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 \(\lambda = 1\) be an eigenvalue of the matrix \[A=\matrix{13 & 8 \\ -12 & -7}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=1\vec{v}\), that is, \[(A-I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-I\).
We can do this through row reduction of the matrix \[A-I = \matrix{13 & 8 \\ -12 & -7} - \matrix{1 & 0 \\0& 1 }=\matrix{ 12 & 8 \\ -12 & -8}\] This can be done as follows:
\[\begin{aligned}
\matrix{12&8\\-12&-8\\}&\sim\matrix{1&{{2}\over{3}}\\-12&-8\\}&{\blue{\begin{array}{c}{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{2}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) 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 \(1\) equals \(\left\langle\cv{-2\\3}\right\rangle\)
We can do this through row reduction of the matrix \[A-I = \matrix{13 & 8 \\ -12 & -7} - \matrix{1 & 0 \\0& 1 }=\matrix{ 12 & 8 \\ -12 & -8}\] This can be done as follows:
\[\begin{aligned}
\matrix{12&8\\-12&-8\\}&\sim\matrix{1&{{2}\over{3}}\\-12&-8\\}&{\blue{\begin{array}{c}{{1}\over{12}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{2}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+12R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 1\) 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 \(1\) equals \(\left\langle\cv{-2\\3}\right\rangle\)
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