14.1 Finding Eigenvalues

In this class, to find the eigenvalues of an $n\times n$ matrix,

1.

Form the matrix $A-\lambda I$. That means, add $-\lambda$ to each diagonal elements. (Don't forget zero diagonal elements.)

2.

Find the determinant of that matrix using the method of minors. (Gaussian elimination is too messy here and should not be used.)

3.

Set this determinant to zero. For an $n\times n$ matrix, the determinant can always be written in the form

$$\vert A\vert = \pm (\lambda - \lambda_1) (\lambda - \lambda_2) \ldots(\lambda - \lambda_n)$$

where $\lambda_1, \lambda_2, \ldots, \lambda_n$ are the eigenvalues. They are found as the roots (or zeros) of the determinant. There are always $n$ eigenvalues. But these $n$ eigenvalues do not necessarily correspond to $n$ different numbers. For example, for some matrix $\lambda_2$ might be the same number as $\lambda_4$. In that case, that number is called a “double eigenvalue”. If three of the eigenvalues are the same number, that number is called a triple eigenvalue, etcetera.

For an $n\times n$ matrix, the determinant is always a polynomial of degree $n$, call it $p_n(\lambda)$. Now finding the roots of quadratics, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, is easy. But if the dimension of the matrix $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, you have to solve a cubic equation. For that the general solution is very and hard to apply, especially if you do not know complex variables. For $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4, the general solution is even worse, and for $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 5 or more, there is no general expression for the roots at all. (It has in fact been proved that such an expression is impossible to find.) To deal with such problems, here are some tricks:

• Do not be too quick to multiply out. Maybe you can recognize a common factor $\lambda-\lambda_i$ in all terms before multiplying out. In that case, $\lambda_i$ is one of your eigenvalues. To find the remaining eigenvalues, take the factor $\lambda-\lambda_i$ out of the entire expression and look at what is left.
• If you can guess a root $\lambda_i$ (by trying, say, 0, $\pm1$, $\pm2$, ..., and seeing whether the determinant is zero for that value of $\lambda$), then you can write the original characteristic polynomial $p_n(\lambda)$ as $(\lambda-\lambda_i)p_{n-1}(\lambda)$ where $p_{n-1}(\lambda)$ is one degree less that $p_n(\lambda)$. So its roots are easier to find. You find $p_{n-1}(\lambda)$ by long division of $p_n(\lambda)$ by $(\lambda-\lambda_i)$.
• Sometimes a fourth order polynomial is really just a quadratic when written in terms of $\lambda^2$, like $(\lambda^2)^2-a(\lambda^2)-b$. In that case, first find the values of $\lambda^2$, then from those the values of $\lambda$.