Are the powers of a diagonalizable matrix with distinct eigenvalues linearly independent? I'm trying to demonstrate that if A is an n by n diagonalizable matrix with distinct eigenvalues $\lambda_1, ..., \lambda_n$ and corresponding eigenvectors $\vec{v}_1, ..., \vec{v}_n$, then {$I, A, A^2, ..., A^{n-1}$} is linearly independent.
I thought about saying:
$$\sum_{k=0}^{n-1}c_kA^k=0$$
$$\sum_{k=0}^{n-1}c_kA^k\vec{v}_i=0$$ (choosing a $\lambda_i$-eigenvector $\vec{v}_i$ such that $\lambda_i\neq0$)
Therefore, $$\sum_{k=0}^{n-1}c_k\lambda_i^k\vec{v}_i=0$$
But I'm not sure how to proceed (or if I'm even heading in the right direction).
Any advice would be greatly appreciated. Thanks so much!
 A: Since you are assuming that $A$ is diagonalizable and that its eigenvalues are $\lambda_1,\lambda_2,\ldots,\lambda_n$, then this is the same thing as asking whether the set$$\left\{\begin{bmatrix}\lambda_1^{\,k}&0&0&\ldots&0\\0&\lambda_2^{\,k}&0&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&\lambda_n^{\,k}\end{bmatrix}\,\middle|\,k\in\{0,1,\ldots,n-1\}\right\}$$is linearly independent. All these matrices belong to the space $D$ of all diagonal matrices and the coordinates of its $k$th element with respect to the basis of $D$ which consists of$$\left\{\begin{bmatrix}1&0&\ldots&0\\0&0&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&0\end{bmatrix},\begin{bmatrix}0&0&\ldots&0\\0&1&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&0\end{bmatrix},\ldots,\begin{bmatrix}0&0&\ldots&0\\0&0&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&1\end{bmatrix}\right\}$$are $\lambda_1^{\,k},\lambda_2^{\,k},,\ldots,\lambda_n^{\,k}$. Now, use the fact that the Vandermond determinant is different from $0$ when that $\lambda_j$'s are distinct.
A: Using your sum and the coefficients $c_k$, we find that for each eigenvector $v_i$ we find:
$$
\sum_{k=0}^{n-1} c_kA^kv_i \;\; =\;\; \left (c_0 + c_1\lambda_i + c_2\lambda_i^2 + \ldots + c_{n-1} \lambda_i^{n-1}\right )v_i \;\; =\;\; 0
$$
which only holds if the polynomial $c_0 + c_1\lambda_i + c_2\lambda_i^2 + \ldots + c_{n-1} \lambda_i^{n-1} = 0$.   Arranging all of these polynomials together we arrange the $n\times n$ matrix equation
$$
\left [\begin{array}{ccccc}
1 & \lambda_1 & \lambda_1^2 & \ldots & \lambda_1^{n-1} \\
1 & \lambda_2 & \lambda_2^2 & \ldots & \lambda_2^{n-1} \\
\vdots & \vdots & & \vdots \ddots & \vdots \\
1 & \lambda_{n-1} & \lambda_{n-1}^2 & \ldots & \lambda_{n-1}^{n-1} \\
\end{array} \right ] \left [ \begin{array}{c}
c_0 \\
c_1 \\
\vdots \\
c_{n-1} \\
\end{array} \right ] \;\; =\;\; \textbf{0}.
$$
It should be evident that if each $\lambda_i\neq \lambda_j$ then the columns are linearly independent, hence the nullspace of this matrix only contains the zero vector.
