- #1
Punkyc7
- 420
- 0
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[itex]_{1}[/itex]...x[itex]_{n}[/itex] 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[itex]_{0}[/itex] = c[itex]_{1}[/itex] x[itex]_{1}[/itex] + . . .+c[itex]_{n}[/itex] x[itex]_{n}[/itex]
for some constants c[itex]_{1}[/itex] ..c[itex]_{n}[/itex] with c[itex]_{1}[/itex] [itex]\neq[/itex]0.
Why does c[itex]_{1}[/itex] [itex]\neq[/itex]0.?
I am assuming that we can get any vector because the eigenvectors form a basis for the space we are in.
My question is why is it if we have x[itex]_{1}[/itex]...x[itex]_{n}[/itex] 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[itex]_{0}[/itex] = c[itex]_{1}[/itex] x[itex]_{1}[/itex] + . . .+c[itex]_{n}[/itex] x[itex]_{n}[/itex]
for some constants c[itex]_{1}[/itex] ..c[itex]_{n}[/itex] with c[itex]_{1}[/itex] [itex]\neq[/itex]0.
Why does c[itex]_{1}[/itex] [itex]\neq[/itex]0.?
I am assuming that we can get any vector because the eigenvectors form a basis for the space we are in.