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!
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.