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 \[A=\matrix{8 & -5 \\ 10 & -7}\] 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 \[A+2 I = \matrix{8 & -5 \\ 10 & -7} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 10 & -5 \\ 10 & -5}\] This can be done as follows:
\[\begin{aligned}
\matrix{10&-5\\10&-5\\}&\sim\matrix{1&-{{1}\over{2}}\\10&-5\\}&{\blue{\begin{array}{c}{{1}\over{10}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-10R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{1\\2} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(-2\) equals \(\left\langle\cv{1\\2}\right\rangle\)
We can do this through row reduction of the matrix \[A+2 I = \matrix{8 & -5 \\ 10 & -7} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 10 & -5 \\ 10 & -5}\] This can be done as follows:
\[\begin{aligned}
\matrix{10&-5\\10&-5\\}&\sim\matrix{1&-{{1}\over{2}}\\10&-5\\}&{\blue{\begin{array}{c}{{1}\over{10}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-10R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -2\) equals \(\left\{ r \cv{1\\2} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(-2\) equals \(\left\langle\cv{1\\2}\right\rangle\)
Unlock full access
