For this if and only if statement, I need to prove in both directions. I’ll start with the forward implication. For this direction of the proof, I must assume the matrix is invertible.
Since the matrix is invertible, it row reduces to the identity matrix. (I know this from the inversion algorithm, where I must row reduce the left side of the extended matrix to the identity.) That means it has rank \(n\text{,}\) since each row has a leading one. This means that the image has dimension \(n\) since the dimension of the image is equal to the rank. Using the previous proposition, this fact implies that the kernel has dimension \(0\text{.}\) The only thing with dimension zero is a point, so the kernel is a point. The kernel always includes zero, since zero is also always sent to zero, so the kernel must just be the zero vector.
I have to prove the reverse implication as well. Luckily in this case, I can essentially reverse the same logic. A zero kernel has dimension zero. Therefore the image has to have dimension \(n\text{.}\) Therefore, there must be \(n\) leading ones. Therefore, the matrix must reduce to the identity matrix (the only \(n \times n\) matrix with \(n\) leading ones). Thus, the matrix is invertible.
