If $AB=I$ show that $A$ has full rank.

Question

Let $A,B\in \mathbb R^{n\times n}$. Show that if $AB=I$ then $B$ has full rank.

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In fact, I show that $A$ has full rank, which is quite obvious but I really have difficulties to show that $B$ has full rank. I tried by contradiction : there is $u\in \mathbb R^n$ s.t. $Bv\neq u$ for all $v\in \mathbb R^n$. So $v=ABv\neq Au$ which contradict $A$ has full rank.

It doesn't look correct. What do you think ?

Answer 1

Because if $rank(B)<n$ so $\det B=0$ and $$1=\det(I)=\det(AB)=\det(A)\det(B)=0,$$ which is a contradiction.

Answer 2

There is a missing thing. How do you know that $Bv\neq u\implies ABv\neq Au$? At this point, you need to use the fact that $A\text{ surjective}\implies A\text{ injective}$. Then, since $AB=\operatorname{Id}$, your argument is correct.

Answer 3

$AB = I$ means $B = A^-1$, so $A$ is invertible. A matrix is only invertible if it is full rank, otherwise when you are in the process of inverting it, you will eventually find a row or colum of zeros.

Answer 4

I'll provide a dimensional argument.

Let $(e_1,\dots,e_n)$ be some basis of $\mathbb R^n$. Then $$ rank(B)=dim(Span(Be_1,\dots,Be_n)), $$ and $$ rank(AB)=dim(Span(ABe_1,\dots,ABe_n)). $$ This only uses the linearity of $A$ and $B$. For example if $u$ is such that $Bv=u$ for some $v$, then you can write $u=Bv=B(a_1\cdot e_1 + \dots + a_n\cdot e_n) = a_1\cdot Be_1 + \dots + a_n\cdot Be_n$, where $u_1,\dots,u_n$ are some scalars.

Now you know that $rank(AB)=n$, so $(ABe_1,\dots,ABe_n)$ are linearly independent (there is $n$ of them). That necessarily means that $(Be_1,\dots,Be_n)$ is also linearly independent, since if not, i.e., if $\sum_{i=1}^n(a_iBe_i)=0$ is some non-trivial linear combination yielding $0$, then $0=A0=A(\sum_{i=1}^n(a_iBe_i))=\sum_{i=1}^n(a_iABe_i)$, so also $(ABe_1,\dots,ABe_n)$ would be linearlt dependent. Hence, B has full rank.

$A$ must clearly have full rank, since if there is some vector $u$ not in the image of $A$, then $u$ in particular cannot be in the image of $AB$, because $B$ could only "feed" less then everything to $A$.

Answer 5

If you show that $A$ has full rank, then you are done. Because, $$AB=I\Longrightarrow B^TA^T=I$$ Hence, $B^T$ has full rank (because that is first matrix here) and hence $B$ (as row rank=column rank).

Your argument is correct

Because injectivity of $A$ is implied by its surjectivity. ($A$ is surjective as $$\forall x\in \mathbb R^n\exists Bx :A(Bx)=x$$)

Alternative direct approach

Suppose that $B$ doesn't have full rank. Then, $$\exists x\neq 0 : Bx=0$$ $$\Longrightarrow ABx=0$$ $$\Longrightarrow Ix=0$$ which is a contradicton.

Hope it helps:)