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A matrix A is said to be diagonalizable if there exists an invertible matrix X and a diagonal matrix D such that AX = XD. In this case, X is called the diagonalizing matrix and the diagonal elements of D are the eigenvalues of A corresponding to the eigenvectors in X.

We know that AX = XD. To proceed, let's consider the i-th column vector of X, denoted by \(x_i\), which corresponds to a nonzero eigenvalue of A (diagonal element in D). Then: \(A \cdot x_i = \lambda_i \cdot x_i\), where \(\lambda_i\) is the i-th nonzero eigenvalue.

Let’s show that the column vectors corresponding to nonzero eigenvalues are linearly independent. Assume that a linear combination of these vectors equals the zero vector: \(\sum_{i=1}^{k} c_i x_i = 0\), where \(c_i\) are constants and k is the number of nonzero eigenvalues. Multiply both sides of the equation by A: \(A \sum_{i=1}^{k} c_i x_i = 0\). Now apply the equation we found in Step 2: \(\sum_{i=1}^{k} c_i (\lambda_i x_i) = 0\). Finally, we can factor out the eigenvalue term and get: \(\sum_{i=1}^{k} (c_i \lambda_i) x_i = 0\). Since \(\lambda_i\) are nonzero for the corresponding vectors \(x_i\), and we have a linear combination equal to the zero vector, we conclude that \(c_i \lambda_i = 0\) for all \(i\), which implies that \(c_i = 0\) for all \(i\). Therefore, the column vectors of X corresponding to nonzero eigenvalues are linearly independent.

We need to show that any vector in R(A) can be written as a linear combination of the column vectors corresponding to nonzero eigenvalues. Let y be any vector in R(A). Then, there exists some vector x such that A x = y. We can write x as a linear combination of the eigenvectors (columns of X), since they form a basis for the whole space: \(x = \sum_{i=1}^{n} b_i x_i\), where \(b_i\) are constants. Now apply A to both sides: \(A \cdot x = A \sum_{i=1}^{n} b_i x_i = \sum_{i=1}^{n} b_i (A \cdot x_i)\). Using equation from Step 2 and ignoring the terms corresponding to zero eigenvalues, we get: \(y = \sum_{i=1}^{k} (b_i \lambda_i) x_i\), which means that y can be written as a linear combination of the column vectors in X corresponding to nonzero eigenvalues.

We have shown that the column vectors of the diagonalizing matrix X corresponding to nonzero eigenvalues of A are linearly independent (Step 3) and span R(A) (Step 4). Therefore, these column vectors form a basis for R(A).