TITLE: Is matrix diagonalization unique?
QUESTION [9 upvotes]: From the following statement, it seems matrix diagonalization is just eigen decomposition. 

Diagonalizing a matrix is also equivalent to finding the matrix's eigenvalues, which turn out to be precisely the entries of the diagonalized matrix. Similarly, the eigenvectors make up the new set of axes corresponding to the diagonal matrix. 

http://mathworld.wolfram.com/MatrixDiagonalization.html
However, from what I have learned, Spectral Theorem is closest to this conclusion. But how the spectral theorem is related to it, or is there some other theorem grants this statement?
Spectral Theorem: Suppose that $V$ is a complex inner-product space and $T \in L(V)$. Then $V$ has an orthonormal basis consisting of eigenvectors of $T$ if and only if $T$ is normal.

REPLY [6 votes]: While the diagonal matrix $D$ is determined up to permutation, the change of basis $U$ is only determined up to scalar multiples of the columns. So we may indeed have diagonalizations
$$A = U_1 D_1 U_1^{-1} = U_2 D_2 U_2^{-1}$$
where $D_1 \neq D_2$, $U_1 \neq U_2$. So diagonalization is not unique in general, but unique up to permutation of the diagonal entries in $D$, and multiples of the columns of $U$.