# Linear Algebra: finding eigenvalues and eigenvectors

Problem number 15:

I tried doing this problem without using any software or graphing calculator. I got my eigenvalues as 1 and -1. and got the eigenvector for $-1$ to be a $2\times 1$ matrix $[0, 0]$. I checked the book at this point and it says that my eigenvalues are $-3$ and $1$.

I solved for my eigenvalues by using the characteristic equation, and for the eigenvectors by using (characteristic equation) $x=0$. What am I doing wrong?

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There's only one 'n' in "eigenvalue". :-) – Stefan Jun 12 '11 at 18:54
In order to figure out what you were doing wrong, we would need to see how you derived the characteristic equation and how you solved it. It's certainly the case that you got the wrong answer, but since you don't show your work, we cannot say where you went wrong. – Arturo Magidin Jun 12 '11 at 20:06
An eigenvector is by definition not the zero vector, so as soon as you find the eigenvector for $-1$ to be $(0,0)$, you know it's time to check your arithmetic. – Gerry Myerson Jun 12 '11 at 21:57

## 3 Answers

In addition to using the characteristic polynomial, when you have small matrices (especially $2\times 2$), the following two facts are often useful:

1. The product of the eigenvalues of $A$ equals $\det(A)$.
2. The sum of the eigenvalues of $A$ equals $\mathrm{trace}(A)$.

With $2\times 2$ matrices, both the trace and determinant can be calculated easily in your head, so if you can find two numbers that multiply to the determinant and add to the trace, you've got your eigenvalues. This is essentially the same as factoring the characteristic polynomial "by eye" rather than via the quadratic formula.

For the matrix $$\left(\begin{array}{rr} 2 & 5\\-1 & -4 \end{array}\right),$$ the trace is $-2$ and the determinant is $-8+5 = -3$. So you want two numbers that add up to $-2$ and multiply to $-3$; the answer is $-3$ and $1$, so the two eigenvalues are $-3$ and $1$.

For the matrix $$\left(\begin{array}{rr} 4 & 3\\-3 & -2 \end{array}\right),$$ the trace is $2$ and the determinant is $-8+9 = 1$, so having both eigenvalue equal to $1$ will do it.

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Hint: The eigenvalues of $A$ are given as the roots of the polynomial $\det\left(A-\lambda I\right)$.

Lets go over number $15$, I'll leave $16$ for you. Then $$A-\lambda I=\left[\begin{array}{cc} 2-\lambda & 5\\ -1 & -4-\lambda\end{array}\right]$$ so that $$\det\left(A-\lambda I\right)=(2-\lambda)(-4-\lambda)+5.$$Expanding then factoring we get $$\lambda^{2}+2\lambda-3=\left(\lambda+3\right)\left(\lambda-1\right),$$so the eigenvalues are $-3$ and $1$.

Hope that helps,

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My book didnt teach me about the polynomial determinant you are talking about. It says to use abs(lambda*I-A) as the characteristic equation – chaver Jun 12 '11 at 20:19
@Virtuoso: That is the same thing, just a different symbol for determinant. (Absolute value bars instead of the word "det") – Eric Naslund Jun 12 '11 at 20:26
@Virtuoso: what do you think the 'abs' of a matrix is? – wildildildlife Jun 12 '11 at 20:48

For the $15$ the characteristic polynomial $\text{det}(A -\lambda I)$ is: $(2-x)(-4-x) + 5 =0$ this says $x^{2}+2x-3=0$ which gives $(x+3)(x-1)=0$ which says $x=-3$ or $x=1$.

Similarly work out the next one. Now for finding the eigenvectors (one corresponding to $3$) you simply have to solve $$\left[\begin{array}{cc} 2 & 5 \\\ -1 & -4 \end{array}\right] \cdot \left[\begin{array}{c} x_{1} \\\ x_{2} \end{array}\right] = -3 \left[\begin{array}{c} x_{1} \\\ x_{2}\end{array}\right]$$

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