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## Homework Statement

I have a matrix A = [3 -2; -3 2],

then I peform det(A-λI) and find solutions λ=0 and λ=5.

For the first case λ=0 I perform A-0I, which of course is just A [3 -2; -3 2]. I then reduce to row echelon form and find the values V1 = 2 and V2 = 3, therefore for λ=0 my eigenvector is [2; 3]

For the second case λ=5, I perform A-5I, which gives [-2 -2; -3 -3]. Reducing to row echelon form yields [-6 -6; 0 0][V1; V2] = [0; 0]

In linear form my, the equation is as follows, -6V1 - 6V2 = 0

The answer provided is [V1; V2] = [-1; 1], my question is, how do you know for a case where the coeffients are the same, that the eigenvector will be [-1; 1] rather than [1;-1] as -6(1) - 6(-1) = 0 and still satifies the equation.

Thanks