Eigenvalues and eigenvectors: Eigenvalues and eigenvectors
Computing eigenvectors for a given eigenvalue
We start with examples to compute the eigenspace of an eigenvalue of a matrix.
Let \(\lambda = -2\) be an eigenvalue of the matrix \[A=\matrix{13 & -6 \\ 45 & -20}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-2\vec{v}\), that is, \[(A+2 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+2 I\).
We can do this through row reduction of the matrix \[A+2 I = \matrix{13 & -6 \\ 45 & -20} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 15 & -6 \\ 45 & -18}\] This can be done as follows:
\[\begin{aligned}
\matrix{15&-6\\45&-18\\}&\sim\matrix{1&-{{2}\over{5}}\\45&-18\\}&{\blue{\begin{array}{c}{{1}\over{15}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{2}\over{5}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-45R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{-2\\-5} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(-2\) equals \(\left\langle\cv{-2\\-5}\right\rangle\)
We can do this through row reduction of the matrix \[A+2 I = \matrix{13 & -6 \\ 45 & -20} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 15 & -6 \\ 45 & -18}\] This can be done as follows:
\[\begin{aligned}
\matrix{15&-6\\45&-18\\}&\sim\matrix{1&-{{2}\over{5}}\\45&-18\\}&{\blue{\begin{array}{c}{{1}\over{15}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{2}\over{5}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-45R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{-2\\-5} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(-2\) equals \(\left\langle\cv{-2\\-5}\right\rangle\)
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