Linear Algebra – Eigenvalues and eigenvectors – Nonreal eigenvalues

Until now we have only allowed real numbers to be eigenvalues of a matrix. However, it is quite easy to extend the definition to complex numbers. Then a consequence is that the corresponding eigenvectors also have complex coordinates. Instead of vectors in \(\mathbb{R}^n\) we then consider vectors in \(\mathbb{C}^n\):

Definition: A vector \(\mathbf{x}\in\mathbb{C}^n\) with \(\mathbf{x}\neq\mathbf{0}\) is called an eigenvector of an \((n\times n)\)-matrix \(A\) if \(A\mathbf{x}=\lambda\mathbf{x}\) for certain \(\lambda\in\mathbb{C}\). Such a number \(\lambda\in\mathbb{C}\) is called an eigenvalue of \(A\). A vector \(\mathbf{x}\neq\mathbf{0}\) with \(A\mathbf{x}=\lambda\mathbf{x}\) is called an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\).

We will only consider real matrices (with real entries). This implies that the characteristic polynomial only has real coefficients and that nonreal roots (so: eigenvalues) can only occur in complex conjugate pairs.

Example: Suppose that \(A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), then we have: \(\det(A-\lambda I)=\begin{vmatrix}-\lambda&-1\\1&-\lambda\end{vmatrix} =\lambda^2+1\). So the (complex) eigenvalues of \(A\) are \(\lambda_1=i\) and \(\lambda_2=-i\). For the corresponding eigenvectors we obtain:

\[\lambda_1=i:\quad\begin{pmatrix}-i&-1\\1&-i\end{pmatrix}=\begin{pmatrix}1&-i\\0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_i=\text{Span}\left\{\begin{pmatrix}i\\1\end{pmatrix}\right\}\]

and

\[\lambda_2=-i:\quad\begin{pmatrix}i&-1\\1&i\end{pmatrix}=\begin{pmatrix}1&i\\0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_{-i}=\text{Span}\left\{\begin{pmatrix}-i\\1\end{pmatrix}\right\}.\]

Since \(A\) has no real eigenvalues, \(A\) is not diagonizable.

Definition: If \(\mathbf{x}\in\mathbb{C}^n\) then \(\overline{\mathbf{x}}\in\mathbb{C}^n\) is the complex conjugate of \(\mathbf{x}\), the vector for which all coordinates are the complex conjugates of the corresponding coordinates of \(\mathbf{x}\). Eech vector \(\mathbf{x}\in\mathbb{C}^n\) can be written in the form \(\text{Re}(\mathbf{x})+i\,\text{Im}(\mathbf{x})\), where \(\text{Re}(\mathbf{x})\) and \(\text{Im}(\mathbf{x})\) are vectors in \(\mathbb{R}^n\); respectively, the real and the imaginary part of the vector \(\mathbf{x}\).
If \(A\) is any \((m\times n)\)-matrix with complex entries, then \(\overline{A}\) is the matrix that arises from \(A\) by replacing each entry by its complex conjugate.

Now if \(A\) is a real \((n\times n)\)-matrix, then we have that \(A=\overline{A}\) and therefore: \(\overline{A\mathbf{x}}=\overline{A}\overline{\mathbf{x}}=A\overline{\mathbf{x}}\). Hence: if \(A\mathbf{x}=\lambda\mathbf{x}\), then we have that \(A\overline{\mathbf{x}}=\overline{A\mathbf{x}}=\overline{\lambda\mathbf{x}}=\overline{\lambda}\overline{\mathbf{x}}\). So: if \(\mathbf{x}\in\mathbb{C}^n\) is an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\in\mathbb{C}\), then the complex conjugate \(\overline{\mathbf{x}}\in\mathbb{C}^n\) is an eigenvector of \(A\) as well and corresponding to the eigenvalue \(\overline{\lambda}\in\mathbb{C}\). Hence, real matrices can only have nonreal eigenvalues occurring in complex conjugate pairs and the corresponding eigenvectors are complex conjugates as well.

Example: Suppose that \(A=\begin{pmatrix}5&-2\\1&3\end{pmatrix}\), then we have: \(\det(A-\lambda I)=\begin{vmatrix}5-\lambda&-2\\1&3-\lambda\end{vmatrix} =\lambda^2-8\lambda+17=(\lambda-4)^2+1\). So the (complex) eigenvalues are \(\lambda=4\pm i\). For the computation of the eigenvectors we choose one of the two complex conjugate eigenvalues:

\[\lambda=4+i:\quad\begin{pmatrix}1-i&-2\\1&-1-i\end{pmatrix}.\]

Since \(\lambda=4+i\) is an eigenvalue, we know that this matrix has only one pivot position. So, there is no need to check this by row reduction (which is a tedious computation with complex numbers). However, we just determine a solution by looking at the first or the second row. All other eigenvectors will be multiples of that solution. Hence:

\[\text{E}_{4+i}=\text{Span}\left\{\begin{pmatrix}2\\1-i\end{pmatrix}\right\}\quad\text{or}\quad\text{E}_{4+i}=\text{Span}\left\{\begin{pmatrix}1+i\\1\end{pmatrix}\right\}.\]

The eigenvectors of \(A\) corresponding to the eigenvalue \(\lambda=4-i\) now easily follow by taking the complex conjugates (everywhere replace \(i\) by \(−i\)):

\[\text{E}_{4ii}=\text{Span}\left\{\begin{pmatrix}2\\1+i\end{pmatrix}\right\}\quad\text{or}\quad\text{E}_{4-i}=\text{Span}\left\{\begin{pmatrix}1-i\\1\end{pmatrix}\right\}.\]

It is not always that easy. However, we try to minimize the computations. Many computations involving complex numbers might easily give rise to difficulties (mistakes).

Example: Suppose that \(A=\begin{pmatrix}1&2&2\\-1&3&3\\0&-2&-1\end{pmatrix}\), then we have:

\begin{align*} \det(A-\lambda I)&=\begin{vmatrix}1-\lambda&2&2\\-1&3-\lambda&3\\0&-2&-1-\lambda\end{vmatrix}=(1-\lambda)\begin{vmatrix}3-\lambda&3\\-2&-1-\lambda\end{vmatrix} +\begin{vmatrix}2&2\\-2&-1-\lambda\end{vmatrix}\\[2.5mm] &=(1-\lambda)(\lambda^2-2\lambda+3)-2-2\lambda+4=-\lambda^3+3\lambda^2-7\lambda+5. \end{align*}

Now we should try to factorize this characteristic polynomial in (complex) factors. A third-degree polynomial (with real coefficients) can have at most two nonreal roots, because these can only occur in complex conjugate pairs. This implies that there is at least one real zero. By trial and error we find that \(\lambda=1\) is an eigenvalue. Hence:

\[\det(A-\lambda I)=-(\lambda^3-3\lambda^2+7\lambda-5)=-(\lambda-1)(\lambda^2-2\lambda+5)=-(\lambda-1)\left[(\lambda-1)^2+4\right].\]

So the other two (complex) eigenvalues are: \(\lambda=1\pm2i\).

Sometimes it might help to compute the determinant \(|A − \lambda I|\) by using a clever row reduction, such that the characteristic polynomial will be factorized more or less ’automatically’. However, it is not always easy to find the most clever elementary row operations:

\begin{align*} |A-\lambda I|&=\begin{vmatrix}1-\lambda&2&2\\-1&3-\lambda&3\\0&-2&-1-\lambda\end{vmatrix}=\begin{vmatrix}1-\lambda&0&1-\lambda\\-1&3-\lambda&3\\0&-2&-1-\lambda\end{vmatrix} =\begin{vmatrix}1-\lambda&0&0\\-1&3-\lambda&4\\0&-2&-1-\lambda\end{vmatrix}\\[2.5mm] &=(1-\lambda)\begin{vmatrix}3-\lambda&4\\-2&-1-\lambda\end{vmatrix}=(1-\lambda)(\lambda^2-2\lambda+5)=-(\lambda-1)\left[(\lambda-1)^2+4\right]. \end{align*}

The determination of the eigenspace \(\text{E}_1\) corresponding to the eigenvalue \(\lambda=1\) is as usual:

\[\lambda=1:\quad\begin{pmatrix}0&2&2\\-1&2&3\\0&-2&-2\end{pmatrix}\sim\begin{pmatrix}0&1&1\\-1&0&1\\0&0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_1=\text{Span}\left\{\begin{pmatrix}1\\-1\\1\end{pmatrix}\right\}.\]

For the eigenspaces corresponding to \(\lambda=1\pm2i\) we proceed as follows. First we choose one of the two eigenvalues and try to find one corresponding eigenvector using a minimal number of computations. All other eigenvectors are multiples of that one. The eigenvectors corresponding to the other eigenvalues are the complex conjugates. Hence:

\[\lambda=1+2i:\quad\begin{pmatrix}-2i&2&2\\-1&2-2i&3\\0&-2&-2-2i\end{pmatrix}\sim\begin{pmatrix}-i&1&1\\-1&2-2i&3\\0&1&1+i\end{pmatrix}.\]

Using a minimal number of simplifications (and sometimes a simple elementary row operation) we are able to find the eigenvectors. Look at the first and the last row:

\[\left\{\begin{array}{r}-ix_1+x_2+x_3=0\\[2.5mm]x_2+(1+i)x_3=0\end{array}\right.\quad\Longrightarrow\quad x_2=1+i,\quad x_3=-1,\quad ix_1=x_2+x_3=i\quad\Longrightarrow\quad x_1=1.\]

Now we easily obtain:

\[\text{E}_{1+2i}=\text{Span}\left\{\begin{pmatrix}1\\1+i\\-1\end{pmatrix}\right\}\quad\text{and therefore}\quad\text{E}_{1-2i}=\text{Span}\left\{\begin{pmatrix}1\\1-i\\-1\end{pmatrix}\right\}.\]

For complex numbers \(z=x+iy\) with \(x,y\in\mathbb{R}\) we also know the (polar) form \(z=re^{i\varphi}\) with \(r=|z|=\sqrt{x^2+y^2}\) and \(\varphi=\text{arg}(z)\). In matrix notation this can be written as:

\[C=\begin{pmatrix}x&-y\\y&x\end{pmatrix}=r\begin{pmatrix}\frac{x}{r}&-\frac{y}{r}\\\frac{y}{r}&\frac{x}{r}\end{pmatrix} =\begin{pmatrix}r&0\\0&r\end{pmatrix}\begin{pmatrix}\cos(\varphi)&-\sin(\varphi)\\\sin(\varphi)&\cos(\varphi)\end{pmatrix}.\]

The eigenvalues of the matrix \(C\) are \(\lambda=x\pm iy\). Check this!

The matrix \(\begin{pmatrix}\cos(\varphi)&-\sin(\varphi)\\\sin(\varphi)&\cos(\varphi)\end{pmatrix}\) is the standard matrix of the rotation about the origin through an angle \(\varphi\) in the positive direction (counterclockwise). Since the multiplication by the matrix \(\begin{pmatrix}r&0\\0&r\end{pmatrix}\) is a simple scalar multiplication by a scalar \(r=|\lambda|\) we have that the linear transformation \(T:\mathbb{R}^2\to\mathbb{R}^2\) with \(T(\mathbf{x})=C\mathbf{x}\) denotes a rotation about the origin through an angle \(\varphi\) (in positive direction), followed by a scalar multiplication (scaling) by a scalar \(r=|\lambda|\). Somewhat more general we have:

Theorem: If \(A\) is a \((2\times2)\)-matrix with a (complex) eigenvalue \(\lambda=x-iy\) with \(x,y\in\mathbb{R}\) and \(y\neq0\), then we have:

\[A=PCP^{-1}\quad\text{with}\quad C=\begin{pmatrix}x&-y\\y&x\end{pmatrix}\quad\text{and}\quad P=\Bigg(\text{Re}(\mathbf{v})\;\text{Im}(\mathbf{v})\Bigg),\]

where \(\mathbf{v}\in\mathbb{C}^2\) is an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda=x-iy\).

Example: Earlier we have seen that the eigenvalues of the matrix \(A=\begin{pmatrix}5&-2\\1&3\end{pmatrix}\) are equal to \(\lambda=4\pm i\). We have also seen that an eigenvector corresponding to \(\lambda=4-i\) is for instance equal to \(\mathbf{v}=\begin{pmatrix}1-i\\1\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}+i\begin{pmatrix}-1\\0\end{pmatrix}\). Dus:

\[\text{Re}(\mathbf{v})=\begin{pmatrix}1\\1\end{pmatrix}\quad\text{and}\quad\text{Im}(\mathbf{v})=\begin{pmatrix}-1\\0\end{pmatrix} \quad\Longrightarrow\quad P=\begin{pmatrix}1&-1\\1&0\end{pmatrix}\quad\text{en}\quad P^{-1}=\begin{pmatrix}0&1\\-1&1\end{pmatrix}.\]

Now we indeed have (check this!):

\[PCP^{-1}=\begin{pmatrix}1&-1\\1&0\end{pmatrix}\begin{pmatrix}4&-1\\1&4\end{pmatrix}\begin{pmatrix}0&1\\-1&1\end{pmatrix} =\begin{pmatrix}5&-2\\1&3\end{pmatrix}=A.\]

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Last modified on April 5, 2021