Solvability of an equation with a square matrix
Everywhere in this section we assume that is square of size
The main problem is about solvability of the equation
.
Dissecting the problem
Exercise 1. If is invertible, then
and
Proof. If then
so
(you could also say that
is one-to-one). From
we have
This tells us that
with
We see that any
belongs to the image of
so
Exercise 2. If then
is invertible.
Proof. Since is one-to-one, we can use the inverse
of
in the general sense:
. It is defined on the image of
We know that
is given by some matrix
:
for all
Hence,
for
Putting these equations side by side we get
(1)
(In detail: the identity matrix is partitioned as so
). Thus,
is the left inverse of
and we've seen before that for square matrices this implies existence of the right inverse, invertibility of
and the equation
.
Exercise 3. If then
is invertible.
Proof. If then by the second characterization of matrix image
Applying Exercise 2 to
we see that the transpose is invertible and hence
This implies invertibility of
Collecting the pieces together
Exercise 4. The following conditions are equivalent:
a) b)
c)
d)
is invertible.
Proof. The equivalence c) d) has been established before. For the implication d)
a)+b) see Exercise 1. By Exercise 2 we have a)
d). By Exercise 3, b)
d.
Conclusion. The condition is the easiest to check. If it holds, then the equation
has a unique solution for any
Remark 1. By negating Exercise 4, we see that the following conditions are equivalent:
does not exists.
Remark 2. By going through the proof one can check that the result of Exercise 4 holds if is replaced by
(the set of vectors with complex coordinates).
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