Prove that S is diagonal Let $S: V\rightarrow\ V$ be an operator on an $n$-dimensional real vector space with an eigenvalue that has geometric multiplicity equal to $n-1$. Prove that $S$ is diagonal. Give an example of such an operator on $\mathbb R³$. 
I'm not sure how to approach this, I know that if I was guaranteed another eigenvalue, then its geometric multiplicity would be $1$, same as its algebraic multiplicity, which would therefore prove that $S$ is diagonal. But I don't understand how to proceed if I only know of the existence of one eigenvalue. Does it have something to do with $S$ being in a real vector space?
 A: I think you're correct in your initial post- you need to know that the given eigenvalue has geometric multiplicity equal to its algebraic multiplicity (which is $n-1$). Then the matrix must have another eigenvalue, with multiplicity equal to 1, which implies that $S$ is diagonal.
If you don't use this assumption, you can get something like:
$$S = \left(\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$$
This has an eigenvalue $1$ with geometric multiplicity $2$, but is not diagonal.
A: Let denote $\lambda$ the eigenvalue with geometric multiplicity $n-1$ and since  the algebraic multiplicity of $\lambda$ is greater than or equal to the geometric multiplicity then there's two cases


*

*the algebraic multiplicity of $\lambda$ is $n$ hence the characteristic polynomial of $S$ is
$$\chi_S(x)=(x-\lambda)^n$$
so $S-\lambda \operatorname{id}$ is nilpotent and $S$ is similar to
$$\begin{pmatrix}\lambda&0&\cdots&\cdots&0\\0&\lambda&\cdots&\cdots&0\\\vdots&&\ddots&&\vdots\\0&&&\lambda&1\\0&&&&\lambda\end{pmatrix}$$

*the algebraic multiplicity of $\lambda$ is $n-1$ hence the characteristic polynomial of $S$ is
$$\chi_S(x)=(x-\lambda)^{n-1}(x-\mu),\; \lambda\ne\mu$$
and in this case $S$ is similar to $\operatorname{diag}(\lambda,\ldots,\lambda,\mu)$.

