Question

The characteristic equation for a second-order matrix $M$ is a quadratic equation. We have considered in detail the case in which M is a real symmetric matrix and the roots of the characteristic equation (eigenvalues) are real, positive, and unequal. Discuss some other possibilities as follows: (a) $\mathrm{M}$ real and symmetric, eigenvalues real, one positive and one negative. Show that the plane is reflected in one of the eigenvector lines (as well as stretched or shrunk). Consider as a simple special case $$M=\left(\begin{array}{rr}1& 0\\ 0& -1\end{array}\right)$$ (b) $\mathrm{M}$ real and symmetric, eigenvalues equal (and therefore real). Show that $\mathrm{M}$ must be a multiple of the unit matrix. Thus show that the deformation consists of dilation or shrinkage in the radial direction (the same in all directions) with no rotation (and reflection in the origin if the root is negative). (c) $M$ real, not symmetric, eigenvalues real and not equal. Show that in this case the eigenvectors are not orthogonal. Hint: Find their dot product. (d) $\mathrm{M}$ real, not symmetric, eigenvalues complex. Show that all vectors are rotated, that is, there are no (real) eigenvectors which are unchanged in direction by the transformation. Consider the characteristic equation of a rotation matrix as a special case.

Short answer

Real, symmetric M with one positive and one negative eigenvalue reflects and stretches planes along the eigenvectors. With equal eigenvalues, M scales isotropically. For non-symmetric M with real, unequal eigenvalues, eigenvectors are not orthogonal. For non-symmetric M with complex eigenvalues, all vectors are rotated.

Step-by-step solution

- Introduction

We are given four different cases for the real matrix M, each having different characteristics for its eigenvalues. Let’s solve each part step by step.

- Part (a): Reflecting and Stretching

Consider matrix $M=\left(\begin{array}{ccc}1& 00& -1\end{array}\right)$. We need to find its eigenvalues and eigenvectors. The characteristic equation is given by $det(M-\lambda I)=0$.

- Finding Eigenvalues for Part (a)

The characteristic equation is: $$det(\left(\begin{array}{ccc}1& 00& -1\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 00& 1\end{array}\right))=0⟹(1-\lambda )(-1-\lambda )=0⟹\lambda =1,-1.$$

- Finding Eigenvectors for Part (a)

For $\lambda =1$: $$(M-I)\mathbf{v}=0⟹\left(\begin{array}{ccc}0& 00& -2\end{array}\right)\left(\begin{array}{c}{v}_1{v}_2\end{array}\right)=\left(\begin{array}{c}00\end{array}\right)⟹{\mathbf{v}}_1=\left(\begin{array}{c}10\end{array}\right).$$ For $\lambda =-1$: $$(M+I)\mathbf{v}=0⟹\left(\begin{array}{ccc}2& 00& 0\end{array}\right)\left(\begin{array}{c}{v}_1{v}_2\end{array}\right)=\left(\begin{array}{c}00\end{array}\right)⟹{\mathbf{v}}_2=\left(\begin{array}{c}01\end{array}\right).$$

- Interpretation for Part (a)

The plane is reflected in the line of the first eigenvector ${\mathbf{v}}_1$ and stretched/shrunk in the directions of the eigenvectors ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$.

- Part (b): Eigenvalue Equal

If M is real and symmetric and has equal real eigenvalues, then it must be of the form $\lambda I$, where $I$ is the identity matrix. The deformation is isotropic scaling (dilation or shrinkage) in all directions with no rotation.

- Matrix Confirmation for Part (b)

Let’s consider $\lambda$, then: $M=\lambda I⟹M=\lambda \left(\begin{array}{ccc}1& 00& 1\end{array}\right).$ All directions are scaled by $\lambda$.

- Part (c): Non-Symmetric Matrix with Real Eigenvalues

For a real non-symmetric matrix with real, unequal eigenvalues, the eigenvectors are not orthogonal. Assume eigenvalues ${\lambda }_{1}eq{\lambda }_2$, with eigenvectors ${\mathbf{v}}_1,{\mathbf{v}}_2$. These eigenvectors will not satisfy ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=0$.

- Example for Part (c)

Example: Consider $M=\left(\begin{array}{ccc}2& 10& 3\end{array}\right)$. Solve for eigenvalues and eigenvectors.

- Part (d): Complex Eigenvalues

For a real non-symmetric matrix with complex eigenvalues, every vector is rotated. Consider a rotation matrix: $R(\theta )=\left(\begin{array}{ccc}\cos \theta & -\sin \theta \sin \theta & \cos \theta \end{array}\right)$. The characteristic equation $${\lambda }^2-2\lambda \cos \theta +1=0$$ gives complex eigenvalues when $0<\theta <\pi$.

- Conclusion for Part (d)

There are no real eigenvectors fixed in direction by $R(\theta )$. Every vector is rotated by the angle $\theta$.

Key concepts

Characteristic Equation

The characteristic equation of a matrix is fundamental in finding its eigenvalues. For a matrix $M$, the characteristic equation is derived from the determinant of the difference between the matrix and a scalar multiple of the identity matrix. This can be expressed as: $\text{det}(M-\lambda I)=0$. Here, $\lambda$ stands for the eigenvalues of $M$.
For a second-order matrix $M$, the characteristic equation is a quadratic equation, typically resulting in two eigenvalues. Different properties and characteristics of the matrix are reflected through these eigenvalues.
For example, if the matrix $M$ has complex eigenvalues, it indicates that the transformation includes rotation.

Matrix Reflection

A reflection matrix in two dimensions will flip vectors over a certain line. Consider the following simple reflection matrix:

$$M=\left(\begin{array}{ccc}1& 00& -1\end{array}\right)$$
The characteristic equation for the above matrix is $\text{det}(M-\lambda I)=0$, which expands to: $(1-\lambda )(-1-\lambda )=0$. Solving this results in the eigenvalues $\lambda =1$ and $\lambda =-1$.
Eigenvalues of 1 and -1 suggest that the matrix reflects any vector across the $x$-axis. Vectors aligned with the $x$-axis remain unchanged, while any vector along the $y$-axis is flipped to its negative counterpart. This includes a stretch or shrink when the eigenvalues are different.

Isotropic Scaling

Isotropic scaling refers to a transformation where all directions are uniformly scaled by the same factor. A matrix exhibiting isotropic scaling has equal eigenvalues.

If $M$ is real and symmetric with equal real eigenvalues, it can be written as $M=\lambda I$. The identity matrix $I$ scales every vector equally in all directions. For example,

$$M=\lambda \left(\begin{array}{ccc}1& 00& 1\end{array}\right)$$
Here, $\lambda$ is the scaling factor.
This means no rotation occurs, just a uniform dilation or shrinkage. If the eigenvalues are negative, this also incorporates a reflection in the origin.

Non-Symmetric Matrix Properties

Non-symmetric matrices can exhibit more complex behavior compared to symmetric matrices. When considering the eigenvalues and eigenvectors of these matrices, the properties vary significantly.

For a real, non-symmetric matrix with real, distinct eigenvalues, the eigenvectors corresponding to these eigenvalues are not orthogonal. Orthogonality depends on the symmetric nature of a matrix. For instance,
$$M=\left(\begin{array}{ccc}2& 10& 3\end{array}\right)$$
has real, distinct eigenvalues, but its eigenvectors are not orthogonal.
When a non-symmetric matrix has complex eigenvalues, the transformation will involve rotation, and there will be no real eigenvectors that remain in the same direction. Rotation matrices are perfect examples of such transformations, with complex eigenvalues indicating a circular motion effect in the transformation.