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
(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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