这是一份leeds利兹大学MATH106001作业代写的成功案例
$$
T=\left(\begin{array}{ll}
3 & 2 \
0 & 1
\end{array}\right)
$$
we take it as the representation of a transformation with respect to the standard basis $T=\operatorname{Rep} \varepsilon_{2}, \mathcal{\varepsilon}{2}(t)$ and we look for a basis $B=\left\langle\vec{\beta}{1}, \vec{\beta}{2}\right\rangle$ such that $$ \operatorname{Rep}{B, B}(t)=\left(\begin{array}{cc}
\lambda_{1} & 0 \
0 & \lambda_{2}
\end{array}\right)
$$
that is, such that $t\left(\vec{\beta}{1}\right)=\lambda{1} \vec{\beta}{1}$ and $t\left(\vec{\beta}{2}\right)=\lambda_{2} \vec{\beta}{2}$. $$ \left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) \vec{\beta}{1}=\lambda_{1} \cdot \vec{\beta}{1} \quad\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) \vec{\beta}{2}=\lambda_{2} \cdot \vec{\beta}_{2}
$$
We are looking for scalars $x$ such that this equation
$$
\left(\begin{array}{ll}
3 & 2 \
0 & 1
\end{array}\right)\left(\begin{array}{l}
b_{1} \
b_{2}
\end{array}\right)=x \cdot\left(\begin{array}{l}
b_{1} \
b_{2}
\end{array}\right)
$$
has solutions $b_{1}$ and $b_{2}$, which are not both zero. Rewrite that as a linear system.
$$
(3-x) \cdot b_{1}+\begin{array}{r}
2 \cdot b_{2}=0 \
(1-x) \cdot b_{2}=0
\end{array}
$$
MATH106001 COURSE NOTES ：
$$
S=\left(\begin{array}{ll}
\pi & 1 \
0 & 3
\end{array}\right)
$$
(here $\pi$ is not a projection map, it is the number $3.14 \ldots$ ) then
$$
\left|\left(\begin{array}{cc}
\pi-x & 1 \
0 & 3-x
\end{array}\right)\right|=(x-\pi)(x-3)
$$
so $S$ has eigenvalues of $\lambda_{1}=\pi$ and $\lambda_{2}=3$. To find associated eigenvectors, first plug in $\lambda_{1}$ for $x$ :
$$
\left(\begin{array}{cc}
\pi-\pi & 1 \
0 & 3-\pi
\end{array}\right)\left(\begin{array}{l}
z_{1} \
z_{2}
\end{array}\right)=\left(\begin{array}{l}
0 \
0
\end{array}\right) \quad \Longrightarrow \quad\left(\begin{array}{l}
z_{1} \
z_{2}
\end{array}\right)=\left(\begin{array}{l}
a \
0
\end{array}\right)
$$
for a scalar $a \neq 0$, and then plug in $\lambda_{2}$ :
$$
\left(\begin{array}{cc}
\pi-3 & 1 \
0 & 3-3
\end{array}\right)\left(\begin{array}{l}
z_{1} \
z_{2}
\end{array}\right)=\left(\begin{array}{l}
0 \
0
\end{array}\right) \quad \Longrightarrow \quad\left(\begin{array}{l}
z_{1} \
z_{2}
\end{array}\right)=\left(\begin{array}{c}
-b /(\pi-3) \
b
\end{array}\right)
$$
where $b \neq 0$.