Jordan normal form: 1’s above or below diagonal?

Given a square complex matrix A, the Jordan normal form of A is a matrix J such that

$P^{-1}A P = J$

and J has a particular form. The eigenvalues of A are along the diagonal of J, and the elements above the diagonal are 0s or 1s. There’s a particular pattern to the 1s, giving the matrix J a block structure, but that’s not the focus of this post.

Some books say a Jordan matrix J has the eigenvalues of A along the diagonal and 0s and 1s below the diagonal.

So we have two definitions. Both agree that the non-zero elements of J are confined to the main diagonal and an adjacent diagonal, but they disagree on whether the secondary diagonal is above or below the main diagonal. It’s my impression that placing the 1s below the main diagonal is an older convention. See, for example, [1]. Now I believe it’s more common to put the 1s above the main diagonal.

How are these two conventions related and how might you move back and forth between them?

It’s often harmless to think of linear transformations and matrices as being interchangeable, but for a moment we need to distinguish them. Let T be a linear transformation and let A be the matrix that represents T with respect to the basis

$e_1, e_2, e_3, \ldots, e_N$

Now suppose we represent T by a new basis consisting of the same vectors but in the opposite order.

$\bar{e}_i = e_{N-i+1}$

If we reverse the rows and columns of A then we have the matrix for the representation of T with respect to the new basis.

So if J is a matrix with the eigenvalues of A along the diagonal and 0s and 1s above the diagonal, and we reverse the order of our basis, then we get a new matrix J′ with the eigenvalues of A along the diagonal (though in the opposite order) and 0s and 1s below the diagonal. So J and J′ represent the same linear transformation with respect to different bases.

Matrix calculations

Let R be the matrix formed by starting with the identity matrix I and reversing all the rows. So while I has 1s along the NW-SE diagonal, R has 1s along the SW-NE diagonal.

Reversing the rows of A is the same as multiplying A by R on the right.

Reversing the columns of A is the same as multiplying A by R on the left.

Here’s a 3 by 3 example:

$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} i & h & g \\ f & e & d \\ c & b & a \end{pmatrix}$

Note that the matrix R is its own inverse. So if we have

$P^{-1}A P = J$

then we can multiply both sides on the left and right by R.

$RP^{-1}APR = RJR$

If J has 1s above the main diagonal, then RJR has 1s below the main diagonal. And if J has 1’s below the main diagonal, RJR has 1s above the main diagonal.

Since R is its own inverse, we have

$RP^{-1}APR = R^{-1}P^{-1}APR = (PR)^{-1}A(PR)$

This says that if the similarity transform by P puts A into Jordan form with 1’s above (below) the diagonal, then the similarity transform by PR puts A into Jordan form with 1’s below (above) the diagonal.

[1] Hirsch and Smale. Differential Equations, Dynamical Systems, and Linear Algebra. 1974.

One thought on “Jordan normal form: 1’s above or below diagonal?”

Jonathan

5 September 2023 at 10:59

This post seems like a good place to mention something that was only just pointed out to me: everyone always says “matrix w.r.t. a basis”, but it really is a “matrix w.r.t. an *ordered* basis”.

One could try to argue that a basis comes with an implicit order, but it’s certainly not part of the definition. And we do often sneak one in when we write the basis out as an ordered list, indexed by {1,2,….n}.

It was kind of shocking to me to realize I’ve been making this implicit assumption of an order for decades. But the readers should go back and read the sentence “Now suppose we represent T by a new basis consisting of the same vectors but in the opposite order.” and ask themselves if and why that constitutes a “new basis”.

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