For matrix \(A = \left[\begin{array}{cc} 0 & 4 & -1 & 1\\-2 & 6 & -1 & 1\\ -2 & 8 & -1 & -1\\-2 & 8 & -3 & 1\end{array}\right]\)

the characteristic polynomial of \(A\) is \(p^{A}(x) = (x + 2)(x − 2)^{2}(x − 4)\)

Find the eigenvalues and corresponding eigenspaces of A.

Following the steps shown in this week’s Khan Acad. tutorials:

To find the eigenvalues, we solve for any root \(x\) that makes the polynomial equal 0.

In this case, the 3 eigenvalues are -2, 2, and 4.

Then to find the eigenspaces corresponding to each eigenvalue,
we calculate the associated eigenvectors for each eigenvalue.

First set of eigenvectors, a.k.a. “first eigenspace”…

\(x_{1} = -2\) :

\((x_{1}\cdot I_{4} - A)\cdot\overrightarrow{v} = \overrightarrow{0}\), meaning \(\left(\left[\begin{array}{cc} -2 & 0 & 0 & 0\\0 & -2 & 0 & 0\\ 0 & 0 & -2 & 0\\0 & 0 & 0 & -2\end{array}\right] - \left[\begin{array}{cc} 0 & 4 & -1 & 1\\-2 & 6 & -1 & 1\\ -2 & 8 & -1 & -1\\-2 & 8 & -3 & 1\end{array}\right]\right)\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

which is \(\left[\begin{array}{cc} -2 & -4 & 1 & -1\\2 & -8 & 1 & -1\\ 2 & -8 & -1 & 1\\2 & -8 & 3 & -3\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

row-reducing to \(\left[\begin{array}{cc} -6 & 0 & 0 & 0\\0 & -6 & 0 & 0\\ 0 & 0 & 1 & -1\\0 & 0 & 0 & 0\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

making \(v_{1} = v_{2} = 0\) and \(v_{3} = v_{4}\) so that the eigenspace for \(x=-2\) is
the set of all vectors that are a scalar multiple of \(\overrightarrow{v}=\left[\begin{array}{cc} 0\\0\\1\\1\end{array}\right]\)

Now for the second eigenspace/set of eigenvectors…

\(x_{2} = 2\) :

\((x_{2}\cdot I_{4} - A)\cdot\overrightarrow{v} = \overrightarrow{0}\), meaning \(\left(\left[\begin{array}{cc} 2 & 0 & 0 & 0\\0 & 2 & 0 & 0\\ 0 & 0 & 2 & 0\\0 & 0 & 0 & 2\end{array}\right] - \left[\begin{array}{cc} 0 & 4 & -1 & 1\\-2 & 6 & -1 & 1\\ -2 & 8 & -1 & -1\\-2 & 8 & -3 & 1\end{array}\right]\right)\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

which is \(\left[\begin{array}{cc} 2 & -4 & 1 & -1\\2 & -4 & 1 & -1\\ 2 & -8 & 3 & 1\\2 & -8 & 3 & 1\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

row-reducing to \(\left[\begin{array}{cc} 2 & 0 & -1 & -3\\0 & 0 & 0 & 0\\ 0 & 2 & -1 & -1\\0 & 0 & 0 & 0\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

The top row tells us that \(v_{1} = \frac{1}{2}v_{3} + \frac{3}{2}v_{4}\)

and the third row gives us \(v_{2} = \frac{1}{2}v_{3} + \frac{1}{2}v_{4}\)

If we set \(v_{3} =\) any scalar \(\alpha\) and \(v_{4}\) = any scalar \(\beta\) then we get

the eigenspace for \(x=2\), the set of all vectors that are the sum of:

\(\alpha \cdot\left[\begin{array}{cc} \frac{1}{2}\\\frac{1}{2}\\1\\0\end{array}\right] + \beta \cdot \left[\begin{array}{cc} \frac{3}{2}\\\frac{1}{2}\\0\\1\end{array}\right]\)

and third and final eigenspace…

\(x_{3} = 4\) :

\((x_{3}\cdot I_{4} - A)\cdot\overrightarrow{v} = \overrightarrow{0}\), meaning \(\left(\left[\begin{array}{cc} 4 & 0 & 0 & 0\\0 & 4 & 0 & 0\\ 0 & 0 & 4 & 0\\0 & 0 & 0 & 4\end{array}\right] - \left[\begin{array}{cc} 0 & 4 & -1 & 1\\-2 & 6 & -1 & 1\\ -2 & 8 & -1 & -1\\-2 & 8 & -3 & 1\end{array}\right]\right)\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

which is \(\left[\begin{array}{cc} 4 & -4 & 1 & -1\\2 & -2 & 1 & -1\\ 2 & -8 & 5 & 1\\2 & -8 & 3 & 3\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

row-reducing to \(\left[\begin{array}{cc} 1 & 0 & 0 & -1\\0 & 1 & 0 & -1\\ 0 & 0 & -1 & 1\\0 & 0 & 0 & 0\end{array}\right]\cdot\left[\begin{array}{cc} v_{1}\\v_{2}\\v_{3}\\v_{4}\end{array}\right] = \left[\begin{array}{cc} 0\\0\\0\\0\end{array}\right]\)

No matter what you set \(v_{4}\) to be, you can see that all the others have to equal \(v_{4}\)

which means that the eigenspace for the eigenvalue \(x = 4\) is any scalar multiple of \(\left[\begin{array}{cc} 1\\1\\1\\1\end{array}\right]\)