I found this site: http://people.math.harvard.edu/~knill/teaching/math21b2004/exhibits/2dmatrices/index.html

Which shows a very fast and simple way to get Eigen vectors for a 2x2 matrix. While harvard is quite respectable, I want to understand how this quick formula works and not take it on faith

Part 1 calculating the Eigen values is quite clear, they are using the characteristic polynomial to get the Eigen values.

Part 2, where they calculate the Eigen vectors is what I don't understand and have tried to prove but cannot.

I understand that that what matters with Eigen vectors is the ratio, not the value. For example, an Eigen value of 2, with vector 3, 4, I could have any other vector, example 6, 8, or 12, 16, etc... any scalar multiple.

In their example, given a matrix in the form a b c d, if b & c are zero, then the vectors are 1 0 and 0 1, which makes sense as you can scale these to any other size.

I don't understand the other two cases (when b=0, or c=0), or I presume the case when b & c are non-zero.

Can somebody offer an explanation or proof of this?

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    $\begingroup$ See the Trace Method $\endgroup$ – Amzoti May 18 '13 at 17:41
  • $\begingroup$ @Amzoti, nice reference. $\endgroup$ – Andreas Caranti May 18 '13 at 17:46
  • $\begingroup$ @AndreasCaranti: Thank you Andreas! Regards $\endgroup$ – Amzoti May 18 '13 at 17:47
  • $\begingroup$ The quadratic formula is actually wrong in the Harvard site. It should be L1 = (T + (T^2 - 4D)^1/2) / 2 $\endgroup$ – Jp_ Feb 13 at 18:34

The Harvard class page isn't actually using the trace method, as that computes each eigenvector from the other eigenvalue(s). It's just solving the equations directly. And since it took me way too long to realize that...

Given an eigenvalue $\lambda$ of the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the associated eigenvector(s) are the vectors $\vec{v} = \begin{bmatrix}x\\y\end{bmatrix}$which satisfy $(\lambda I - A)\vec{v} = 0$.

So we have:

$$\begin{bmatrix} \lambda - a & -b \\ -c & \lambda - d \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}(\lambda - a)x - by \\ (\lambda - d)y - cx \end{bmatrix} = 0$$

This gives us two (equivalent) ratios for the vector elements:

$$y = \frac {\lambda - a} b x = \frac c {\lambda - d} x$$

Choosing $x$ to be the denominator of each fraction gives us the solutions on the web site:

$$\vec{v} = \begin{bmatrix}b \\ \lambda - a \end{bmatrix}, \begin{bmatrix}\lambda - d \\ c \end{bmatrix}$$

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Computing the eigenvalues comes down to finding the roots of $\lambda^2 -(a+d)\lambda + (ad-bc) = 0$. That part you know already.

So if the eigenvalues are $\lambda_1$ and $\lambda_2$, then assume $c\neq 0$ and then the claim is that the eigenvectors are $v_i = (\lambda_i-d,c)$. Then $$Av = (a\lambda_i-ad + bc, c\lambda_i - cd + cd) = (a\lambda_i - (ad-bc),c\lambda_i),$$ and $$\lambda_iv = (\lambda_i^2 - d\lambda_i,c\lambda_i).$$
The last coordinates are clearly equal, and we know that $\lambda_i^2 -(a+d)\lambda_i + (ad-bc) = 0$, which implies $\lambda_i^2 - d\lambda_i = a\lambda_i - (ad-bc)$, so the first coordinates are equal too.

If $c=0$ but $b\neq 0$, then the math works out the same way for $v_i=(b,\lambda_i-a)$.

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  • $\begingroup$ So - next.. In other ways that I have calculated the Eigen vectors I get other values. For example: for (lambda=2), I might get the vector (3,4) - I get a different vector value, ie: (6,8). $\endgroup$ – duane May 18 '13 at 21:18
  • $\begingroup$ Sorry, I did that wrong ... Eigen vectors are in the form c(a,b) where c is a constant. Thus if I come up with (2,-3) using this method, and (-4,6) using another method both vectors are valid... because what matters is the ratio of (a/b) must be identical. Is that correct? $\endgroup$ – duane May 18 '13 at 21:29
  • $\begingroup$ Yes. Any multiple of an eigenvector $v$ is also an eigenvector, because $A(cv) = cAv = c\lambda v = \lambda (cv)$. $\endgroup$ – Josephine Moeller May 18 '13 at 22:29

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