Elegant Solution or Cryptic Hint, depending on who's reading (given by David Ullrich)

Question

David has given me this solution:

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 $T^2=0$.

My question is

In the third line from below: "Now say $p$ is the characteristic polynomial of $T$. Then $p$ is quadratic, ..." why he said then the characteristic polynomial is quadratic, why it is quadratic, could anyone explains this for me, please?

Answer 1

The characteristic polynomial of an $n \times n$ matrix is always of degree $n$. Here $n=2$.

Answer 2

The characteristic polynomial is just the function $p(x)=\det(A-x I)$ where $A$ is the matrix of $T$. For an $n \times n$ matrix this is a polynomial of degree $n$.