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Mathematical Expression Editor
Describing Eigenvalues and Eigenvectors
At several places in this course it has been valuable to restrict ourselves
to square matrices, and we do so again when discussing eigenvalues and
eigenvectors.
In Theorem th:matrixtran, we proved that any \(n \times n\) matrix induces a linear transformation from \(\RR ^n\) to
itself. For our first few examples, let us consider the case \(n = 2\).
Let \(A=\begin{bmatrix} 2& 1\\ 1&2 \end{bmatrix}\). The following interactive helps us to visualize the matrix transformation
associated with \(A\). Given a vector \(\vec {x}\) in \(\RR ^2\) (left panel), its image, \(A\vec {x}\), is also in \(\RR ^2\) (right panel).
Drag the head of the vector \(\vec {x}\) to see its image.
What do you observe about the relationship between vectors along the red line \(y=-x\) and
their images?
For many vectors \(\vec {x}\) of \(\RR ^2\), \(A\vec {x}\) does not point in the same direction as \(\vec {x}\) but if we look at the
vectors along the line \(y=-x\), we notice that the transformation does not seem to affect
their magnitude and direction. Such vectors are sometimes called fixed vectors of \(A\).
For example \(\begin{bmatrix}1\\-1\end{bmatrix}\) is a fixed vector.
What do you observe about the relationship between vectors along the blue line \(y=x\) and
their images?
We observe that the magnitudes of these vectors are changed by the transformation,
but the direction in which the vectors point is unchanged.
In Exploration init:eignintro we found that the images of vectors \(\begin{bmatrix}1\\1\end{bmatrix}\) and \(\begin{bmatrix}-1\\1\end{bmatrix}\) stay parallel to the original
vectors under the linear transformation induced by matrix \(A\). Such vectors are
examples of eigenvectors of \(A\).
In general, any nonzero vector whose image under a matrix transformation is parallel
to the original vector is called an eigenvector of the matrix that induced the
transformation. The following definition captures this idea algebraically.
Let \(A\) be an \(n \times n\) matrix. We say that a non-zero vector \(\vec {x}\) is an eigenvector of \(A\)
if
\[A\vec {x} = \lambda \vec {x}\]
for some scalar \(\lambda \). We say that \(\lambda \) is an eigenvalue of \(A\) associated with the eigenvector \(\vec {x}\).
Let’s revisit Exploration init:eignintro in light of Definition def:eigen. In the exploration, we observed
visually that vectors parallel to \(\begin{bmatrix} 1\\ 1 \end{bmatrix}\) were eigenvectors associated with \(A=\begin{bmatrix} 2& 1\\ 1&2 \end{bmatrix}\), as these vectors
changed length but remained parallel to the original vector under the linear
transformation induced by \(A\). To verify this algebraically, observe that all vectors
parallel to \(\begin{bmatrix} 1\\ 1 \end{bmatrix}\) can be written in the form \(\begin{bmatrix} a\\ a \end{bmatrix}\), \((a\neq 0)\). We compute
\[\begin{bmatrix} 2& 1\\ 1&2 \end{bmatrix} \begin{bmatrix} a\\ a \end{bmatrix} = \begin{bmatrix} 3a\\ 3a \end{bmatrix}= 3 \begin{bmatrix} a\\ a \end{bmatrix}\]
This shows that any non-zero scalar multiple of \(\begin{bmatrix} 1\\ 1 \end{bmatrix}\) is an eigenvector of \(A\) which has a
corresponding eigenvalue of 3.
Fixed vectors of Exploration init:eignintro are also eigenvectors. For example,
This shows that \(\begin{bmatrix} 1\\ -1 \end{bmatrix}\) is a fixed vector and an eigenvector of \(A\) which has a corresponding
eigenvalue of \(1\).
The above discussion leads us to the following result.
If \(\vec {x}\) is an eigenvector of matrix \(A\) and \(\lambda \) is the corresponding eigenvalue, then every
non-zero scalar multiple of \(\vec {x}\) is also an eigenvector of \(A\) and \(\lambda \) is the corresponding
eigenvalue.
A couple of finer points of Definition def:eigen require clarification.
The definition requires that eigenvectors be non-zero. Imagine what would
happen if we allowed \(\vec {x}=\vec {0}\) to be an eigenvector of \(A\). Clearly \(A\vec {0}=\lambda \vec {0}\) for all scalars \(\lambda \). This
means that every number would be an eigenvalue of every matrix. Because
eigenvalues are supposed to capture certain information about the matrix,
allowing every number to be an eigenvalue of every matrix would defeat
the purpose.
Until now, we had talked about eigenvectors as vectors whose images
under a matrix transformation are parallel to the original vectors. But
the algebraic definition allows non-zero vectors that map to zero to be
considered eigenvectors. (What would an eigenvalue of such an eigenvector
be?) The zero vector has no direction, so we cannot say that the image
of such an eigenvector is parallel to the original vector. Example ex:eigen will
illustrate this point.
Let \(P=\begin{bmatrix} 1& 0\\ 0&0\end{bmatrix}\). Note that \(P\) takes a vector in \(\RR ^2\) and projects it onto the \(x\)-axis, as we learned in
Practice Problem prob:standardmatrix3. Which vectors in \(\RR ^2\) would be the eigenvectors, and what are the
corresponding eigenvalues?
Since \(P\vec {x}\) is the projection of \(\vec {x}\) onto the \(x\)-axis, in many cases \(P\vec {x}\)
and \(\vec {x}\) are not parallel. Notice, however, that all of the red vectors located along the
\(x\)-axis in the diagram are fixed by \(P\). So, for any of the red vectors we have \(P\vec {x}=\vec {x}=1\vec {x}\), which
means that each of the red vectors is an eigenvector of \(P\) with the corresponding
eigenvalue of \(1\).
The blue vectors along the y-axis are also eigenvectors. To see this, note that each of
the blue vectors is of the form \(\vec {x}=\begin{bmatrix}0\\x_2\end{bmatrix}\). But then
So each of the blue vectors is an eigenvector of \(P\) with the corresponding eigenvalue of
\(0\).
Let \(A=\begin{bmatrix}1&6\\1&0\end{bmatrix}\). The GeoGebra interactive below shows vector \(\vec {x}\), and its image, \(A\vec {x}\) in the
same coordinate plane. (Move the head of \(\vec {x}\) to see the effect of \(A\) on different
vectors.)
Note that vectors \(\vec {x}_1=\begin{bmatrix}3\\1\end{bmatrix}\) and \(\vec {x}_2=\begin{bmatrix}-2\\1\end{bmatrix}\) (and their scalar multiples) remain positioned along the
same lines even as they change magnitude and direction. This indicates that \(\vec {x}_1\) and \(\vec {x}_2\),
along with all of their scalar multiples, are eigenvectors of \(A\). What are the eigenvalues
associated with these eigenvectors?
Eigenvalue associated with \(\vec {x}_1\) is \(\lambda _1=\answer {3}\).
Eigenvalue associated with \(\vec {x}_2\) is \(\lambda _2=\answer {-2}\).
A natural question is this: does every square matrix have eigenvalues and
eigenvectors? We will see later that the answer to this question is “yes", provided that
we permit eigenvalues and entries of eigenvectors to be complex numbers. The next
example illustrates this.
Let \(M=\begin{bmatrix} \frac {\sqrt {2}}{2} & -\frac {\sqrt {2}}{2}\\ \frac {\sqrt {2}}{2} & \frac {\sqrt {2}}{2} \end{bmatrix}\). Note that \(M\) takes any vector \(\vec {x}\) in \(\RR ^2\) and rotates it \(45^{\circ }\), as we saw in Example
ex: rotate45.
Since \(M\) rotates every vector in \(\RR ^2\), every nonzero vector changes direction, so there are
no eigenvectors in the plane. It turns out that \(M\) does have eigenvectors and
eigenvalues, but in order to find them we need to work with vectors whose
entries are complex numbers. Since these vectors are not in \(\RR ^2\), we cannot see
them.
To follow the computation below, you need to recall that the imaginary unit \(i\) is
defined by \(i^2=-1\).
Consider the vector \(\begin{bmatrix} \frac {\sqrt {2}}{2}\\ \frac {\sqrt {2}}{2} i \end{bmatrix}\). We compute:
so \(\begin{bmatrix} \frac {\sqrt {2}}{2}\\ \frac {\sqrt {2}}{2} i \end{bmatrix}\) is an eigenvector of \(M\). Its corresponding eigenvalue is \(\frac {\sqrt {2}}{2}-\frac {\sqrt {2}}{2}i\).
We will continue to work with complex numbers as we study eigenvalues and
eigenvectors.
Why All the Fuss About Eigenvalues and Eigenvectors?
The first in-depth study of eigenvalues can probably be attributed to Fourier as he
studied partial differential equations early in the nineteenth century, and in
particular when he studied what is known as the heat equation. [Trefethen and
Embree] By the twentieth century mathematicians understood the connections
between differential equations and eigenvalues. Systems of differential equations are
often best represented by matrices, especially in the context of using computers to
find numerical solutions. Most algorithms to solve these systems work by iterating
some process, and eigenvalues along with their corresponding eigenvectors indicate
what will happen to such a process after many repetitions.
The most famous modern example of a large-scale eigenvalue problem is the Google
PageRank algorithm, which helped set Google apart from its competitors as a search
engine. Some of the relevant mathematics can be learned by working through the
paper, “The $25,000,000,000 Eigenvector, The Linear Algebra Behind Google”, by
Kurt Bryan and Tanya Leise.
Show that \(\begin{bmatrix} 1\\1\\1 \end{bmatrix}\) is an eigenvector of \(C\). What is its corresponding eigenvalue? \(\answer {4}\)
Show that \(\begin{bmatrix} 1\\-1\\0 \end{bmatrix}\) is an eigenvector of \(C\). What is its corresponding eigenvalue? \(\answer {1}\)
Show that \(\begin{bmatrix} 1\\1\\-2 \end{bmatrix}\) is an eigenvector of \(C\). What is its corresponding eigenvalue? \(\answer {1}\)
Let \(Q=\begin{bmatrix} 0& 0\\ 0&1\end{bmatrix}\). Note that \(Q\) takes any vector in \(\RR ^2\) and projects it onto the \(y\)-axis. Which
vectors in \(\RR ^2\) would be eigenvectors, and what are the corresponding eigenvalues?
Returning to Example 7, let \(M=\begin{bmatrix} \frac {\sqrt {2}}{2} & -\frac {\sqrt {2}}{2}\\ \frac {\sqrt {2}}{2} & \frac {\sqrt {2}}{2} \end{bmatrix}\). Show that \(\begin{bmatrix} \frac {\sqrt {2}}{2}\\ -\frac {\sqrt {2}}{2} i \end{bmatrix}\) is an eigenvector of \(M\). What is its
corresponding eigenvalue?
Please enter an exact answer; no decimal approximations. To enter square roots, click
inside the answer box, then use the “Math Editor" button that appears at the top of
your XIMERA window.
Arguing geometrically, identify the linear transformation whose standard matrix
has eigenvalues \(\lambda _1=1\) and \(\lambda _2=-1\).
Vertical Shear Horizontal Shear Counterclockwise
Rotation through a \(90^\circ \) angleReflection About the line \(y=mx\)Horizontal Stretch Vertical Stretch
Let \(A=\begin{bmatrix} 3&0&0\\0&3&0\\0&0&3\end{bmatrix}\). Can you find an eigenvector and its corresponding eigenvalue? Can you find
another “eigenpair”? Can you find all of the eigenvectors of \(A\)?
The rotation matrix in Example 7 has complex eigenvectors and eigenvalues. Think
geometrically to find an example of a (non-identity) rotation matrix with real
eigenvectors and eigenvalues.
Enter degree measure between 0 and 360.
Answer: Rotation through \(\answer {180}\) degrees.
Can an eigenvalue have multiple eigenvectors associated with it?
Yes No
Can an eigenvector have multiple eigenvalues associated with it?
Yes No
Prove that if \(\vec {x}\) is an eigenvector of matrix \(A\) and \(\lambda \) is the corresponding eigenvalue,
then every non-zero scalar multiple of \(\vec {x}\) is also an eigenvector of \(A\) and \(\lambda \) is the
corresponding eigenvalue. (Theorem 2)
Bibliography
[Trefethen and Embree] Trefethen, Lloyd and Embree, Mark, Spectra and
Pseudospectra, Princeton University Press, 2005, p. 5-6
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