Diagonalizing a matrix with eigenvectors as the bases

Question

I understand that for a n x n matrix, if there are n linearly independent eigenvectors, you can diagonalize the standard matrix by setting the bases as the eigenvectors and the diagonal of the matrix would be the eigenvalues. This is mathematically sound for me. However, as I'm learning linear algebra, I'm trying to visualize the concepts to create a solid foundation. I was wondering if anyone could give me an intuitive explanation for why transforming a standard n x n matrix to a matrix with n linearly independent eigenvectors as the basis serves to create a diagonalized matrix. Thank you.

Answer 1

This stems from the very definition of the Eigenvectors and values.

If

$$Ax_k=\lambda x_k,$$

in the diagonalized world

$$A\begin{pmatrix}x_1&x_2&x_3&\cdots&x_n\end{pmatrix} =\begin{pmatrix} \lambda_1&0&0&\cdots&0 \\0&\lambda_2&0&\cdots&0 \\&&\cdots \\0&0&0&\cdots&\lambda_n \\\end{pmatrix}\begin{pmatrix}x_1&x_2&x_3&\cdots&x_n\end{pmatrix}. $$

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More intuitively, an arbitrary linear transform will transform a unit sphere in an ellipsoid. The ellipsoid axis correspond to the Eigenvectors and their length are the Eigenvalues.

If you set yourself in a reference frame attached to the axis, the transformation is a mere scaling along every axis.