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.


2 Answers 2


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 $A$.

Edit: Given the comments on my answer, I'm including a link to a thread about left/right inverses:

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

  • 2
    $\begingroup$ 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, $AB = I = BA$. $\endgroup$
    – Decaf-Math
    Jan 9, 2019 at 19:58
  • $\begingroup$ Either that or append a proof that $AB=I\to BA=I$ for square matrices on finite-dimensional spaces. $\endgroup$
    – J.G.
    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 $$ 0=\det[a_1\ \dots\ a_{n-1}\ 0]= \det A -c_1\det[a_1\ \dots\ a_{n-1}\ a_1] -\dots -c_{n-1}\det[a_1\ \dots\ a_{n-1}\ a_{n-1}] $$

  • $\begingroup$ 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. $\endgroup$ Jan 9, 2019 at 23:02
  • $\begingroup$ @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. $\endgroup$
    – egreg
    Jan 9, 2019 at 23:11

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