
A vector which satisfies $M\stackrel{\u27f6}{v}=x\stackrel{\u27f6}{v}$ for some number $x$ is called an eigenvector of the matrix $M$ and $x$ is called the eigenvalue of $M$ corresponding to $\stackrel{\u27f6}{v}$ . ( $\stackrel{\u27f6}{v}$ is called an eigenvector corresponding to $x$ .)
The condition $M\stackrel{\u27f6}{v}=x\stackrel{\u27f6}{v}$ can be rewritten as $(MxI)\stackrel{\u27f6}{v}=\stackrel{\u27f6}{0}$ . This equation says that the matrix $(MxI)$ takes $\stackrel{\u27f6}{v}$ into the $\stackrel{\u27f6}{0}$ vector, which implies that $(MxI)$ cannot have an inverse so that its determinant must be 0.
The equation $\mathrm{det}(MxI)=0$ is a polynomial equation in the variable $x$ for given $M$ . It is called the characteristic equation of the matrix $M$ . You can solve it to find the eigenvalues $x$ , of $M$ .
The trace of a square matrix $M$ , written as $\mathrm{Tr}(M)$ , is the sum of its diagonal elements.
The characteristic equation of a 2 by 2 matrix $M$ takes the form
Once you know an eigenvalue $x$ of $M$ , there is an easy way to find a column eigenvector corresponding to $x$ (which works when $x$ is not a multiple root of the characteristic equation). We will describe it for 3 by 3 matrices, but it can be generalized to apply to any size square matrices. To do so, take the cross product of any two distinct rows of $(MxI)$ .
If it is not the $\stackrel{\u27f6}{0}$ vector, it is a column eigenvector!
Why does this work?
The condition that $\stackrel{\u27f6}{v}$ is a column eigenvector of $M$ is the condition that $(MxI)\stackrel{\u27f6}{v}=\stackrel{\u27f6}{0}$ .
The components of $(MxI)\stackrel{\u27f6}{v}$ are the dot products of the rows of $(MxI)$ with $\stackrel{\u27f6}{v}$ .
If $\stackrel{\u27f6}{v}$ is the vector product of two rows of $(MxI)$ it certainly has dot product 0 with those two rows.
On the other hand, its dot product with the other row of $(MxI)$ is the determinant of $(MxI)$ , which is also 0.
We can deduce then that the vector product of any two rows of $(MxI)$ has 0 dot product with every row of $(MxI)$ , which is the condition that $\stackrel{\u27f6}{v}$ is an eigenvector of $M$ corresponding to eigenvalue $x$ .
What can go wrong? Well, the vector product could be $\stackrel{\u27f6}{0}$ . This will happen if one of the rows is a multiple of the other. If it happens for two different row pairs, this means all the rows are multiples of each other, which means every vector perpendicular to any row that is not all zeroes, is an eigenvector.
Exercises:
4.12 Write the characteristic equation for the matrix with rows $(1,2)$ and $(3,4)$ .
4.13 Do the same for the matrix with rows $(1,2,5)$ , $(3,1,3)$ and $(4,2,8)$ .
4.14 Find an eigenvalue of this matrix. (Hint: there is one that is a pretty simple number.)
4.15 Find a column eigenvector corresponding to it.
