Having found the eigenvalues of the example matrix \(A=\begin{pmatrix}1\amp 2\\9\amp 4
\end{pmatrix}\) in (4.2.2) to be \(7\) and \(-2\text{,}\) we can now ask what the corresponding eigenvectors are. We must therefore solve the equation
\begin{equation}
A \left|v\right> = \lambda \left|v\right>\tag{4.3.1}
\end{equation}
in the two cases \(\lambda=7\) and \(\lambda=-2\text{.}\) In the first case, we have
\begin{equation}
\begin{pmatrix}1\amp 2\\9\amp 4
\end{pmatrix}
\begin{pmatrix}x\\ y
\end{pmatrix}
= 7\begin{pmatrix}x\\ y
\end{pmatrix}\tag{4.3.2}
\end{equation}
Thus, we must solve this system of two equations. But we quickly discover that these equations are reduntant, since the first implies \(2y=6x\text{,}\) while the second implies \(3y=9x\text{,}\) which is the same condition. This is a good thing! (Why?)
We conclude that any vector \(\begin{pmatrix}x\\ y
\end{pmatrix}\) with \(y=3x\) is an eigenvector of \(A\) with eigenvalue \(7\text{.}\) To check an explicit example, choose any value for \(x\text{,}\) such as \(x=1\text{,}\) yielding the vector
and the first equation now yields \(2y=-3x\text{,}\) while the second yields \(6y=-9x\text{.}\) Again, these two equations are redundant; the eigenvectors with eigenvalue \(-2\) satisfy \(y=-\frac32\text{.}\) An explicit example is
and you should again check by explicit computation that \(A\left|v_{-2}\right>=-2\left|v_{-2}\right>\text{.}\)
To Remember.
The set of equations that you solve for the eigenvectors will always be redundant (or trivial). So you will always have to choose one of the components. This freedom leads to the useful property that multiples of eigenvectors are still eigenvectors, see Section 4.4.