Two by two matrix example

Find the eigenvalues and eigenvectors for the matrix [latex]A=\left(\begin{array}{c}-2&3\\-4&5\end{array}\right)[/latex].

1. Start by expanding the characteristic polynomial of A: [latex]\det(\lambda I-A)=\det\left(\begin{array}{c}\lambda+2&-3\\4&\lambda-5\end{array}\right)[/latex][latex]=(\lambda+2)(\lambda-5)+12=\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2)[/latex].
2. Next, solve the characteristic equation of A: [latex]\det(\lambda I-A)=0[/latex], so [latex](\lambda-1)(\lambda-2)=0[/latex], so [latex]\lambda_1=1[/latex] and [latex]\lambda_2=2[/latex].
3. The eigenvectors corresponding to the eigenvalues are the solutions of [latex](\lambda I-A)\mathbf{x}=\mathbf{0}[/latex] for each eigenvalue. So, for [latex]\lambda_1=1[/latex] use Gauss-Jordan elimination to row reduce [latex](\lambda_1I-A)=\left(\begin{array}{c}1+2&-3\\4&1-5\end{array}\right)=\left(\begin{array}{c}3&-3\\4&-4\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&-1\\0&0\end{array}\right)[/latex].
4. Therefore, from the previous step, [latex]x-y=0[/latex], and if we set [latex]y=t[/latex] so that [latex]x=t[/latex], the eigenvector corresponding to [latex]\lambda_1=1[/latex] is [latex]\left(\begin{array}{c}1\\1\end{array}\right)[/latex].
5. Finally, for [latex]\lambda_2=2[/latex], use Gauss-Jordan elimination to row reduce [latex](\lambda_2I-A)=\left(\begin{array}{c}2+2&-3\\4&2-5\end{array}\right)=\left(\begin{array}{c}4&-3\\4&-3\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&-3/4\\0&0\end{array}\right)[/latex].
6. Therefore, from the previous step, [latex]x-(3/4)y=0[/latex], and if we set [latex]y=t[/latex] so that [latex]x=(3/4)t[/latex], the eigenvector corresponding to [latex]\lambda_2=2[/latex] is [latex]\left(\begin{array}{c}3/4\\1\end{array}\right)[/latex]. Equivalently, to avoid fractions we can write this eigenvector as [latex]\left(\begin{array}{c}3\\4\end{array}\right)[/latex].

Three by three matrix example with three distinct eigenvalues

Find the eigenvalues and eigenvectors for the matrix [latex]A=\left(\begin{array}{c}1&2&-1\\1&-1&2\\1&0&1\end{array}\right)[/latex].

1. Start by expanding the characteristic polynomial of A: [latex]\det(\lambda I-A)=\det\left(\begin{array}{c}\lambda-1&-2&1\\-1&\lambda+1&-2\\-1&0&\lambda-1\end{array}\right)[/latex][latex]=\ldots=(\lambda+1)\lambda(\lambda-2)[/latex].
2. Next, solve the characteristic equation of A: [latex]\det(\lambda I-A)=0[/latex], so [latex](\lambda+1)\lambda(\lambda-2)=0[/latex], so [latex]\lambda_1=-1[/latex], [latex]\lambda_2=0[/latex], and [latex]\lambda_3=2[/latex].
3. Next, use Gauss-Jordan elimination to row reduce [latex](\lambda_1I-A)=\left(\begin{array}{c}-2&-2&1\\-1&0&-2\\-1&0&-2\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&0&2\\0&1&-5/2\\0&0&0\end{array}\right)[/latex].
4. Therefore, from the previous step, [latex]x+2z=0[/latex] and [latex]y-(5/2)z=0[/latex]. If we set [latex]z=t[/latex] so that [latex]x=-2t[/latex] and [latex]y=(5/2)t[/latex], the eigenvector corresponding to [latex]\lambda_1=-1[/latex] is [latex]\left(\begin{array}{c}-2\\5/2\\1\end{array}\right)[/latex]. Equivalently, to avoid fractions we can write this eigenvector as [latex]\left(\begin{array}{c}-4\\5\\2\end{array}\right)[/latex].
5. Next, use Gauss-Jordan elimination to row reduce [latex](\lambda_2I-A)=\left(\begin{array}{c}-1&-2&1\\-1&1&-2\\-1&0&-1\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&0&1\\0&1&-1\\0&0&0\end{array}\right)[/latex].
6. Therefore, from the previous step, [latex]x+z=0[/latex] and [latex]y-z=0[/latex]. If we set [latex]z=t[/latex] so that [latex]x=-t[/latex] and [latex]y=t[/latex], the eigenvector corresponding to [latex]\lambda_2=0[/latex] is [latex]\left(\begin{array}{c}-1\\1\\1\end{array}\right)[/latex].
7. Finally, use Gauss-Jordan elimination to row reduce [latex](\lambda_3I-A)=\left(\begin{array}{c}1&-2&1\\-1&3&-2\\-1&0&1\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&0&-1\\0&1&-1\\0&0&0\end{array}\right)[/latex].
8. Therefore, from the previous step, [latex]x-z=0[/latex] and [latex]y-z=0[/latex]. If we set [latex]z=t[/latex] so that [latex]x=t[/latex] and [latex]y=t[/latex], the eigenvector corresponding to [latex]\lambda_3=2[/latex] is [latex]\left(\begin{array}{c}1\\1\\1\end{array}\right)[/latex].

Three by three matrix example with two distinct eigenvalues

Find the eigenvalues and eigenvectors for the matrix [latex]A=\left(\begin{array}{c}-2&2&-1\\2&1&-2\\-3&-6&0\end{array}\right)[/latex].

1. Start by expanding the characteristic polynomial of A: [latex]\det(\lambda I-A)=\det\left(\begin{array}{c}\lambda+2&-2&1\\-2&\lambda-1&2\\3&6&\lambda\end{array}\right)[/latex][latex]=\ldots=(\lambda+3)^2(\lambda-5)[/latex].
2. Next, solve the characteristic equation of A: [latex]\det(\lambda I-A)=0[/latex], so [latex](\lambda+3)^2(\lambda-5)=0[/latex], so [latex]\lambda_1=-3[/latex] (with multiplicity 2) and [latex]\lambda_2=5[/latex].
3. Next, use Gauss-Jordan elimination to row reduce [latex](\lambda_1I-A)=\left(\begin{array}{c}-1&-2&1\\-2&-4&2\\3&6&-3\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&2&-1\\0&0&0\\0&0&0\end{array}\right)[/latex].
4. Therefore, from the previous step, [latex]x+2y-z=0[/latex]. If we set [latex]y=s[/latex] and [latex]z=t[/latex] so that [latex]x=-2s+t[/latex], the eigenvectors corresponding to [latex]\lambda_1=-3[/latex] are [latex]\left(\begin{array}{c}-2\\1\\0\end{array}\right)[/latex] and [latex]\left(\begin{array}{c}1\\0\\1\end{array}\right)[/latex].
5. Next, use Gauss-Jordan elimination to row reduce [latex](\lambda_2I-A)=\left(\begin{array}{c}7&-2&1\\-2&4&2\\3&6&5\end{array}\right)[/latex] to [latex]\left(\begin{array}{c}1&0&1/3\\0&1&2/3\\0&0&0\end{array}\right)[/latex].
6. Therefore, from the previous step, [latex]x+(1/3)z=0[/latex] and [latex]y+(2/3)z=0[/latex]. If we set [latex]z=t[/latex] so that [latex]x=-(1/3)t[/latex] and [latex]y=-(2/3)t[/latex], the eigenvector corresponding to [latex]\lambda_2=5[/latex] is [latex]\left(\begin{array}{c}-1/3\\-2/3\\1\end{array}\right)[/latex]. Equivalently, to avoid fractions we can write this eigenvector as [latex]\left(\begin{array}{c}-1\\-2\\3\end{array}\right)[/latex].