so my basis instead of being $\{v,w\}$ would now be $\{cv,w \}$ for $c\neq 0$. –  fpqc Oct 13 '11 at 6:27
@Benjamin: Whether a vector is an eigenvector of an operator is basis-independent. I didn't look at what you wrote about eigenvectors in detail; I understood your problem to be that you could reach the form $\left[\begin{array}{cc} \lambda & c \\ 0 &\lambda \end{array}\right]$ but not $\left[\begin{array}{cc} \lambda & 1 \\ 0 &\lambda \end{array}\right]$, so that's what I addressed. –  joriki Oct 13 '11 at 7:21
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