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 \[A=\matrix{-2 & 0 \\ 14 & 5}\] 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 \[A+2 I = \matrix{-2 & 0 \\ 14 & 5} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 0 & 0 \\ 14 & 7}\] This can be done as follows:
\[\begin{aligned}
\matrix{0&0\\14&7\\}&\sim\matrix{14&7\\0&0\\}&{\blue{\begin{array}{c}R_2\\R_1\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}{{1}\over{14}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] 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 \[A+2 I = \matrix{-2 & 0 \\ 14 & 5} - \matrix{-2 & 0 \\0& -2 }=\matrix{ 0 & 0 \\ 14 & 7}\] This can be done as follows:
\[\begin{aligned}
\matrix{0&0\\14&7\\}&\sim\matrix{14&7\\0&0\\}&{\blue{\begin{array}{c}R_2\\R_1\end{array}}}\\\\ &\sim\matrix{1&{{1}\over{2}}\\0&0\\}&{\blue{\begin{array}{c}{{1}\over{14}}R_1\\\phantom{x}\end{array}}} \end{aligned}\] 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 \(\left\langle\cv{1\\-2}\right\rangle\)
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