Eigenvectors of linear transformations

The following examples attempt to show linear transformations at work, represented as multiplication by matrices \(A\), \(B\), \(C\) and \(D\), respectively. We will indicate a vector \(v\) (in black) together with its image (in yellow). Moreover, a triangle (pale green) is shown together with its image under the transformation. Note that we follow Poole (Linear algebra, Boston, 2011) in calling every linear map a `transformation', irrespective of its invertibility.
In the examples below, we will give the effect of our geometric transformation by applying matrices to columns. These columns describe the position vectors with respect to a coordinate system that will be indicated if necessary.

The one-dimensional space \(\text{span}(v)\) is indicated by a very thick gray line.

You may use your mouse to move \(v\) (drag its tip); the image of \(v\) under the respective transformation will move automatically.

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Reflection, followed by scaling

Our black vector \(v\) is mapped to the yellow vector \(Av\) by reflecting in the thin red line and then scaling by the factor \(\frac12\).

If we introduce cartesian coordinates with axes as indicated by the red and blue line, this transformation is affected by multiplication with the matrix (product)

\( A = \left[\begin{array}{cc} \frac12 & 0 \\ 0 & -\frac12 \end{array}\right] \) \( = \left[ \begin{array}{cc} \frac12 & 0 \\ 0 &\frac12 \end{array}\right] \left[ \begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right] \).

It is just a coincidence (in fact, a consequence of the fact that \(A\) is symmetric) that the eigenspaces are orthogonal to each other).

Projection (that is, a non-invertible transformation)

Matrix examples for such projections include the following.

• \( B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right] \)

  (if the green and red lines form the first and second axis of a cartesian coordinate system, respectively).

• \( B = \left[ \begin{array}{c} \frac35 \\[1ex] \frac45 \end{array}\right] \left[ \begin{array}{cc} \frac35 & \frac45 \end{array}\right] \) \( = \left[ \begin{array}{cc} \frac9{25} & \frac{12}{25} \\[1ex] \frac{12}{25} & \frac{16}{25} \end{array}\right] \)

  (if the red line is spanned by the vector \(u = \left[ \begin{array}{c} \frac35 \\[1ex] \frac45 \end{array}\right] \), and the green line is perpendicular to the red one).

• \( B = \left[ \begin{array}{cc} 5 & -2 \\ 10 & -4 \end{array}\right] \)

  (here the red line is spanned by \(\left[ \begin{array}{c} 1 \\ 2 \end{array}\right] \) and the green one by \(\left[ \begin{array}{c} 2 \\ 5 \end{array}\right] \); these lines are not perpendicular).

There are many more examples of \(2\times2\) matrices with eigenvalues \(0\) and \(1\); in fact, these are just those matrices \(\left[ \begin{array}{cc} a & b \\ c & d \end{array}\right] \) where \(a+d=1\) and \(ad-bc=0\).

In other words, the trace and determinant of \(B\) have to satisfy \(\mathrm{tr}(B)=1\) and \(\det(B)=0\). This may also be phrased as:

The characteristic polynomial \(\det(B-\lambda I)\) of \(B\) equals \(\lambda^2-\lambda\).

Shearing (algebraic and geometric multiplicities differ)

In the following example, we do not find (in fact, we do not have) eigenvectors in different directions.

Among the matrices describing such shears, we have:

• \( C = \left[ \begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array}\right] \) (here the red line is the second coordinate axis).

• \( C = \left[ \begin{array}{cc} -1 & 2 \\ -2 & 3 \end{array}\right] \)
  (the red line is spanned by \( \left[ \begin{array}{cc} 1 \\ 1 \end{array}\right] \).)

The characteristic property of these matrices describing shears is the fact that they have the eigenvalue \(1\) with algebraic multiplicity \(2\) while the geometric multiplicity is only \(1\).

There are many more examples of \(2\times2\) matrices with this property; in fact, these are just those matrices \(\left[ \begin{array}{cc} a & b \\ c & d \end{array}\right] \) that are not diagonal (i.e., at least one of the entries \(b\), \(c\) is different from zero) with \(a+d=2\) and \(ad-bc=1\).

In other words, the trace and determinant of \(C\) have to satisfy \(\mathrm{tr}(C)=2\) and \(\det(C)=1\). This may also be phrased as follows:

The characteristic polynomial \(\det(C-\lambda I)\) of \(C\) equals \(\lambda^2-2\lambda+1 = (\lambda-1)^2 \).

Rotation (no real eigenvalues, but imaginary ones)

Finally, we discuss an example of a transformation that can be described by a matrix with real entries but having no real eigenvalue at all:

In its start configuration, the drawing indicates a counterclockwise rotation by \(60^\circ\), or \(\frac\pi3\) radians. That rotation can be described by the matrix

\( D = \left[ \begin{array}{cc} \frac12 & \frac{-\sqrt{3}}2 \\[1ex] \frac{\sqrt{3}}2 & \frac12 \end{array}\right] \)

The characteristic polynomial is \(\det(D-\lambda I) = \lambda^2-\lambda+1 \); the eigenvalues are \(\frac12+\frac{\sqrt3}2i\) and \(\frac12-\frac{\sqrt3}2i\).

There are many more examples of \(2\times2\) matrices without real eigenvalues; in fact, these are just those matrices \( D = \left[ \begin{array}{cc} a & b \\ c & d \end{array}\right] \) where the characteristic equation \( 0 = \det(D-\lambda I) \) \( = \lambda^2 -(a+d)\lambda + (ad-bc) \) has no solution in the field of real numbers.

This happens precisely if the discriminant \( (a+d)^2-4(ad-bc) \) is negative; i.e., if

\( (a-d)^2 \lt -4bc \).

(For the rotation in the drawing, we have \(a=d\) and \(bc\lt0\), and the condition is satisfied.)

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