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 \(\lambda = -5\) be an eigenvalue of the matrix \[A=\matrix{-32 & -18 \\ 54 & 31}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-5\vec{v}\), that is, \[(A+5 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+5 I\).
We can do this through row reduction of the matrix \[A+5 I = \matrix{-32 & -18 \\ 54 & 31} - \matrix{-5 & 0 \\0& -5 }=\matrix{ -27 & -18 \\ 54 & 36}\] This can be done as follows:
\[\begin{aligned}
\matrix{-27&-18\\54&36\\}&\sim\matrix{1&{{2}\over{3}}\\54&36\\}&{\blue{\begin{array}{c}-{{1}\over{27}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{2}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-54R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{2\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(-5\) equals \(\left\langle\cv{2\\-3}\right\rangle\)
We can do this through row reduction of the matrix \[A+5 I = \matrix{-32 & -18 \\ 54 & 31} - \matrix{-5 & 0 \\0& -5 }=\matrix{ -27 & -18 \\ 54 & 36}\] This can be done as follows:
\[\begin{aligned}
\matrix{-27&-18\\54&36\\}&\sim\matrix{1&{{2}\over{3}}\\54&36\\}&{\blue{\begin{array}{c}-{{1}\over{27}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{2}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-54R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -5\) equals \(\left\{ r \cv{2\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(-5\) equals \(\left\langle\cv{2\\-3}\right\rangle\)
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