# Prove that $A$ is invertible iff $\det(A)\neq 0$ with Cauchy-Binet theorem.

Let $$A$$ a matrix $$n\times n$$ over $$\mathbb{R}$$. I'm trying to prove that A is invertible if and only if $$\det(A)\neq0$$ using the Cauchy-Binet theorem.

I know that the Cauchy -Binet theorem is $$\det(A B)=\det(A)\cdot \det(B)$$

But for now, I couldn't think of any solutions to solve the proof.

Suppose that $$A$$ is invertible. Then there is a matrix $$B$$ such that $$AB=I$$. The Cauchy-Binet theorem then implies that $$1=det(I)=det(AB)=det(A)det(B)$$ so that $$det(A)\neq 0\neq det(B)$$.

Conversely, suppose $$det(A)\neq 0$$. Then $$B=\frac{1}{detA}adj(A)$$ satisfies $$AB=I$$, so that $$A$$ is invertible. Here $$adj(A)$$ is the adjugate of the matrix $$A$$.

If $$AB = I$$ then $$BA = I$$

• In my experience, some professors are incredibly pedantic with the definition of matrix inverse; you need not only specify that there exists $B$ such that $AB = I$, but also that $BA = I$ as well. That is, $AB = I = BA$. Commented Jan 9, 2019 at 19:58
• Either that or append a proof that $AB=I\to BA=I$ for square matrices on finite-dimensional spaces.
– J.G.
Commented Jan 9, 2019 at 21:12

If $$A$$ is invertible, then $$AA^{-1}=I$$ (identity matrix), so $$1=\det I=\det(AA^{-1})=\det A\det(A^{-1})$$ and therefore $$\det A\ne0$$.

The converse doesn't follow from Binet's theorem, but rather from the fact that the determinant is multilinear and alternating on the columns of a matrix.

Fact 1. If $$A$$ has a zero column, then $$\det A=\det A+\det A$$, so $$\det A=0$$.

Fact 2. If $$A$$ has two identical columns, then $$\det A=-\det A$$, by swapping them, so $$\det A=0$$.

Fact 3. If $$A$$ is not invertible, then $$\det A=0$$.

Since we want to show that $$\det A=0$$, possibly with a column swap we can assume that the last column is a linear combination of the other $$n-1$$ columns. Say $$A=[a_1\ \dots\ a_{n-1}\ a_n]$$, with $$a_n=c_1a_1+\dots+c_{n-1}a_{n-1}$$ Then, by multilinearity and facts 1 and 2, we have $$0=\det[a_1\ \dots\ a_{n-1}\ 0]= \det A -c_1\det[a_1\ \dots\ a_{n-1}\ a_1] -\dots -c_{n-1}\det[a_1\ \dots\ a_{n-1}\ a_{n-1}]$$

• A somewhat similar approach for the converse uses the fact that row and column operations can be achieved by multiplying (on the left or right) by elementary matrices. Commented Jan 9, 2019 at 23:02
• @CheerfulParsnip Indeed, in my course I don't mention multilinearity, but rather define the determinant to be “invariant” by elementary column operations (or, equivalently, row operations): multiplying a column by $c$ multiplies the determinant by $c$; adding to a column another column multiplied by $d$ doesn't change the determinant; swapping two columns multiplies the determinant by $-1$. I find this better because the course puts great emphasis on elimination. Commented Jan 9, 2019 at 23:11