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To answer the first question, the determinant (or equivalently, elementary row operations) give an upper bound on the machinery required. If $AB = I$, then

$\displaystyle \det(AB) = \det(A)\det(B) = 1.$

In particular, $\det(A) \neq 0$ so $A^{-1}$ exists. Thus,

$\displaystyle B = IB = (A^{-1}A)B = A^{-1}(AB) = A^{-1}I = A^{-1}.$

From the definition of inverse, this implies $BA = I$. The determinant is a bit high-powered, though. Is there a more elementary proof? Somewhat surprisingly, there is! (Kudos to Bill Dubuque, whose answer on the relevant math stackexchange post was the nicest.)

Let $A, B \in M_n(\mathbb{R}) = V$. Since $\dim(V) = n^2$, there is a dependent combination of the $n^2 + 1$ matrices

$\displaystyle I, B, B^2, \ldots, B^{n^2}.$

In other words, there is a polynomial $p \in \mathbb{R}[x]$ of order $\leq n^2$ with $p(B) = 0$. If $p(x) = xq(x)$ for some $q$, then, using $AB = I$, we have

$\displaystyle 0 = p(B) = Bq(B) \quad \Longrightarrow \quad 0 = A\cdot 0 = ABq(B) = q(B).$

So $q(B) = 0$. Continuing in this fashion, we can choose $p(x) = \sum_{k=0}^{n^2}a_kx^k$ such that $p(0) = a_0 \neq 0$. Now, $AB = I$ also implies, for any $n > 0$,

$\displaystyle  (BA - I)B^n = B(AB)B^{n-1} - B^n = B^n - B^n = 0.$

We combine these two as follows:

$\displaystyle \begin{align*} (BA - I)p(B) & = (BA - I)\sum_{k=0}^{n^2}a_kB^k \\ & = (BA - I)a_0I + \sum_{k=1}^{n^2}a_k(BA - I)B^k \\ & = a_0(BA - I) = 0. \end{align*}$

For the last equality, we have used $p(B) = 0$. Since $a_0 \neq 0$, it follows that $BA = I$. $\square$

Pete Clark's post on the same question shows that the finite-dimensionality of $M_n(\mathbb{R})$ is essential. Let's look at the (infinite-dimensional) vector space of polynomials with real coefficients, $\mathbb{R}[x]$. We have the familiar derivative map $D: \mathbb{R}[x] \to \mathbb{R}[x]$, and the "anti-derivative" map $A: \mathbb{R}[x] \to \mathbb{R}[x]$, where we choose $A(0) = 0$ (no constant term). Both maps are linear, and $D \circ A = I$ since


However, $A \circ D \neq I$, since $A \circ D$ kills the constant term:

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I found a great answer to the second question in another StackExchange post. To see the geometric significance of the transpose, we need something called the singular value decomposition (SVD). I won't prove it, but for a square matrix $C \in M_n(\mathbb{R})$, we can always find orthogonal matrices $U, V \in M_n(\mathbb{R})$ such that


for some diagonal matrix $\Lambda = \mbox{diag}(\lambda_1, \ldots, \lambda_n)$. So we can think of the action of $C$ on $\mathbb{R}^n$ as follows:

• Rotate by $V$.
• Scale by factors $\lambda_i$ along the coordinate axes.
• Rotate by $U$.

• Rotate by $U^{-1}$.
• Scale by factors $\lambda_i$ along the coordinate axes.
• Rotate by $V^{-1}$.