# Quest about free variables and eigenvalues

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• Apr 8th 2014, 06:59 AM
Latsabb
Quest about free variables and eigenvalues
In class today, our professor was going over examples of how to find eigenvalues and eigenvectors. This was very straight forward for his 2x2 matrices, but there was something different with the 3x3's. Rather than work out the three eingenvalues to his 3x3 matrices, he was giving a single eigenvalue, and then calculating the eigenvector for just that value. He showed in three different examples that you may end up with two free variables, and then showed how you would find two eigenvectors with that. I was playing with those matrices, and put them into wolframalpha to find the other eigenvalues, when I noticed a pattern. All of those that had two free variables, had only two solutions to the cubic equation, rather than three.

So my question is this: Was this a coincidence, or is that actually a property/rule? If I get two free variables, can I assume that there are only two solutions, and vice versa? And if so, can someone give me the quick and dirty on why that works out? And what happens with a 2x2 if the quadratic only has one solution? Wouldnt that only give you one eigenvector, instead of two?

Thank you in advance for your time.
• Apr 14th 2014, 04:49 AM
HallsofIvy
Re: Quest about free variables and eigenvalues
Well, the "eigenvalue equation" for an n by n matrix is an nth degree polynomial equation. It is far easier to solve a quadratic equation than a cubic equation! That was the only reason, I suspect, to starting with one of the eigenvalues "given". Dividing a cubic equation by "x- a", where a is a solution, reduces it to a quadratic equation for the other two solutions.

I'm not clear what you mean by "free variables" here. That term is used in a number of different ways in different mathematical problems. It is true that if a given eigenvalue is a "double" root of the eigenvalue equation, then the space of eigenvectors corresponding to that eigenvalue may be of dimension 1 or 2. I suspect that it is in the case that a double root has eigenspace of dimension only 1 that your "free variables" arise.

As a rather trivial example, the matrix $\begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}$ has "2" as a double eigenvalue and both <1, 0, 0> and <0, 1, 0> are independent eigenvectors.

On the other hand, $\begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}$ also has "2" as a double eigenvalue but now the space of eigenvectors corresponding to eigenvalue 2 is only one dimensional and is spanned by <1, 0, 0>.