Finding Eigenvectors

Skip to main content

\(\newcommand{\vf}[1]{\mathbf{\boldsymbol{\vec{#1}}}} \renewcommand{\Hat}[1]{\mathbf{\boldsymbol{\hat{#1}}}} \let\VF=\vf \let\HAT=\Hat \newcommand{\Prime}{{}\kern0.5pt'} \newcommand{\PARTIAL}[2]{{\partial^2#1\over\partial#2^2}} \newcommand{\Partial}[2]{{\partial#1\over\partial#2}} \newcommand{\tr}{{\rm tr\,}} \newcommand{\RR}{{\mathbb R}} \newcommand{\HR}{{}^*{\mathbb R}} \renewcommand{\AA}{\vf A} \newcommand{\BB}{\vf B} \newcommand{\CC}{\vf C} \newcommand{\EE}{\vf E} \newcommand{\FF}{\vf F} \newcommand{\GG}{\vf G} \newcommand{\HH}{\vf H} \newcommand{\II}{\vf I} \newcommand{\JJ}{\vf J} \newcommand{\KK}{\vf K} \renewcommand{\SS}{\vf S} \renewcommand{\aa}{\VF a} \newcommand{\bb}{\VF b} \newcommand{\ee}{\VF e} \newcommand{\gv}{\VF g} \newcommand{\iv}{\vf\imath} \newcommand{\rr}{\VF r} \newcommand{\rrp}{\rr\Prime} \newcommand{\uu}{\VF u} \newcommand{\vv}{\VF v} \newcommand{\ww}{\VF w} \newcommand{\grad}{\vf\nabla} \newcommand{\zero}{\vf 0} \newcommand{\Ihat}{\Hat I} \newcommand{\Jhat}{\Hat J} \newcommand{\nn}{\Hat n} \newcommand{\NN}{\Hat N} \newcommand{\TT}{\Hat T} \newcommand{\ihat}{\Hat\imath} \newcommand{\jhat}{\Hat\jmath} \newcommand{\khat}{\Hat k} \newcommand{\nhat}{\Hat n} \newcommand{\rhat}{\HAT r} \newcommand{\shat}{\HAT s} \newcommand{\xhat}{\Hat x} \newcommand{\yhat}{\Hat y} \newcommand{\zhat}{\Hat z} \newcommand{\that}{\Hat\theta} \newcommand{\phat}{\Hat\phi} \newcommand{\LL}{\mathcal{L}} \newcommand{\DD}[1]{D_{\textrm{$#1$}}} \newcommand{\bra}[1]{\langle#1|} \newcommand{\ket}[1]{|#1/rangle} \newcommand{\braket}[2]{\langle#1|#2\rangle} \newcommand{\LargeMath}[1]{\hbox{\large$#1$}} \newcommand{\INT}{\LargeMath{\int}} \newcommand{\OINT}{\LargeMath{\oint}} \newcommand{\LINT}{\mathop{\INT}\limits_C} \newcommand{\Int}{\int\limits} \newcommand{\dint}{\mathchoice{\int\!\!\!\int}{\int\!\!\int}{}{}} \newcommand{\tint}{\int\!\!\!\int\!\!\!\int} \newcommand{\DInt}[1]{\int\!\!\!\!\int\limits_{#1~~}} \newcommand{\TInt}[1]{\int\!\!\!\int\limits_{#1}\!\!\!\int} \newcommand{\Bint}{\TInt{B}} \newcommand{\Dint}{\DInt{D}} \newcommand{\Eint}{\TInt{E}} \newcommand{\Lint}{\int\limits_C} \newcommand{\Oint}{\oint\limits_C} \newcommand{\Rint}{\DInt{R}} \newcommand{\Sint}{\int\limits_S} \newcommand{\Item}{\smallskip\item{$\bullet$}} \newcommand{\LeftB}{\vector(-1,-2){25}} \newcommand{\RightB}{\vector(1,-2){25}} \newcommand{\DownB}{\vector(0,-1){60}} \newcommand{\DLeft}{\vector(-1,-1){60}} \newcommand{\DRight}{\vector(1,-1){60}} \newcommand{\Left}{\vector(-1,-1){50}} \newcommand{\Down}{\vector(0,-1){50}} \newcommand{\Right}{\vector(1,-1){50}} \newcommand{\ILeft}{\vector(1,1){50}} \newcommand{\IRight}{\vector(-1,1){50}} \newcommand{\Partials}[3] {\displaystyle{\partial^2#1\over\partial#2\,\partial#3}} \newcommand{\Jacobian}[4]{\frac{\partial(#1,#2)}{\partial(#3,#4)}} \newcommand{\JACOBIAN}[6]{\frac{\partial(#1,#2,#3)}{\partial(#4,#5,#6)}} \newcommand{\ii}{\ihat} \newcommand{\jj}{\jhat} \newcommand{\kk}{\khat} \newcommand{\dS}{dS} \newcommand{\dA}{dA} \newcommand{\dV}{d\tau} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} \)

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 \(Av_{-2}=-2v_{-2}\text{.}\)