Semi-inverting a Non-diagonalizable Matrix

Donald R. Burleson, Ph.D.

© Copyright June 2017 by Donald R. Burleson. All rights reserved.

In an article titled "Semi-inversion of a Diagonalizable Nonsingular
Matrix," found at www.blackmesapress.com/Diagonalizable.htm , after showing
that a diagonalizable (and invertible) matrix A can rather easily be semi-inverted,
i.e. transformed to A^{i }, which if similarly transformed again yields

, by exponentiating the eigenvalues by way of either the canonical diagonalization or the spectral (principal idempotent) decomposition, I mentioned that while for nonsingular matrices the condition of diagonalizability is a sufficient condition for semi-invertibility, it is not in general a necessary condition, as one may find examples of non-diagonalizable matrices that are still semi-invertible. The example I gave there was a simple 2X2 matrix. For larger non-diagonalizable matrices the process of semi-inversion is often more laborious, but I wish here to suggest an approach that can yield the desired results at least in reasonably straightforward circumstances.

Basically this approach consists of taking a non-diagonalizable (and
nonsingular) matrix A and examining its powers A^{2}, A^{3}, etc. to find a pattern, in
terms of functions of n, that one may generalize (and affirm by mathematical
induction) in order to write A^{n} , where one then extends the matrix form to the
condition n = i to form the semi-inverse A^{i}. Since essentially a matrix-valued
semi-inversion operator has to have the property , one then
(to verify the result) repeats the process on A^{i} , examining powers (A^{i})^{2} , (A^{i})^{3}, etc.
to find a pattern by which one may write the matrix (A^{i})^{n} in general as an array of
functions of n, where then one puts n = i to formally produce (A^{i})^{i} = A^{-1} (A to the
power -1) and check to see that this process actually produces the inverse matrix,
which of course one has computed beforehand for reference.

**EXAMPLE 1:**

Let A = with inverse

and with an (algebraic multiplicity 3) eigenvalue

A is not diagonalizable since the triple eigenvalue 1 generates only eigenvectors of
the form (x, 0, 0)^{T} for an eigenspace that is only one-dimensional, where the
dimensions of whatever eigenspaces might have belonged to the matrix A would
have had to add up to 3 for the 3X3 matrix to be diagonalizable. (A is of course
nonsingular since it has no zero eigenvalue.)

We examine a few powers of A:

etc., where one discerns (and easily proves by mathematical induction) that the general pattern gives

so that the natural identification of the desired semi-inverse would formally be

Then to verify that the result, when the semi-inversion operator is applied again,
does indeed produce the ordinary inverse of A, we examine powers of A^{i} , where
by hindsight the elements in the first row, second column of each matrix will be
written in a style that shows the emerging pattern:

so that the pattern gives

The substitution of i for n to produce the expected corresponding elements in the
matrix A^{-1} gives results that are obvious except that one must examine the element
in the position 1,2 when n = i:

so that altogether (A^{i})^{i} = A^{-1} as required.

Clearly this example works rather smoothly because the matrix powers here
do not produce any functions of n that are difficult to discern, but with many other
non-diagonalizable matrices the patterns would be less transparent. Still, this
approach does in principle allow us to compute semi-inverses for many non-diagonalizable matrices, by direct verification that defining the operator as
that operator that maps A to the result of formally replacing n with i in the general
power A^{n} , does in fact lead to More precisely, if the matrix
power A^{n} is made up of functions of n that one may denote as the
matrix elements, e.g. as in the above example, then one
defines the image of the semi-inversion operator functionally as the matrix whose
u,v-th element is provided that for each such function of n it is possible
to evaluate the function at n = i. In the same notation, when one applies the
operator a second time in succession, one produces in each elemental position

, the u,v-th element of A^{-1}. Thus in
practice a non-diagonalizable matrix can be semi-inverted so long as one can write
each of the elements of A^{n} as a particular function of n for which it is possible to
evaluate that function in all cases at n = i.

In the case of a diagonalizable matrix, while the semi-inverse can be computed by diagonalization or principal idempotent decomposition, the "matrix powers pattern identification" method illustrated here in the above example will in principle work as well, yielding the same result.

**EXAMPLE 2:**

Let with eigenvalues

and respective eigenvectors (1, 0, 0)^{T}, (0,1, 0)^{T}, and (2, 0, 1)^{T} which as the columns
will form a modal matrix M, where M and its inverse are then

so that the canonical diagonalization of A is

** **and replacing the diagonal
eigenvalues with their exponentiations 1^{i} = 1, (-1)^{i} = , 2^{i} = 2^{i} and multiplying
the expression out gives the semi-inverse as

**. **But even though the matrix is diagonalizable,
we still could alternatively have employed the "pattern identification" approach.
Observing the first few powers of A we see:

where one readily discerns the elemental patterns as functions of n to be:

and formally putting n = i gives a semi-inverse exactly corresponding to the one obtained by diagonalization. But although the patterns here are easy to determine, they would be more subtle in most cases, and when the matrix is diagonalizable, the easier way by far is to semi-invert it by way of diagonalization and operations on the eigenvalues.

A case in point is the following.

**EXAMPLE 3:**

Consider the matrix A = which is apparently simple in form and only slightly different from a matrix previously examined, and is in fact diagonalizable. In the earlier article at www.blackmesapress.com/Diagonalizable.htm this matrix was easily shown by diagonalization to have a semi-inverse equal to

The "pattern identification" method ultimately works on this matrix but requires identifying functions of n a bit more problematic than the previous example. The first few powers of the matrix A are:

and while some of the patterns for writing A^{n} are obvious, e.g. A^{n}_{2,2} = 2^{n} and

A^{n}_{3,3} = (-1)^{n}, others are not quite so immediate. For the position row 1, column 2,
with successive values 2, 4, 10, 20, 42, 84, ... we need to note that these may be
written respectively 2^{1}, 2^{2}, 2^{1} + 2^{3}, 2^{2} + 2^{4}, 2^{1} + 2^{3} + 2^{5}, 2^{2} + 2^{4} + 2^{6}, etc., or

when n is even and when n is odd, these expressions having been derived using the familiar algebraic identity

We need a single expression for both these cases (even, odd), in order to have a
single function of n where we may later put n = i, and to resolve this parity-generated difference we may write, for both cases together, the common
expression , and when n = i, the expression matches the
corresponding element in A^{i} as determined by diagonalization; in this case
that result is already known and serves as a check.

Similarly, for the elements in position row 3, column 2, with successive
values 1, 3, 5, 11, 21, 43, etc. we may write in general and when
n = 1 this matches with the corresponding element in A^{i}. Ironically, these patterns
as functions of n are suggested by the results produced by diagonalization and one
need only check them to see that they generate the successive sets of values
observed in the "pattern identification" approach, but of course when a matrix is
not diagonalizable to start with, this is not an available option.

The matrix just examined, which fortunately is diagonalizable, is much
more readily semi-inverted by diagonalization and exponentiation of its
eigenvalues. Non-diagonalizable matrices present a less felicitous computational
prospect, and one is encouraged by the fact that with regard to nXn matrices, in the space C^{nXn} with the standard topology, the set of all
non-diagonalizable matrices has Lebesgue measure zero.