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 \[A=\matrix{5 & -6 \\ 3 & -4}\] 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 \[A-2 I = \matrix{5 & -6 \\ 3 & -4} - \matrix{2 & 0 \\0& 2 }=\matrix{ 3 & -6 \\ 3 & -6}\] This can be done as follows:
\[\begin{aligned}
\matrix{3&-6\\3&-6\\}&\sim\matrix{1&-2\\3&-6\\}&{\blue{\begin{array}{c}{{1}\over{3}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-3R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 2\) equals \(\left\{ r \cv{2\\1} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(2\) equals \(\left\langle\cv{2\\1}\right\rangle\)
We can do this through row reduction of the matrix \[A-2 I = \matrix{5 & -6 \\ 3 & -4} - \matrix{2 & 0 \\0& 2 }=\matrix{ 3 & -6 \\ 3 & -6}\] This can be done as follows:
\[\begin{aligned}
\matrix{3&-6\\3&-6\\}&\sim\matrix{1&-2\\3&-6\\}&{\blue{\begin{array}{c}{{1}\over{3}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-3R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 2\) equals \(\left\{ r \cv{2\\1} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(2\) equals \(\left\langle\cv{2\\1}\right\rangle\)
Unlock full access
