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Section 5.3 Finding Eigenvectors

Having found the eigenvalues of the example matrix \(A=\begin{pmatrix}1\amp 2\\9\amp 4 \end{pmatrix}\) in the last section 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{5.3.1} \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{5.3.2} \end{equation}

or equivalently

\begin{equation} \begin{pmatrix}x+2y\\ 9x+4y \end{pmatrix} = \begin{pmatrix}7x\\ 7y \end{pmatrix}\text{.}\tag{5.3.3} \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

\begin{equation} v_7 = \begin{pmatrix}1\\ 3 \end{pmatrix}\tag{5.3.4} \end{equation}

and check by explicit computation that \(Av_7=7v_7\text{,}\) as expected.

Turning to the case \(\lambda=-2\text{,}\) a similar construction yields

\begin{equation} \begin{pmatrix}1\amp 2\\9\amp 4 \end{pmatrix} \begin{pmatrix}x\\ y \end{pmatrix} = -2\begin{pmatrix}x\\ y \end{pmatrix}\tag{5.3.5} \end{equation}

or equivalently

\begin{equation} \begin{pmatrix}x+2y\\ 9x+4y \end{pmatrix} = \begin{pmatrix}-2x\\ -2y \end{pmatrix}\text{,}\tag{5.3.6} \end{equation}

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

\begin{equation} v_{-2} = \begin{pmatrix}2\\ -3 \end{pmatrix}\tag{5.3.7} \end{equation}

and you should again check by explicit computation that \(Av_{-2}=-2v_{-2}\text{.}\)

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