Raising the bar

Solvability of an equation with a square matrix

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

Dissecting the problem

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

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

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

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

(1) $latex BA=BAI=I.$

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

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

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

Collecting the pieces together

Exercise 4. The following conditions are equivalent:

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

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

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

Remark 1. By negating Exercise 4, we see that the following conditions are equivalent:

$latex a^\prime)\ N(A)\neq {0},$  $latex b^\prime)\ \text{Img}(A)\neq R^k,$  $latex c^\prime)\ \det A=0,$  $latex d^\prime)\ A^{-1}$ does not exists.

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

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