Every linear operator on $\mathbb{R}^5$ has an invariant 3-dimensional subspace I am trying to determine whether the following statement is true or false:
Every linear transformation on $\mathbb{R}^5$ has an invariant 3-dimensional subspace.
Since $\dim(\mathbb{R}^5)=5$ then given any linear operator $T$ on $\mathbb{R}^5$ I know that $\deg(\text{char}_T(x))=5$, and hence,$\text{char}_T(x)$ has at least one real root, meaning that $T$ has at least one real eigenvalue, $\lambda$. Thus,
$$\text{char}_T(x)=(x-\lambda)f(x),$$
where $f$ can be factored as the product of irreducibles into two quadratics, a quadratic and two linear factors, or 4 linear factors. I don't know where to go from there. Perhaps the statement is false? Thank you for your help!
 A: This is an easy consequence of the existence of the real Jordan normal form of the matrix of the endomorfism. That matrix is similar to a block diagonal matrix, with each block being a real Jordan block. There are several cases to be considered. For instance, if you endomorfism has one and only one real eigenvalue (with multiplicity $1$) and four complex non-real eigenvalues, then the real Jordan normal form will be of the type$$\begin{pmatrix}1&0&0&0&0\\0&a&-b&0&0\\0&b&a&0&0\\0&0&0&c&-d\\0&0&0&d&c\end{pmatrix}$$and therefore the span of the first three vectors of the correspondeng basis will be invariant. If you endomorfism has one and only one real eigenvalue (with multiplicity $1$) and two complex non-real eigenvalues (each with multiplicity $2$), then either the real Jordan normal form will be like the previous one (with $c=1$ and $d=b$) or will have the form$$\begin{pmatrix}1&0&0&0&0\\0&a&-b&1&0\\0&b&a&0&1\\0&0&0&a&-b\\0&0&0&b&a\end{pmatrix},$$but again you can consider the span of the first three vectors of the corresponding basis. And so on.
