Showing that Column Vectors Corresponding to Nonzero Eigenvectors form a basis for R(*A*)

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A study buddy and I were going through this question and while we made some progress, we were ultimately unsure if what we were doing was correct or not.

The question is:

Let A be a diagonalizable matrix and let X be the diagonalizing matrix. Show that the column vectors of X correspond to the nonzero eigenvalues of R(A).

What we did was this:

Say that A is n x n. Therefore X must be n x n and a matrix D must be n x n since if A is diagonalizable and X is the diagonalizing matrix:

A = X D X $^{-1}$

Since X is invertible, its eigenvectors must be linearly independent. Therefore, because X is n x n, there are n linearly independent eigenvectors. These eigenvectors span $\mathbb{R}$$^n$.

Because of this, R(A) $\subset$ $\mathbb{R}$$^n$, we've shown that the the column vectors of X correspond to the nonzero eigenvalues of R(A)... except we haven't. My buddy and I got stuck here and we weren't sure what to do since we haven't accounted for the eigenvalues that are equal to 0.

Thanks for any help in advance.

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Hint: Let $\lambda_1,\dots\lambda_n$ be the eigenvalues of $A$ with associated eigenvectors $\mathbf v_1,\dots,\mathbf v_n$. Collect the equations $A\mathbf v_i=\lambda_i\mathbf v_i$ into a single equation.