# 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?

• You have to diagonalize the matrix: the eigenvectors form a base where the matrix is diagonal. Here the matrix is certainly diagonalizable since every eigenvalue has multiplicity $1$. Thus, you have to solve the two systems $(A-\lambda I)(x_1,x_2)^T=0$. – Jean-Claude Arbaut Mar 27 '16 at 21:31
• @Jean-ClaudeArbaut my second basis is $(-1,1)^T$ which is correct but then a answer for the first basis ($\lambda = -4$)is $(7,5)^T$. I reduce $A + 4.I$ and get the following equation: $5x_1 -7x_2 = 0$. How do you get the 7 and 5 from that equation? I guess if you do $5x_1 = 7$ by letting $x_2 = 1$, but then shouldnt $x_1 = 7/5$, or are both answers correct i.e $(7/5,5/7)^T$ and $(7,5)^T$ – coders Mar 28 '16 at 11:17
• Once you have an eigenvector $u$, any $ku$ with real $k$ is also an eigenvector: there is actually an eigenspace. So both are correct, yes. However, your first is not correct, it should be $(7/5,1)^T$. – Jean-Claude Arbaut Mar 28 '16 at 11:40

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: $P^{-1}AP=\left(\begin{matrix}8&0\\0&-4\end{matrix}\right)$.