Drawing a conclusion from equal determinants

Question

If, for two square matrices $A$ and $B$ with equal dimension, $A=B$, then clearly $\det(A)=\det(B)$. Other than just providing a counter example, why can we not conclude from $\det(C)=\det(D)$ that $C=D$?

Answer 1

an amusing way to look at this:

suppose that the determinant was injective, then you'd have, with $C:=XY$ and $D:=YX$

$C=XY \mapsto \det\big(XY\big) = \det\big(X\big)\det\big(Y\big) = \det\big(Y\big)\det\big(X\big)= \det\big(YX\big)$ but $D=YX \mapsto \det\big(YX\big) $. Since we assume the determinant is injective (really an isomrophism) and the selection of $\text{n x n}$ $X$ and $Y$ was arbitrary, we've "proven" that all matrix multiplication must commute (which is the contradiction)

Answer 2

Here's some insight that might be helpful. Geometrically, the determinant can be thought of how much a linear transformation "stretches" the space. Let's consider $\mathbb{R}^2$. If we have the matrix $$\begin{bmatrix} 2&0\\ 0&2 \end{bmatrix}$$ it has a determinant of $4$. The matrix can be interpreted as stretching the $x$ and $y$ axes by a factor of two, which means that an original $1\times 1$ "unit" square becomes a $2\times 2$ square.

Now, you can probably think of many linear transformations that have this property. Indeed, consider the matrix $$\begin{bmatrix} -2&0\\ 0&-2 \end{bmatrix}$$ which "flips" the axes and then does the same stretching as the matrix above. The area of a $1\times 1$ square is still going to be $2\times 2$, so the determinant is also $4$.

You might be wondering why $$\begin{bmatrix} -2&0\\ 0&2 \end{bmatrix}$$ doesn't work. This is because only one axis is flipped, so the "orientation" of the space is "odd", which means that the determinant should have a negative sign in front of it, giving us $-4$.

This idea generalizes to the change in area of the unit parallelepiped in higher dimensions.

Answer 3

The space of $n \times n$ real matrices is essentially $\mathbb R^{n^2}$, i.e. $n^2$-tuples of real numbers, just written in a particular arrangement. $\det$ thus corresponds to a continuous function from $\mathbb R^{n^2}$ to $\mathbb R$. There are no one-to-one continuous functions from $\mathbb R^{n^2}$ to $\mathbb R$ if $n > 1$.