19.2: Solving Eigenproblems - A 2x2 Example
- Page ID
- 67887
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider calculating eigenvalues for any \(2 \times 2\) matrix. We want to solve:
\[|A - \lambda I_2 | = 0 \nonumber \]
\[\begin{split}
\left|
\left[
\begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{matrix}
\right]
- \lambda \left[
\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}
\right]
\right|
=
\left|
\left[
\begin{matrix}
a_{11}-\lambda & a_{12} \\
a_{21} & a_{22}-\lambda
\end{matrix}
\right]
\right|
=0
\end{split} \nonumber \]
We know this determinant:
\[(a_{11}-\lambda)(a_{22}-\lambda) - a_{12} a_{21} = 0 \nonumber \]
If we expand the above, we get:
\[a_{11}a_{22}+\lambda^2-a_{11}\lambda-a_{22}\lambda - a_{12} a_{21} = 0 \nonumber \]
and
\[\lambda^2-(a_{11}+a_{22})\lambda+a_{11}a_{22} - a_{12} a_{21} = 0 \nonumber \]
This is a simple quadratic equation. The roots of \(A\lambda^2 + B\lambda + C = 0 \) can be solved using the quadratic formula:
\[ \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \nonumber \]
Using the above equation. What are the eigenvalues for the following \(2 \times 2\) matrix. Try calculating this by hand and then store the lower value in a variable namede1
and the larger value in e2
to check your answer:
\[\begin{split}A =
\left[
\begin{matrix}
-4 & -6 \\
3 & 5
\end{matrix}
\right]
\end{split} \nonumber \]
Find a numpy
function that will calculate eigenvalues and verify the answers from above.
What are the corresponding eigenvectors to the matrix \(A\)? This time you can try calculating by hand or just used the function you found in the previous answer. Store the eigenvector associated with the e1
value in a vector named v1
and the eigenvector associated with the eigenvalue e2
in a vector named v2
to check your answer.
Both sympy and numpy can calculate many of the same things. What is the fundamental difference between these two libraries?