# A family of vectors is linearly independent.

Let $K$ be a field and $E$ be a $K$-vector space of dimension $n$. Let $\phi$ be an endomorphism of $E$.

Let $(\lambda_1,\cdots,\lambda_n)$ be a family of distinct scalars and $(x_1,\cdots,x_n)$ be a family of vectors such that $\phi(x_i)=\lambda_ix_i$ for each $i$.

I want to show that the family $(x_1,\cdots,x_n)$ is linearly independent. I have two clear hints

1) for each positive integer $k$, $\phi^k(x_i)=\lambda_i^kx_i$

2) If there exists a family of scalars $(\alpha_1,\cdots,\alpha_n)$ such that $\sum_{i=1}^{n}\alpha_ix_i=0$ then for each positive integer $k$, $\sum_{i=1}^{n}\alpha_i\lambda_i^kx_i=0$

I can see easily the two hints but I don't see how to proceed to show that the family of vectors is linearly independent. Possibly with the second hint if we set the equation $\sum_{i=1}^{n}\alpha_ix_i=0$ and want to show that all the scalars $\alpha_i$ must be zero but how to do that, thank you for your help!

You have to suppose the vectors $x_i$ are non-zero.
Consider the linear map from $E^n$ to $E^n$ which maps $\begin{bmatrix}x_1\\\dots\\x_n\end{bmatrix}$ to $\begin{bmatrix}\sum_{i=1}^n\alpha_i x_i\\\phi(\sum_{i=1}^n\alpha_i x_i)\\\vdots\\\phi^{n-1}(\sum_{i=1}^n\alpha_i x_i)\end{bmatrix}$.
Write the matrix of the system obtained for $k=1,2,\dots,n$. What is its determinant?
• The matrix in which basis ? remember that the $x_i$ are vectors not scalars so turning the hint 2 into a matrix does not seem to be a valid operation. I think you want to compute the determinant of $(x_1, \cdots,x_n)$ in some basis and show that it is not zero, is it true ? what would be that basis ? May 5 '16 at 18:37