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Assuming I've tested for diagonalization, can I just take the eigenvalues and arbitrarily place them in in the i,j cells to produce a diagonal matrix?

Say I have a matrix $M$ with eigenvalues $\lambda_1 = 4,\; \lambda_2 = \lambda_3 = -2.$

$$M =\begin{bmatrix}{1} & {-3} & {3} \\ {3} & {-5} & {3} \\ {6} & {-6} & {4}\end{bmatrix}$$

And let,

$$A = \begin{bmatrix}4&0&0\\0&-2&0\\0&0&-2\end{bmatrix} , \quad B = \begin{bmatrix}-2&0&0\\0&4&0\\0&0&-2\end{bmatrix} , \quad C = \begin{bmatrix}-2&0&0\\0&-2&0\\0&0&4\end{bmatrix}$$

Are $A , B , C$ valid diagonal matrices , or does the order of the eigenvalues matter?

Hanul Jeon
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Cody Smith
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  • Did my best to add a reasonable example. – Cody Smith Apr 21 '16 at 01:41
  • Thanks for editing! So yes, you can input the eigenvalues in whichever order you want along the diagonal. However if you are diagonalizing, you need to keep in mind that the eigenvectors have to match the placement of the eigenvalues. – Cameron Williams Apr 21 '16 at 01:42
  • Such diagonal matrices are orthogonally similar, so they equally demonstrate the diagonalizability of the original matrix. – hardmath Apr 21 '16 at 01:54
  • So if I want to diagonalize a matrix, I don't actually have to compute P inverse, I can just find the eigenvalues and generate one of those matrices? – Cody Smith Apr 21 '16 at 02:02
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    Yep. You got it. That is one thing I do not like about the way diagonalization is often presented in texts. The diagonal matrix is the diagonal matrix of eigenvalues. – Cameron Williams Apr 21 '16 at 02:06

2 Answers2

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Yes. Assuming that your matrix is in fact diagonalizable (which will happen if all of the eigenvalues are distinct, but can also sometimes happen when you have repeated eigenvalues), then your matrix will be similar to ANY diagonal matrix that has the eigenvalues (with proper multiplicities) along the diagonal.

One way to see this is to look at what happens when you conjugate a matrix by a permutation matrix, that is, a matrix where every row and every column has exactly one nonzero entry, and that entry is equal to 1. Doing so just swaps rows and columns around (and doesn't change the values of the entries of the matrix), and does so in a way that are along the diagonal remain on the diagonal.

If you play around with conjugation by permutation matrices, you should be able to come up with an explicit way to conjugate $\pmatrix{a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c}$ into $\pmatrix{c & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & b}$, for example, and if you can figure out that, you will see how every diagonal matrix with a given set of entries is similar to each other.

Aaron
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    It may be worth mentioning that, by convention, when you have repeated eigenvalues, people generally put the repeated values consecutively, and when you have non-diagonalizable matrices (so that you use some other normal form, for example Jordan normal form, you will be forced to have at least some of the repeated diagonal entries be consecutive. However, for the diagonalizable matrices, this is only convention and not strictly necessary. – Aaron Apr 21 '16 at 02:17
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    Additionally, when your eigenvalues are all real (for example, with a symmetric matrix), convention states that you should order the eigenvalues by size. Again, this is only a convention. – Aaron Apr 21 '16 at 02:18
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Yes.

Let $A : V \to V$ be a linear transformation that has diagonal matrix representation $$D = \left( \begin{array}{ccccc} \lambda_1 & & \huge0 \\ & \ddots & \\ \huge0 & & \lambda_n \\ \end{array} \right)$$

with respect to some basis $e_1, ..., e_n$. Then for all $k$, $$ D e_k = \lambda_k e_k$$ that is each $\lambda_k$ is an eigenvalue.

Furthermore the ordering of the $\lambda$s is determined by the ordering of the basis which is arbitrary.

Bysshed
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