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!
Best Answer
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.