Invertability of a linear transformation Given $T : \mathbb{R}^3 \to \mathbb{R}^3$ such that $T(x_1,x_2,x_3) = (3x_1,x_1-x_2,2x_1+x_2+x_3)$ Show that $(T^2-I)(T-3I) = 0$.
Solution 1: I can very easily write down the matrix representing $T$, calculate each of the terms in each set of parenthesis, and multiply the two matrixes to show that they equal zero.
Though this would show the desired statement, I think its cumbersome and perhaps might be missing the point of the exercise. So:
Other Attempt: I can multiply out the above expression to have that $T^3-IT+(T^2)(-3I)+3I^2=0$. Now we can easily verify that $T$ is invertible by simply looking at the kernel and seeing its $(x_1,x_2,x_3)=0$. Now what I want to conclude is that the matrix product above is also in the kernel since we have that each linear component is invertible, and so their sum should be, and the statement is almost trivially proven. 
The problem is i'm not really comfortable with saying that since each linear component is invertible its sum must be, and am having trouble showing this concretely.
Could anyone provide me with a small insight?
Thanks.

Let $A=T^2-I$ and $B=T-3I$. Proving that $A\cdot B = 0$ is the same as showing that $\text{ker}(A\cdot B)=\text{ker}(A)\cup\text{ker}(B)=\Bbb R^3$.
An easy calculation shows that $\text{ker}(B)=\langle (1,-2,3) \rangle$. Now observe that
$$
\{v_1=(1,-2,3),v_2=(0,1,0),v_3=(0,0,1)\}
$$
is a basis for $\Bbb R^3$. Then it is enough to show that $A(v_i)=T^2v_i-v_i=0$, i.e. $T^2v_i=v_i$, for $i=2,3$.
$$
\begin{align*}
T^2v_2 &= T(0,-1,1) = (0,1,-1+1) = v_2\\
T^2v_3 &= T(0,0,1) = (0,0,1) = v_3
\end{align*}
$$
