Setting Free variables when finding eigenvectors

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When finding eigenvectors after determining eigenvalues, the choice of free variable can lead to different representations of the same eigenvector. Setting either 'a' or 'b' as a free variable is valid, resulting in eigenvectors that differ only by a scalar multiple, such as sign. It's important to note that if one vector is an eigenvector, any non-zero scalar multiple of it is also an eigenvector for the same eigenvalue. Normalization is often applied to ensure the eigenvector has a length of one. Ultimately, both approaches yield correct eigenvectors, highlighting the flexibility in their representation.
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Homework Statement
Confusion in finding eigenvector? (example shown below)
Relevant Equations
matrix multiplication
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.
1597304906485.png


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.
1597305037416.png


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?
 
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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##,
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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