28
Jul 18

## Solvability of an equation with a square matrix ## Solvability of an equation with a square matrix

Everywhere in this section we assume that $A$ is square of size $k\times k.$ The main problem is about solvability of the equation $Ax=y$.

### Dissecting the problem

Exercise 1. If $A$ is invertible, then $N(A)=\{0\}$ and $\text{Img}(A)=R^k.$

Proof. If $Ax=0,$ then $x=A^{-1}Ax=0,$ so $N(A)=\{0\}$ (you could also say that $A$ is one-to-one). From $AA^{-1}=I$ we have $AA^{-1}x=x.$ This tells us that $x=Ay$ with $y=A^{-1}x.$ We see that any $x$ belongs to the image of $A,$ so $\text{Img}(A)=R^k.$

Exercise 2. If $N(A)=\{0\},$ then $A$ is invertible.

Proof. Since $A$ is one-to-one, we can use the inverse $f^{-1}$ of $f(x)=Ax$ in the general sense: $f^{-1}(Ax)=x$. It is defined on the image of $A.$ We know that $f^{-1}$ is given by some matrix $B$: $BAx=x$ for all $x.$ Hence, $BAe_i=e_i$ for $i=1,...,k.$ Putting these equations side by side we get

(1) $BA=BAI=I.$

(In detail: the identity matrix is partitioned as $I=(e_1,...,e_k),$ so $BAI$ $=(BAe_1,...,BAe_k)=(e_1,...,e_k)=I$). Thus, $B$ is the left inverse of $A,$ and we've seen before that for square matrices this implies existence of the right inverse, invertibility of $A$ and the equation $B=A^{-1}$.

Exercise 3. If $\text{Img}(A)=R^k,$ then $A$ is invertible.

Proof. If $\text{Img}(A)=R^k,$ then by the second characterization of matrix image $N(A^T)=(\text{Img}(A))^{\perp}=\{0\}.$ Applying Exercise 2 to $A^T,$ we see that the transpose is invertible and hence $\det A=\det A^{T}\neq 0.$ This implies invertibility of $A.$

### Collecting the pieces together

Exercise 4. The following conditions are equivalent:

a) $N(A)=\{0\},$ b) $\text{Img}(A)=R^k,$ c) $\det A\neq 0,$ d) $A$ is invertible.

Proof. The equivalence c) $\Longleftrightarrow$ d) has been established before. For the implication d) $\Longrightarrow$ a)+b) see Exercise 1. By Exercise 2 we have a) $\Longrightarrow$ d). By Exercise 3, b) $\Longrightarrow$ d.

Conclusion. The condition $\det A\neq 0$ is the easiest to check. If it holds, then the equation $Ax=y$ has a unique solution for any $y\in R^k.$

Remark 1. By negating Exercise 4, we see that the following conditions are equivalent: $a^\prime) N(A)\neq \{0\},$ $b^\prime) \text{Img}(A)\neq R^k,$ $c^\prime) \det A=0,$ $d^\prime) A^{-1}$ does not exists.

Remark 2. By going through the proof one can check that the result of Exercise 4 holds if $R^k$ is replaced by $C^k$ (the set of vectors with complex coordinates).