Eigenvectors, does order matter?

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Eigenvectors maintain their identity based on the specific order of their components, meaning that changing the order of elements results in a different vector. For example, [1 0 2]^t is distinct from [0 1 2]^t or [1 2 0]^t. However, any scalar multiple of an eigenvector is also an eigenvector, so vectors like [2 0 4] or [4 0 8] are valid. The distinction in order is crucial, as it affects the vector's properties and its relationship to the corresponding eigenvalue. Understanding these principles is essential for correctly working with eigenvectors in linear algebra.
charlies1902
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I got 2 questions about eigenvectors.

Let's say you have an eigenvector [1 0 2]^t.


1. Does the order matter? Like can I change the order to [0 1 2]^t or [1 2 0]^t?


2. It can be any scalar multiple of the vector right? Like I could have [2 0 4] or [4 0 8]
 
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charlies1902 said:
I got 2 questions about eigenvectors.

Let's say you have an eigenvector [1 0 2]^t.


1. Does the order matter? Like can I change the order to [0 1 2]^t or [1 2 0]^t?


2. It can be any scalar multiple of the vector right? Like I could have [2 0 4] or [4 0 8]

Scalar multiples are still eigenvectors, sure. Show me why. Does order of the components in the vector matter? What would make you think it doesn't??
 
The vector <1, 0, 0> is NOT the same as the vector <0, 1, 0> which appears to be what you are asking!
 
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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