Eigenvalues and eigenvectors: Eigenvalues and eigenvectors
Solving an eigenvalue problem
When you determine eigenvalues of a matrix and corresponding eigenspaces, then you solve an eigenvalue problem for a matrix. Below is an example.
Solve the eigenvalue problem for the matrix \[
A = \matrix{-25 & 56 \\ -12 & 27}
\]In other words, determine the eigenvalues and vectors.
A = \matrix{-25 & 56 \\ -12 & 27}
\]In other words, determine the eigenvalues and vectors.
The characteristic equation is \[\det\matrix{-25-\lambda & 56 \\ -12 & 27-\lambda }=0\] We first rewrite the characteristic polynomial of \(A\):
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -25-\lambda & 56 \\ -12 & 27-\lambda \end{array} \right\vert &= (-25-\lambda)(27-\lambda)-56\cdot-12 \\
&= (-25-\lambda)(27-\lambda)+672 \\ &= \lambda^2-2\,\lambda-3
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-2\,\lambda-3 = (\lambda-3)(\lambda+1) \] So, the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -1\).
Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-25 & 56 \\ -12 & 27}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), this is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\). We can do this through row reduction of the matrix \[A-3 I = \matrix{-25 & 56 \\ -12 & 27} - \matrix{3 & 0 \\0& 3 }=\matrix{ -28 & 56 \\ -12 & 24}\] This can be done as follows:
\[\begin{aligned}
\matrix{-28&56\\-12&24\\}&\sim\matrix{1&-2\\-12&24\\}&{\blue{\begin{array}{c}-{{1}\over{28}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{2\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{2\\1}\right\rangle\).
Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{-25 & 56 \\ -12 & 27}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-1\vec{v}\), that is, \[(A+I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+I\). We can do this through row reduction of the matrix \[A+I = \matrix{-25 & 56 \\ -12 & 27} - \matrix{-1 & 0 \\0& -1 }=\matrix{ -24 & 56 \\ -12 & 28}\] This can be done as follows:
\[\begin{aligned}
\matrix{-24&56\\-12&28\\}&\sim\matrix{1&-{{7}\over{3}}\\-12&28\\}&{\blue{\begin{array}{c}-{{1}\over{24}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{7}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) equals \(\left\{ r \cv{-7\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-7\\-3}\right\rangle\).
If possible, we avoided fractions in the solution.
\[ \begin{aligned}
\det(A-\lambda I) = \left\vert \begin{array}{cc} -25-\lambda & 56 \\ -12 & 27-\lambda \end{array} \right\vert &= (-25-\lambda)(27-\lambda)-56\cdot-12 \\
&= (-25-\lambda)(27-\lambda)+672 \\ &= \lambda^2-2\,\lambda-3
\end{aligned}\] To solve this quadratic equation, we can factor it in the following way: \[ \lambda^2-2\,\lambda-3 = (\lambda-3)(\lambda+1) \] So, the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -1\).
Let \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-25 & 56 \\ -12 & 27}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), this is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\). We can do this through row reduction of the matrix \[A-3 I = \matrix{-25 & 56 \\ -12 & 27} - \matrix{3 & 0 \\0& 3 }=\matrix{ -28 & 56 \\ -12 & 24}\] This can be done as follows:
\[\begin{aligned}
\matrix{-28&56\\-12&24\\}&\sim\matrix{1&-2\\-12&24\\}&{\blue{\begin{array}{c}-{{1}\over{28}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{2\\1} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{2\\1}\right\rangle\).
Let \(\lambda = -1\) be an eigenvalue of the matrix \[A=\matrix{-25 & 56 \\ -12 & 27}\] Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-1\vec{v}\), that is, \[(A+I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+I\). We can do this through row reduction of the matrix \[A+I = \matrix{-25 & 56 \\ -12 & 27} - \matrix{-1 & 0 \\0& -1 }=\matrix{ -24 & 56 \\ -12 & 28}\] This can be done as follows:
\[\begin{aligned}
\matrix{-24&56\\-12&28\\}&\sim\matrix{1&-{{7}\over{3}}\\-12&28\\}&{\blue{\begin{array}{c}-{{1}\over{24}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{7}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -1\) equals \(\left\{ r \cv{-7\\-3} \middle|\;r\in\mathbb R\right\}=\left\langle\cv{-7\\-3}\right\rangle\).
If possible, we avoided fractions in the solution.
Unlock full access