Find a base for which the matrix of a linear transformation is diagonal Define $T​ : R^2→R^2 $
by $​T(x​)=Ax$
Find a basis $B$ for $R^2$ with the property that $[T]_B$ is diagonal.
$$A=\left(\begin{matrix}1 & -7\\-5&3\end{matrix}\right)$$
I have worked out the eigenvalues: $\lambda =-4, 8$
I am confused as to what they're asking, what should I work out and try to prove? So I must find a vector(basis) $B$ so that $T$ with relation to $B$ is diagonal? Must I find the eigenvalues?
 A: First, to find the eigenvalues, you compute $\det A-\lambda I=(1-\lambda)(3-\lambda)-35=(\lambda-8)(\lambda+4)$. Hence the eigenvalues are $8$ and $-4$.
To find an eigenvector associated with $\lambda=8$, you solve the system
$$\left(\begin{matrix}-7&-7\\-5&-5\end{matrix}\right)\left(\begin{matrix}x_1\\x_2\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
This is equivalent to the single equation $x_1+x_2=0$, so there is an arbitrary parameter, say $x_2$, and $x_1=-x_2$. With $x_2=-1$, that yields the eigenvector $(1,-1)^T$.
For the eigenvalue $-4$, the system is
$$\left(\begin{matrix}5&-7\\-5&7\end{matrix}\right)\left(\begin{matrix}x_1\\x_2\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
This is equivalent to the single equation $5x_1-7x_2=0$, so there is also an arbitrary parameter. With $x_2=5$, you get $x_1=7$, so the eigenvector $(7,5)^T$.
Now, the matrix of your new base is $P=\left(\begin{matrix}1&7\\-1&5\end{matrix}\right)$. In this base, the matrix of the transformation $T$ is diagonal: $P^{-1}AP=\left(\begin{matrix}8&0\\0&-4\end{matrix}\right)$.
