Setting Free variables when finding eigenvectors

Sunwoo Bae · Aug 13, 2020

upon finding the eigenvalues and setting up the equations for eigenvectors, I set up the following equations.

So I took b as a free variable to solve the equation int he following way.

Handwritten working on grid paper: the matrix [4 2; -4 -2] times [a; b] equals [0; 0], giving 4a + 2b = 0, then a = -(1/2)b, so u = [-1/2; 1] times b, and with b = 2 the vector u = [-1; 2].

But I also realized that it would be possible to take a as a free variable, so I tried taking a as a free variable too.

Handwritten working on grid paper: the matrix equation [4 2; -4 -2][a; b] = [0; 0], then 4a + 2b = 0, 4a = -2b, b = -2a, vector u = [1; -2]a, let a = 1, and u = [1; -2].

But now I am confused because this results in vectors that is different in sign. Can anyone explain whether I should use a or b as a free variable?

PeroK · Aug 13, 2020

If ##\vec u## is an eigenvector (corresponding to a certain eigenvalue), then ##-\vec u## is also an eigenvector (corresponding to the same eigenvalue). Both your answers are correct.

In general, ##c \vec u## is also an eigenvector for any number ##c \ne 0##. Often you choose ##c## such that the eigenvector is normalised - i.e. has length ##1##,

Setting Free variables when finding eigenvectors

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