If $A$ be a matrix with $n$ distinct eigenvalues then $u, Au , A^2u ,\cdots, A^{n-1} u$ will be independent. 
If $A$ be a matrix with $n$ distinct eigenvalues then there exists a vector $u$  such that $u,Au,A^2u,\cdots, A^{n-1} u$  will be independent.

Can anyone give me a hint to solve this problem? 
I know that if $A$ has $n$  distinct eigen values then it will have $n$ independent eigen vectors..
 A: Hint:
You must prove there exists $u$ such that $u,Au,A^2u,\cdots,A^{n-1}u$ is linearly independent
Take $u=u_1+u_2+\cdots+u_n$ where $u_i$'s are eigenvectors corresponding to the distinct $n$ eigenvalues and then show this $u$ works!

Added: Start with $$a_0u+a_1Au+\cdots+a_{n-1}A^{n-1}u=0$$
and substitute $u$ to expand this and do some calculations to end with $Bx=0$  where $B$ is a Vandermonde matrix with distinct entries and $x=(a_0,a_1,...,a_{n-1})^t$ and then conclude $x=B^{-1}0=0$

Edit: Now $$Au=Au_1+Au_2+\cdots+Au_n=\lambda_1u_1+\lambda_2u_2+\cdots+\lambda_nu_n$$
$$A^2u=\lambda_1^2u_1+\lambda_2^2u_2+\cdots+\lambda_n^2u_n$$ 
$$\vdots$$ 
$$A^{n-1}u=\lambda_1^{n-1}u_1+\lambda_2^{n-1}u_2+\cdots+\lambda_n^{n-1}u_n$$
Suppose $a_0u+a_1Au+\cdots+a_{n-1}A^{n-1}u=0$.  That is, 
$a_0(u_1+\cdots+u_n)+a_1(\lambda_1u_1+\cdots+\lambda_nu_n)+\cdots+a_{n-1}(\lambda_1^{n-1}u_1+\cdots+\lambda_n^{n-1}u_n)=0$
The above can be re written as $$(a_0+a_1\lambda_1+\cdots+a_{n-1}\lambda_1^{n-1})u_1$$ $$+(a_0+a_1\lambda_2+\cdots+a_{n-1}\lambda_2^{n-2})u_2$$ $$+\cdots+(a_0+a_1\lambda_n+\cdots+a_{n-1}\lambda_n^{n-1})u_n=0$$
since $\{u_i\}$ are linearly independent(How?), $$a_0+a_1\lambda_1+\cdots+a_{n-1}\lambda_1^{n-1}=0$$ $$a_0+a_1\lambda_2+\cdots+a_{n-1}\lambda_2^{n-1}=0$$ $$\vdots$$ $$a_0+a_1\lambda_n+\cdots+a_{n-1}\lambda_n^{n-1}=0$$
The above can be written as $$\begin{pmatrix} 1& \lambda_1 &\lambda_1^2& \cdots&\lambda_1^{n-1}\\ 1& \lambda_2 &\lambda_2^2& \cdots&\lambda_2^{n-1} \\ \vdots\\  1& \lambda_n &\lambda_n^2& \cdots&\lambda_n^{n-1}   \end{pmatrix} \begin{pmatrix} a_0\\a_1\\\vdots\\a_{n-1}\end{pmatrix}=\begin{pmatrix} 0\\0\\\vdots\\0\end{pmatrix}$$ Since the coefficient matrix is Vandermonde, and $\lambda_i$'s are distinct, so its determinant is non zero, so......?  
