AI HL · AHL 5
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Eigenvalues and eigenvectors
AHL 5.6
Menu > Matrix & Vector > Advanced · eigVl(a) and eigVc(a)
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Drag v (orange). The teal arrow is Mv. Find the directions where Mv lies along the same line as v: those v are the eigenvectors.
Dragged close to an eigenvector line, v clicks onto it. Select v (Tab) and use the arrow keys for small steps.
Find the eigenvalues and eigenvectors of \(M = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(\det\begin{pmatrix}4-\lambda & 1 \\ 2 & 3-\lambda\end{pmatrix} = 0\)
Take \(\lambda\) off the leading diagonal.
\((4 - \lambda)(3 - \lambda) - 2 = 0\), so \(\lambda^2 - 7\lambda + 10 = 0\), \(\lambda = 5\) or \(2\)
Check: \(5 + 2 = 7 = 4 + 3\).
\(4x + y = 5x\), so \(y = x\)
Choose \(x = 1\): the eigenvector is \(\begin{pmatrix}1 \\ 1\end{pmatrix}\).
\(4x + y = 2x\), so \(y = -2x\)
Choose \(x = 1\): the eigenvector is \(\begin{pmatrix}1 \\ -2\end{pmatrix}\).
\(\begin{pmatrix}1 \\ 3\end{pmatrix}\) is an eigenvector of \(\begin{pmatrix}5 & 1 \\ 6 & a\end{pmatrix}\). Find \(a\) and both eigenvalues.
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}5 & 1 \\ 6 & a\end{pmatrix}\begin{pmatrix}1 \\ 3\end{pmatrix} = \begin{pmatrix}8 \\ 6+3a\end{pmatrix}\)
Matrix first, vector second.
\(8 = \lambda(1)\), so \(\lambda = 8\)
Top row first.
\(6 + 3a = 8(3)\), so \(a = 6\)
The bottom row must be 8 times the bottom of the eigenvector.
\(5 + 6 - 8 = 3\)
The eigenvalues add up to the leading diagonal, 11.
Show that \(\begin{pmatrix}0 & -2 \\ 2 & 0\end{pmatrix}\) has no real eigenvalues.
Have a go first. Then open the working one line at a time.
\((-\lambda)(-\lambda) - (-2)(2) = 0\), so \(\lambda^2 + 4 = 0\)
Take \(\lambda\) off the leading diagonal, then find the determinant.
\(\lambda^2 = -4\) has no real solutions
No vector keeps its direction: the matrix turns every vector through 90° and doubles it.
\(4\) is an eigenvalue of \(M = \begin{pmatrix}k & 3 \\ 2 & 1\end{pmatrix}\). Find \(k\), then find the other eigenvalue and the eigenvector for \(\lambda = 4\).
Have a go first. Then open the working one line at a time.
\(\det\begin{pmatrix}k-\lambda & 3 \\ 2 & 1-\lambda\end{pmatrix} = (k-\lambda)(1-\lambda) - 6 = 0\)
The eigenvalue 4 must make this true.
\((k-4)(1-4) - 6 = 0\)
So \(-3(k-4) = 6\), \(k - 4 = -2\), \(k = 2\).
\(\lambda^2 - 3\lambda - 4 = 0,\quad \lambda = 4 \text{ or } -1\)
The sum of the eigenvalues is the sum of the leading diagonal: \(2 + 1 = 3\).
\((2-4)x + 3y = 0\)
So \(y = \tfrac{2}{3}x\). Choose \(x = 3\): the eigenvector is \(\begin{pmatrix}3 \\ 2\end{pmatrix}\).