While sitting in a determinant class, our Professor, while describing determinants, coined an observation which made me wonder. He stated that the determinant of a square matrix can be represented as a function in the following way -

Let $M$ be the set of all square matrices, then a function, say X, is defined from $M \to \mathbb{R}$ which returns the determinant of any square matrix $C \in M$ i.e. $$ X : M \to \mathbb{R}$$ which returns $|M|$

As soon as he said this, I began to wonder about the injectivity, surjectivity and bijectivity of this function.
I was told that for a given $C \in M$, there exists only one unique determinant $|M| \in \mathbb{R}$. But the converse is not true according to me. This can be proved by the following argument -

Let's say that $$\begin{vmatrix} x &5 \\ x &x\\ \end{vmatrix} = -6 $$ On solving for $x$, we obtain two values i.e $2$ and $3$, therefore the matrix is not unique for the given determinant i.e. for one image ($-6$), there exist multiple preimages. Hence, according to me, the function so defined is not injective, and consequently not bijective as well.

Now checking for surjectivity, we'll have to look for the equality of the codomain and the range of the function. In my opinion, for infinite square matrices, there will be infinite unique determinants belonging to $\mathbb{R}$. Hence the function should be surjective, but only for real entries into the matrix.

For complex entries, I don't think the function remains surjective, as the range will not coincide with the codomain (i.e. $\mathbb{R}$). However, in both cases, I feel that the function is neither injective not bijective.

I would like to know if I am right or not by knowing the views of the math-geeks sitting out there. Thanks!

  • $\begingroup$ "For complex entries, I don't think the function remains surjective, as the range will not coincide with the codomain (i.e. $\mathbb{R}$)." In fact, if you don't change the codomain, then forget surjectivity or injectivity; your function ceases to be a function. $\endgroup$ – Arthur Mar 22 '18 at 8:43

These are good questions to ask of newly-encountered functions.

If $A$ and $B$ are $n \times n$ matrices, the following properties are valid for the determinant function:

  • $\det(cA) = c^n \det(A)$
  • $\det(AB) = \det(A) \det(B)$
  • Swapping any two columns of $A$ changes the sign of its determinant but not the magnitude.

Using these, we can tackle both injectivity and surjectivity.

  • Surjectivity: note that $\det(I) = 1$. Therefore, to arrive at a matrix with determinant $d$, simply multiply the identity matrix by $\sqrt[n]{d}$ and / or swap two of its columns.
  • Injectivity: Let $B$ be any non-identity matrix of determinant $1$, such as a rotation matrix. Then, according to the second rule above, we'll have $\det(AB) = \det(A)\det(B) = \det(A)$.

You can also approach this question by considering the geometric interpretation of the determinant: it gives the (signed) volume of the $n$-dimensional parallelepiped spanned by the column vectors of the matrix. For instance, working in the $2 \times 2$ case, one can see that the determinant cannot be injective because applying a shear transform (or rotation or any other area-preserving transformation) to a parallelogram does not change its area; hence, we can get two unique parallelograms of equal area which correspond to two unique matrices of equal determinant$^\dagger$. Moreover, the determinant is surjective as we can construct parallelograms of any prescribed area (the area function varies continuously as we vary the length of a side). These ideas generalize easily to higher dimensions.

$^\dagger$If $A$ represents the matrix whose columns span the original parallelogram and $B$ is the matrix representation of the shear transformation, then the columns of $BA$ will span the sheared parallelogram. I.e. $\det(A) = \det(BA)$.

  • $\begingroup$ Thanks a lot. That was of great help. But I fail to understand one little thing. If the determinant represents the signed volume of a $n$- dimensional parallelopiped, then how can this volume be a complex number, since the determinant of a matrix can evaluate to a complex number also? $\endgroup$ – Saksham Mar 22 '18 at 15:47
  • $\begingroup$ Ahh, I was only commenting on real-valued matrices. I believe one loses the geometric interpretation in the complex case, but many determinant properties still hold, particularly most (if not all) of the algebraic ones. If you Google "complex determinant geometric" you can find some threads on here with discussion, such as this one. $\endgroup$ – Kaj Hansen Mar 22 '18 at 21:48

You are right in saying that the function is not injective. To show surjectivity, note that for any real $k$, $k^{\frac{1}{n}}I$ has determinant $k$ (Where $I$ is the $n\times n$ identity matrix)


Clearly there are infite matrices with $\det(A)=0$. (Just scale any singular matrix to get an infinite number). Therefore you are right about injectivity and bijectivity. About surjectivity, you must have a matrix $A$ for every $t\in \mathbb{R}$, that has exactly that value das determinant.

You can just use a $1\times 1$ matrix, to set that determinant.

This holds also true for any complex values


Surjectivity follows from considering the matrix $$ \begin{bmatrix} a & 0 & 0 & \dots & 0 & 0 \\ 0 & 1 & 0 & \dots & 0 & 0 \\ 0 & 0 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & 0 \\ 0 & 0 & 0 & \dots & 0 & 1 \end{bmatrix} $$ that is, the identity with just the top left entry changed to $a$ instead of $1$. This has determinant $a$ by multilinearity.

Injectivity can be shown false by considering the identity with the bottom right entry changed into $a$, which has determinant $a$ as well. If your matrices are not $1\times 1$, this falsifies injectivity.

However, the determinant is indeed injective on $1\times1$ matrices, because it's essentially the identity map.


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