Do there exist two-dimensional subspaces $W$ and $Y$ of $\mathbb{R^4},$ both invariant under $\alpha,$ such that $\mathbb{R^4} = W \oplus Y$?

Question

The question is given below:

Could anyone give me a hint for solving the second part of the question?

Answer 1

Elegant Solution or Cryptic Hint, depending on who's reading:

Previous. If $T:\Bbb R^2\to\Bbb R^2$ is linear and $T^3=0$ then $T^2=0$.

Proof: The minimal polynomial $m(x)$ is a factor of $x^3$, while Cayley-Hamilton shows $\deg(m)\le 2$.

Now since $\alpha^3=0$, if $\Bbb R^4$ were the sum of invariant subspaces of dimension $2$ it would follow that $\alpha^2=0$, which is not so.

Detailed Version

Previous. If $T:\Bbb R^2\to\Bbb R^2$ is linear and $T^3=0$ then $T^2=0$.

Proof: Say $m$ is the minimal polynomial for $T$. By definition $m$ is a non-constant monic polynomial, and if $p$ is any polynomial with $p(T)=0$ then $m$ divides $p$. Since $T^3=0$ this shows that $m$ is either $x$, $x^2$ or $x^3$.

Now say $p$ is the characteristic polynomial of $T$. Then $p$ is quadratic, and $p(T)=0$ implies $m$ divides $p$. So $m$ is $x$ or $x^2$. So either $T=0$ or $T^2=0$, and in either case it follows that $T^2=0$.

Answer to the Question: No.

Note first that $\alpha^2\ne0$ but $\alpha^3=0$.

Now suppose that $\Bbb R^4$ is the sum of two invariant subspaces $X$ and $Y$, each of dimension $2$. Define $$T:X\to X$$ by $$Tx=\alpha x.$$ That is, $T$ is the restriction of $\alpha$ to $X$. Now $\alpha^3=0$ shows that $T^3=0$, and since $X$ has dimension $2$ the previous result shows that in fact $T^2=0$. Similarly for $S$, the restriction of $\alpha$ to $Y$.

Now say $v\in\Bbb R^4$. There exist $x\in X$, $y\in Y$ with $v=x+y$. So $$\alpha^2v=\alpha^2x+\alpha^2y=T^2x+S^2y=0.$$So $\alpha^2=0$, contradiction.