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 \(\lambda = 3\) be an eigenvalue of the matrix \[A=\matrix{-39 & -70 \\ 21 & 38}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=3\vec{v}\), that is, \[(A-3 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A-3 I\).
We can do this through row reduction of the matrix \[A-3 I = \matrix{-39 & -70 \\ 21 & 38} - \matrix{3 & 0 \\0& 3 }=\matrix{ -42 & -70 \\ 21 & 35}\] This can be done as follows:
\[\begin{aligned}
\matrix{-42&-70\\21&35\\}&\sim\matrix{1&{{5}\over{3}}\\21&35\\}&{\blue{\begin{array}{c}-{{1}\over{42}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{5}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{-5\\3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(3\) equals \(\left\langle\cv{-5\\3}\right\rangle\)
We can do this through row reduction of the matrix \[A-3 I = \matrix{-39 & -70 \\ 21 & 38} - \matrix{3 & 0 \\0& 3 }=\matrix{ -42 & -70 \\ 21 & 35}\] This can be done as follows:
\[\begin{aligned}
\matrix{-42&-70\\21&35\\}&\sim\matrix{1&{{5}\over{3}}\\21&35\\}&{\blue{\begin{array}{c}-{{1}\over{42}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&{{5}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = 3\) equals \(\left\{ r \cv{-5\\3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.
In other words, the eigenspace of eigenvalue \(3\) equals \(\left\langle\cv{-5\\3}\right\rangle\)
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