6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The operator defined by the matrix in some basis has a basis of orthonormal eigenvectors. The matrix of the same operator in this basis is the diagonalization. The change of basis gives you the equivalence. $\endgroup$ – OR. Jul 25 '13 at 22:22
  • $\begingroup$ There are many different types of matrix factorization each of which finds use in different problem domains: en.wikipedia.org/wiki/Matrix_decomposition $\endgroup$ – Antinous Feb 13 '18 at 22:35
6
$\begingroup$

Matrix diagonalization is more general than the spectral theorem. For instance, you may not be in an inner product space, and it still may be helpful to diagonalize a matrix. Not every matrix can be diagonalized, though; for instance,

$$\left[\begin{matrix} 1 & 1 \\ 0 & 1 \end {matrix}\right]$$

has eigenvalues 1 and 1, but cannot be diagonalized.

The spectral theorem tells you that in a certain situation, you are guaranteed to be able to diagonalize. Even better, the eigenvectors have some extra structure: they are orthogonal to each other.

If a matrix is diagonalized, its diagonal form is unique, up to a permutation of the diagonal entries. This is because the entries on the diagonal must be all the eigenvalues. For instance,

$$\left[\begin{matrix} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 1 \end {matrix}\right] \text { and }\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 2 \end {matrix}\right]$$

are examples of two different ways to diagonalize the same matrix.

$\endgroup$
  • $\begingroup$ Thanks Eric. I am asking if a matrix can be diagonalized, then if that is unique.. $\endgroup$ – WishingFish Jul 25 '13 at 22:47
  • $\begingroup$ I added to my answer. $\endgroup$ – Eric Auld Jul 25 '13 at 22:54
  • $\begingroup$ What does "up to a permutation of the diagonal entries" mean? You mean up to the $n$, where $n$ is the (maximum) number of permutations? $\endgroup$ – mavavilj Jan 31 '16 at 11:55
  • $\begingroup$ @mavavilj it basically boils down to the fact you can interchange any two eigenvalues on the diagonals. $\endgroup$ – Antinous Feb 13 '18 at 22:30
5
$\begingroup$

Eric's answer is spot on if by "diagonalization" you mean finding invertible $Q$ so that $QAQ^{-1}$ is diagonal. However we may also find unitary $U^\star, V$ so that $U^\star A V$ is diagonal, the singular value decomposition.

Another possibility is if you find $L, D, U$ where $L$ is lower triangular, $D$ is diagonal, $U$ is upper triangular, and $A=LDU$. The LDU decomposition is useful for solving linear systems. If $A$ is not invertible, this doesn't work, but if it is the diagonal $D$ is unique.

$\endgroup$
  • $\begingroup$ And it is uniquely to be the eigenvalues, if the matrix is normal, right? $\endgroup$ – WishingFish Jul 26 '13 at 16:53
  • $\begingroup$ Absolute values of the eigenvalues, yes. $\endgroup$ – vadim123 Jul 26 '13 at 17:28
3
$\begingroup$

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 $U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.