2x2 matrix A has only one eigenvalue λ with eigenvector v

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SUMMARY

The discussion centers on a 2x2 matrix A with a single eigenvalue λ and corresponding eigenvector v, alongside a non-zero vector w that is not an eigenvector. Participants conclude that vectors v and w are linearly independent, as any dependence would imply w is a multiple of v, thus making w an eigenvector. The matrix representation with respect to the basis {v, w} is established as (λ c; 0 λ) for some non-zero constant c, with the possibility of setting c to 1 for a suitable choice of w.

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nlews
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This is a revision problem I have come across,

I have completed the first few parts of it, but this is the last section and it seems entirely unrelated to the rest of the problem, and I can't get my head around it!

Suppose that the 2x2 matrix A has only one eigenvalue λ with eigenvector v, and that w is a non zero vector which is not an eigenvector..show that:

a) v and w are linearly independent
b) the matrix with respect to the basis {v, w} is
(λ c
0 λ)
for some c =not to 0
c) for a suitable choice of w, c = 1I am stuck.
I know how to show that the eigenvalues are linearly independent, but how do I show that these two vectors are linearly independent to each other?
as for b and c i don't know where to start! Please help!
 
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nlews said:
Suppose that the 2x2 matrix A has only one eigenvalue λ with eigenvector v, and that w is a non zero vector which is not an eigenvector..show that:

a) v and w are linearly independent

I know how to show that the eigenvalues are linearly independent, but how do I show that these two vectors are linearly independent to each other?

If v and w are linearly dependent, then w is a multiple of v, so obviously w is also an eigenvector.

Get some sleep! :zzz:​
 


ahh ok..so I can prove by contradiction! thank you that helps massively for part a!
 

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