Question

Find the eigenvalues and the basis for the eigenspace in c^2 for the matrix [3,-3:3,3] I found eigenvalues of 3+3i and 3-3i, but I only get one eigenvector [1,i]

Answer 1

Then [1,-i] should be the other eigenvector, corresponding the eigenvalue 3+3i. When your matrix entries are real, and you matrix has complex eigenvalues, generally (not always), your eigenvectors are complex conjugates of each other.

Answer 2

Thanks!

Answer 3

We wish to determine the eigenvalues of $$\begin{pmatrix}3&-3\\3&3\end{pmatrix}$$ Let our matrix be \(\mathbf{A}\). By definition, an eigenvalue \(\lambda\) satisfies \(\mathbf{Ax}=\lambda\mathbf{x}\) for some (non-trivial) eigenvector \(\mathbf{x}\). Rearranging our equation, we have \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=0\). It follows \(\mathbf{A}-\lambda\mathbf{I}\) must be singular and therefore \(\det(\mathbf{A}-\lambda\mathbf{I})=0\).

We write our matrix \(\mathbf{A}-\lambda\mathbf{I}\) quickly:$$\begin{pmatrix}3-\lambda&-3\\3&3-\lambda\end{pmatrix}$$ Our determinant then yields our characteristic polynomial \((3-\lambda)^2-3(-3)=9-6\lambda+\lambda^2+18=\lambda^2-6\lambda+18\). Our equation is then the characteristic equation \(\lambda^2-6\lambda+18=0\) and its roots \(\lambda=\frac{6\pm\sqrt{6^2-4(9)}}{2}=3\pm\sqrt{-9}=3\pm3i\) is our eigenvalue.

Thus we know for some eigenvectors \(\mathbf{x}\) satisfy \(\mathbf{Ax}=(3+3i)\mathbf{x}\) or \(\mathbf{Ax}=(3-3i)\mathbf{x}\). So let $$\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}$$ and thus $$\mathbf{Ax}=\begin{pmatrix}3x-3y\\3x+3y\end{pmatrix}=(3+3i)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3x+3ix\\3y+3iy\end{pmatrix}$$ and therefore $$3x-3y=3x+3ix\\3x+3y=3y+3iy$$ so \(-3y=3ix\Rightarrow y=-ix\) and \(3x=3iy\Rightarrow x=iy\) and therefore \(k\begin{pmatrix}i\\1\end{pmatrix}\) for non-zero \(k\) is an eigenvector.

A similar approach instead focusing on the eigenvalue \(3-3i\) yields \(k\begin{pmatrix}-i\\1\end{pmatrix}\) for non-zero \(k\) is an eigenvector as well.

Answer 4

*are* our eigenvalue*s*