Let $U$ and $V$ be the subspaces of $\mathbb{R}^3$ defined by $U=\{(x,y,z)^T:2x+3y+4z=0 \}$ and $V=\{(x,y,z)^T:x+2y+5z=0 \}.$
Then consider the following statement:
There exists a linear transformation $T:\mathbb{R}^3 \to \mathbb{R}^3$ such that $T(U) \cap V \ne \{ 0\}$ and the characteristic polynomial of $T$ is not the product of linear polynomials with real coefficients.
My attempt:
We can observe that there exits a non-zero vector $v$ contained in the intersection of subspaces $U$ and $V.$ Extend $\{v\}$ to form a basis $\{v,v_1,v_2 \}$ of $\mathbb{R}^3.$ Construct a linear transformation $T:\mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(v)=v,T(v_1)=T(v_2)=0.$
Now, the characteristic polynomial of $T$ is $x^2(x-1),$ which is not product of distinct linear polynomials with real coefficients. (Note: Please see that the characteristic polynomial of $T$ is product of linear factors but the factors are not distinct). This $T$ satisfies all the required conditions of the statement.
The question I had is this:
Instead of taking a repetition of a linear factor, can I get a linear mapping $T$ such that its characteristic polynomial has an irreducible factor?
The answer is YES. But I am confused as to how it is possible. I am stuck here. Please help.
I got the answer. Extend $\{v \}$ to $ \mathcal{B}=\{v,v_1,v_2 \}$ to form a basis of $\mathbb{R}^3.$ Now define $T: \mathbb{R}^3 \to \mathbb{R}^3$ as $T(v)=v,T(v_1)=v_2$ and $T(v_2)=-v_1.$ Then, with respect to this basis, $T$ has matrix representation,
$$[T]_{\mathcal{B}}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} $$
So, we have that characteristic polynomial of $T$ is $x(x^2+1).$ Also, $T(U) \cap V \ne \{ 0\}.$
I rest!