Exercise 4, Section 3.4 of Hoffman’s Linear Algebra 
Let $V$ be a two-dimensional vector space over the field $F$, and let $B$ be an ordered basis for $V$. If $T$ is a linear operator on $V$ and $[T]_B=
\begin{bmatrix} a & b \\c & d\\ \end{bmatrix}$. Prove that $T^2-(a+d)T+(ad-bc)I=0$.

My attempt: It’s easy to check $T^2-(a+d)\cdot T+(ad-bc)\cdot\text{id}_V\in L(V,V)$. Let $B=\{\alpha_1,\alpha_2\}$ be basis of $V$. Then $T(\alpha_1)=a\cdot \alpha_1+c\cdot \alpha_2$ and $T(\alpha_2)=b\cdot \alpha_1 +d\cdot \alpha_2$. So
\begin{align}
T^2(\alpha_1) &=T(T(\alpha_1)) \\
&=T(a\cdot \alpha_1 +c\cdot \alpha_2) \\
&=a\cdot T(\alpha_1)+c\cdot T(\alpha_2) \\
&=a\cdot (a\cdot \alpha_1+c\cdot \alpha_2)+c\cdot (b\cdot \alpha_1+d\cdot \alpha_2)\\
&=(a^2+cb)\cdot \alpha_1 +(ac+cd)\alpha_2.
\end{align} Then $(T^2-(a+d)\cdot T+(ad-bc)\cdot\text{id}_V)(\alpha_1)$ $=T^2(\alpha_1)-(a+d)\cdot T(\alpha_1)+(ad-bc)\cdot \text{id}_V(\alpha_1)$ $= [(a^2+cb)\cdot \alpha_1 +(ac+cd)\alpha_2]-(a+d)\cdot [a\cdot \alpha_1+c\cdot \alpha_2] +(ad-bc)\cdot \alpha_1$ $=[(a^2+cb)\cdot \alpha_1 +(ac+cd)\alpha_2]$$-$$[(a^2+ad)\cdot \alpha_1+(ac+cd)\cdot \alpha_2]$$+$$[(ad)\cdot \alpha_1-(bc)\cdot \alpha_1]$ $=0_V$. Similarly $(T^2-(a+d)\cdot T+(ad-bc)\cdot\text{id}_V)(\alpha_2)=0_V$. By theorem 1 section 3.1, $T^2-(a+d)\cdot T+(ad-bc)\cdot\text{id}_V=0_{L(V)}$. Is my proof correct? Once can also show $(T^2-(a+d)\cdot T+(ad-bc)\cdot\text{id}_V)(\alpha)=0_V$, $\forall \alpha \in V$. But that is not efficient.
Observation: See $\text{tr}([T]_{B})=a+d$ and $\text{det}([T]_B)=ad-bc$. So $T^2-\text{tr}([T]_{B})\cdot T+ \text{det}([T]_B) \cdot \text{id}_V=0_{L(V)}$. Can we generalize this problem?

Alternative proof: By theorem 13 section 3.4 and this post, $([T]_B)^2-(a+d)\cdot [T]_B+(ad-bc)\cdot I_2$ is matrix of $T^2-(a+d)\cdot T+(ad-bc)\cdot \text{id}_V$ relative to ordered basis $B$. It’s easy to check $([T]_B)^2-(a+d)\cdot [T]_B+(ad-bc)\cdot I_2$ $= \begin{bmatrix} a & b \\c & d\\ \end{bmatrix}^2-(a+d)\cdot \begin{bmatrix} a & b \\c & d\\ \end{bmatrix} +(ad-bc)\cdot \begin{bmatrix} 1 & 0 \\0 & 1\\ \end{bmatrix}$ $= \begin{bmatrix} 0 & 0 \\0 & 0\\ \end{bmatrix}$. By this post, $T^2-(a+d)\cdot T+(ad-bc)\cdot \text{id}_V=0_{L(V)}$.
