Question about a Linear Algebra Proof?

[Punkyc7]

If we have a matrix with n distinct eigenvalues I understand why we have n distinct eigenvectors.

My question is why is it if we have x_{1}...x_{n} eigenvectors of A with the largest eigenvalue equal to 1 and the rest of the eigenvalues are less than or equal to one, why can we assume that for any y_{0} = c_{1} x_{1} + . . .+c_{n} x_{n}

for some constants c_{1} ..c_{n} with c_{1} \neq0.

Why does c_{1} \neq0.?

I am assuming that we can get any vector because the eigenvectors form a basis for the space we are in.

 

[Dick]

What are you actually trying to prove? Just because 1 is the largest eigenvalue there's no reason why the eigenvectors must be a basis. They might not be. And if you are given that x1..xn are a basis there is no reason why c1 must be nonzero. Suppose y0=x2?

 

[Punkyc7]

I wasnt trying to prove anything I was trying to uderstand why the statement is written the way it is. The eigen vectors are linearly independent so since there is n of them doesn't that mean they form a basis?

 

[HallsofIvy]

If an n by n matrix has n independent eigenvectors, then you can take the eigenvectors to be a basis (you can take any n independent vectors to be a basis for an n dimensional space).

But I don't understand what you mean by "for any y 0 = c 1 x 1 + . . .+c n x n".
What does the "for any" mean here? If you just mean that any vector in the space can be written as a linear combination of the basis vectors, of course that is true. But it is NOT true that c1 cannot be 0. For example, if y0 is any vector you can take y0= xn. In that case, cn= 1 while all other coefficients, including c1, are 0.