Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so.
\(B = \begin{bmatrix}-12 & 30 \\ -5 & 13\end{bmatrix}\)
To get the eigenvalues, we start by multiplying the matrix B’s identity matrix by \(\lambda\).
\(det(B-\lambda I) = 0\)
The identity matrix will be 2x2, which becomes:
\(det(B - \lambda \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}) = 0\)
Multiplying the identity matrix by the scalar \(\lambda\) results in
\(det(B - \begin{bmatrix}\lambda & 0 \\ 0 & \lambda\end{bmatrix}) = 0\)
We can substitute in the value of B, now.
\(det(\begin{bmatrix}-12 & 30 \\ -5 & 13\end{bmatrix} - \begin{bmatrix}\lambda & 0 \\ 0 & \lambda\end{bmatrix}) = 0\)
This can also be written as:
\(det(\begin{bmatrix}-12-\lambda & 30 \\ -5 & 13-\lambda\end{bmatrix}) = 0\)
From here, we can reduce the problem into a linear equation.
\((-12\times-\lambda)\times(13\times-\lambda) - (30)\times(-5)\)
If we use the FOIL method, this becomes…
\((-12\times13)+(-12\times-\lambda)+(13\times-\lambda)+(\lambda\times\lambda)-(30)\times(-5)\)
Reducing this, we get:
\(-156+12\lambda-13\lambda+\lambda^{2}+150\)
Which can be reduced and rewritten as…
\(\lambda^{2}-\lambda-6\)
Using polynomial factorization, we get
\((\lambda-3)(\lambda+2) = 0\)
From this, we can determine that the eigenvalues of B are 3 and -2.
Since the eigenvalues are distinct and appear only once, the algebraic multiplicities of 3 and -2 are 1 (\(\alpha_{B}(3) = 1\) and \(\alpha_{B}(-2) = 1\)).
For the eigenspaces, we do the following:
\(\varepsilon_{B}(\lambda) = B - \lambda I\)
We have two separate eigenvalues, so to start with, we’ll use 3.
\(\varepsilon_{B}(\lambda) = B - 3I\)
\(\varepsilon_{B}(\lambda) = B - 3 \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = B - \begin{bmatrix}3 & 0 \\ 0 & 3 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12 & 30 \\ -5 & 13\end{bmatrix} - \begin{bmatrix}3 & 0 \\ 0 & 3 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12 & 30 \\ -5 & 13\end{bmatrix} - \begin{bmatrix}3 & 0 \\ 0 & 3 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12-3 & 30 \\ -5 & 13-3\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15 & 30 \\ -5 & 10\end{bmatrix}\)
We then put it into reduced row echelon form.
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15 & 30 \\ -5\times3 & 10\times3\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15 & 30 \\ -15 & 30\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15 & 30 \\ -15- -15 & 30-30\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15 & 30 \\ -15+15 & 30-30\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15 & 30 \\ 0 & 0\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-15\div-15 & 30\div-15 \\ 0 & 0\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}1 & -2 \\ 0 & 0\end{bmatrix}\)
Written as an equation, this is…
\(x -2y = 0\)
or
\(x = 2y\)
Which can be written as…
\(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2y \\ y\end{pmatrix} = \begin{pmatrix}2 \\ 1\end{pmatrix}\)
So when the eigenvalue is 3, the eigenspace is \(\begin{pmatrix}2 \\ 1\end{pmatrix}\)
\(\varepsilon_{B}(\lambda) = B - -2I\)
\(\varepsilon_{B}(\lambda) = B - -2 \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = B - \begin{bmatrix}-2 & 0 \\ 0 & -2 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12 & 30 \\ -5 & 13\end{bmatrix} - \begin{bmatrix}-2 & 0 \\ 0 & -2 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12 & 30 \\ -5 & 13\end{bmatrix} - \begin{bmatrix}-2 & 0 \\ 0 & -2 \end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12--2 & 30 \\ -5 & 13--2\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-12+2 & 30 \\ -5 & 13+2\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10 & 30 \\ -5 & 15\end{bmatrix}\)
We then put it into reduced row echelon form.
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10 & 30 \\ -5\times2 & 15\times2\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10 & 30 \\ -10 & 30\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10 & 30 \\ -10--10 & 30-30\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10 & 30 \\ -10+10 & 30-30\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10 & 30 \\ 0 & 0\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}-10\div-10 & 30\div-10 \\ 0 & 0\end{bmatrix}\)
\(\varepsilon_{B}(\lambda) = \begin{bmatrix}1 & -3 \\ 0 & 0\end{bmatrix}\)
Written as an equation, this is…
\(x -3y = 0\)
or
\(x = 3y\)
Which can be written as…
\(\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}3y \\ y\end{pmatrix} = \begin{pmatrix}3 \\ 1\end{pmatrix}\)
So when the eigenvalue is -2, the eigenspace is \(\begin{pmatrix}3 \\ 1\end{pmatrix}\)
\(\varepsilon_{B}(\lambda) = (3: \begin{pmatrix}2 \\ 1\end{pmatrix}, -2:\begin{pmatrix}3 \\ 1\end{pmatrix})\)
Since the eigendspaces are one dimensional, the geometric multiplicities of 3 and -2 are 1 (\(\gamma_{B}(3) = 1\) and \(\gamma_{B}(-2) = 1\)).
\(\lambda = (3, -2)\)
\(\alpha_{B} = (3: 1, -2: 1)\)
\(\varepsilon_{B}(\lambda) = (3: \begin{pmatrix}2 \\ 1\end{pmatrix}, -2:\begin{pmatrix}3 \\ 1\end{pmatrix})\)
\(\gamma_{B} = (3: 1, -2: 1)\)