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Let $\mathbb{A}$ be one of $\mathbb{Z}$ (the set of all integers), $\mathbb{Q}$ (the set of all rational numbers), $\mathbb{R}$ (the set of all real numbers), or $\mathbb{C}$ (the set of all complex numbers). (You can of course see $\mathbb{A}$ as an arbitrary commutative ring with unity.)

Let $\mathbb{A}^{m \times n}$ be the set of all $m \times n$ matrices whose entries are in $\mathbb{A}$.

The $(i, j)$-entry of $A$ is denoted as $[A]_{i,j}$.

Suppose that $a_1$, $a_2$, $\dots$, $a_n$ are $m \times 1$ matrices. Then $[a_1, a_2, \dots, a_n]$ is the $m \times n$ matrix whose $(i, j)$-entry is equal to $[a_j]_{i,1}$. The piece of notation allows one to display a matrix using its columns.

Determinants are defined here.


I wish to prove the following theorem:

Theorem D. Suppose that $f$: $\mathbb{A}^{n \times n} \to \mathbb{A}$ has the following two properties:

  • (alternating property) Suppose that $a_1$, $a_2$, $\dots$, $a_n$ are in $\mathbb{A}^{n \times 1}$. If there exist two distinct integers $i$, $j$ such that $a_i = a_j$, then $f ({[a_1, a_2, \dots, a_n]}) = 0$.
  • (multilinear property) Suppose that $j$ is a positive integer less than or equal to $n$. Suppose that the $n-1$ matrices $a_1$, $\dots$, $a_{j-1}$, $a_{j+1}$, $\dots$, $a_n$ are in $\mathbb{A}^{n \times 1}$. Suppose that $x$, $y$ are in $\mathbb{A}^{n \times 1}$. Suppose that $s$, $t$ are in $\mathbb{A}$. Then $$ \begin{aligned} & f {([a_1, \dots, a_{j-1}, sx + ty, a_{j+1}, \dots, a_n])} \\ = {} & s f {([a_1, \dots, a_{j-1}, x, a_{j+1}, \dots, a_n])} + t f {([a_1, \dots, a_{j-1}, y, a_{j+1}, \dots, a_n])}. \end{aligned} $$

Then $f(A) = f(I) \det {(A)}$ for any $A \in \mathbb{A}^{n \times n}$, in which $I$ is the $n \times n$ identity matrix.

It is well known that if $f$ has the alternating property, the multilinear property and the property that $f(I) = 1$, then $f$ is just the determinant function; here is a proof based on the Leibniz Formula $$ \det {(A)} = \sum_{\sigma \in S_n} \operatorname{sgn} {(\sigma)} \prod_{j = 1}^n [A]_{\sigma(j),j}, $$ in which $\operatorname{sgn}$ is the sign function of permutations in the permutations group $S_n$. One is able to learn from the proof that if $f$ has the alternating property and the multilinear property, then $f$ must be $f(I)$ times the determinant function, which proves Theorem D.

One can also find a proof of Theorem D here.

One can also find a proof in some linear algebra textbooks.

However, all the proofs that I have seen make use of the Leibniz formula. I wonder whether it is possible to prove the result without the Leibniz formula and without the sign of permutations, so that Theorem D can be introduced in elementary textbooks or courses of linear algebra.


Theorem D is powerful: one can use it to show that $\det {(AB)} = \det {(A)} \det {(B)}$ for any $A$, $B \in \mathbb{A}^{n \times n}$ in an easy manner; one can use it to show that the determinant of the block matrix $$ \begin{bmatrix} A & C \\ 0 & B \\ \end{bmatrix}, $$ in which $A \in \mathbb{A}^{m \times m}$, $B \in \mathbb{A}^{n \times n}$ and $C \in \mathbb{A}^{m \times n}$, equals $\det {(A)} \det {(B)}$, in an easy manner.

Here is a short proof. Define $$ f(A) = \det {\begin{bmatrix} A & C \\ 0 & B \\ \end{bmatrix}}$$ for any $A \in \mathbb{A}^{m \times m}$. One can verify that $f$ has the alternating property and the multilinear property (because of the same properties of the determinant function), so $$ f(A) = \det {\begin{bmatrix} I & C \\ 0 & B \\ \end{bmatrix}} \det {(A)}.$$ Using the alternating property and the multilinear property, one can show that if $Q$ is the square matrix obtained from the square matrix $P$ by multiplying a column by the scalar $k$ and then adding it to another column, then the determinant of $Q$ is equal to that of $P$. Hence $$ f(A) = \det {\begin{bmatrix} I & 0 \\ 0 & B \\ \end{bmatrix}} \det {(A)}.$$ Define $$ g(B) = \det {\begin{bmatrix} I & 0 \\ 0 & B \\ \end{bmatrix}}$$ for any $B \in \mathbb{A}^{n \times n}$. One can verify that $g$ has the alternating property and the multilinear property, so $$ g(B) = \det {\begin{bmatrix} I & 0 \\ 0 & I \\ \end{bmatrix}} \det {(B)} = \det {(B)}.$$ Hence $$ \det {\begin{bmatrix} A & C \\ 0 & B \\ \end{bmatrix}} = \det {(A)} \det {(B)}.$$


My attempt is just the same as the proofs that I have seen:

Suppose that column $j$ of the $n \times n$ identity matrix $I$ is $e_j$. Choose any $A \in \mathbb{A}^{n \times n}$. Suppose that column $j$ of $A$ is $a_j$. One may write $$ a_{k} = [A]_{1,k} e_{1} + [A]_{2,k} e_{2} + \dots + [A]_{n,k} e_{n} = \sum_{i_k = 1}^{n} {[A]_{i_k,k} e_{i_k}}. $$ Hence $$ \begin{align*} f(A) = {} & f([a_1, a_2, \dots, a_n]) \\ = {} & f\left(\left[ \sum_{i_1 = 1}^{n} {[A]_{i_1,1} e_{i_1}, a_2, \dots, a_n} \right]\right) \\ = {} & \sum_{i_1 = 1}^{n} {[A]_{i_1,1}\, f([ e_{i_1}, a_2, \dots, a_n ])} \\ = {} & \sum_{i_1 = 1}^{n} {[A]_{i_1,1}\, f\left(\left[ e_{i_1}, \sum_{i_2 = 1}^{n} [A]_{i_2,2} e_{i_2}, \dots, a_n \right]\right)} \\ = {} & \sum_{i_1 = 1}^{n} { \sum_{i_2 = 1}^{n} {[A]_{i_1,1} [A]_{i_2,2}\, f([ e_{i_1}, e_{i_2}, \dots, a_n ])}} \\ = {} & \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \\ = {} & \sum_{i_1 = 1}^{n} { \sum_{i_2 = 1}^{n} { \dots \sum_{i_n = 1}^{n} { [A]_{i_1,1} [A]_{i_2,2} \dots [A]_{i_n,n} \, f([e_{i_1}, e_{i_2}, \dots, e_{i_n}])}}}. \end{align*} $$ Because of the alternating property, one may write $$ \begin{aligned} f(A) = \sum_{\substack{ 1 \leq i_1, i_2, \dots, i_n \leq n \\ i_1, i_2, \dots, i_n\,\text{are distinct} }} {[A]_{i_1,1} [A]_{i_2,2} \dots [A]_{i_n,n}\, f([e_{i_1}, e_{i_2}, \dots, e_{i_n}])}. \end{aligned} $$

Now I am stuck. The proofs that I have seen use $$ f([e_{i_1}, e_{i_2}, \dots, e_{i_n}]) = \operatorname{sgn} { \begin{pmatrix} 1 & 2 & \dots & n \\ i_1 & i_2 & \dots & i_n \\ \end{pmatrix} } f(I), $$ which follows from the antisymmetric property, which in turn follows from the alternating property and the multilinear property, to conclude that $$ \begin{aligned} f(A) = f(I) \sum_{\sigma \in S_n} \operatorname{sgn} {(\sigma)} \prod_{j = 1}^n [A]_{\sigma(j),j}. \end{aligned} $$ Finally, one uses the Leibniz formula to show that $f(A)$ is $f(I)$ times $\det {(A)}$.

I appreciate any hints.


A postscript at 05:10 on 2023-06-17:

The question can be solved if one allows the use of column echelon form; here is an answer (note that a matrix with entries in $\mathbb{Z}$ is automatically a matrix with entries in $\mathbb{Q}$). I will accept it if there is no better solution. It is a pity that one cannot "accept" more than one answer.

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  • $\begingroup$ How do you define the determinant without the Leibnitz formula. $\endgroup$
    – copper.hat
    Jun 17, 2023 at 0:01
  • $\begingroup$ @copper.hat Here is a definition. $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 0:05
  • $\begingroup$ That is the Leibnitz formula, just defined recursively. $\endgroup$
    – copper.hat
    Jun 17, 2023 at 0:09
  • $\begingroup$ @copper.hat Maybe the term "the Leibniz formula" has more than one meaning; you can just see "the Leibniz formula" in the description of the question as the formula which "explicitly expresses the determinant of a square matrix, without determinants of other matrices". $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 0:14
  • $\begingroup$ You can prove this without directly appealing to the Leibniz formula using the universal property of $\bigwedge^n\mathbb{A}^n$, for example. But you're just sweeping the Leibniz formula under the rug somewhere. $\endgroup$
    – blargoner
    Jun 17, 2023 at 3:32

3 Answers 3

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If you are willing to reduce $A$ to column echelon form, and accept that $\det$ is multilinear and alternating, then there is a proof that is straightforward and avoids the Leibnitz formula.

The idea is simple, just reduce $A$ to upper triangular form.

Suppose the columns of $A$ are $a_k$. We want to compute $f(a_1,...,a_n)$. Note that we have $f(a_1,...,a_k+ \lambda a_p,...,a_n) = f(a_1,...,a_n)$ for any $p \neq k$ and any $\lambda$. Since $\det$ is also multilinear and alternating, we have $\det(a_1,...,a_k+ \lambda a_p,...,a_n) = \det(a_1,...,a_n)$. Similarly, if any two columns are switched, both $f$ and $\det$ change sign.

Now switch columns and add multiples of another column to a column so that the resulting $a_1',...,a_n'$ are in column echelon form. Then we have $f(a_1',...,a_n') = (-1)^k f(a_1,...,a_n)$ and $\det(a_1',...,a_n') = (-1)^k \det(a_1,...,a_n)$, where $k$ is the number of times a pair of columns was switched.

Since $a_1',...,a_n'$ are in column echelon form, we have $f(a_1',...,a_n') = (-1)^k (a_1')_1 ... (a_n')_n f(e_1,...,e_n)$ and similarly, $\det(a_1',...,a_n') = (-1)^k (a_1')_1 ... (a_n')_n $. In particular, we have $f(a_1,...,a_n) = \det(a_1,...,a_n) f(e_1,...,e_n)$.

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  • $\begingroup$ The proof is good. I did not think of column echelon forms, because I saw $\mathbb{A}$ as a general commutative ring (anneau) with $1$, instead of $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$. $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 4:18
  • $\begingroup$ I did not pay attention to the question. The above only works with a field, so $\mathbb{Z}$ would need some extra work. $\endgroup$
    – copper.hat
    Jun 17, 2023 at 4:28
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    $\begingroup$ Your proof is still fine, since $\mathbb{Z}$ is just a subring of $\mathbb{Q}$, which suffices for an elementary linear algebra course. Of course, a proof that directly works with a ring is better. $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 4:30
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    $\begingroup$ I find something interesting. By examining your proof, I realize that it is possible to state a new list of "axioms" of "a constant times $\det$" (in which entries of matrices are in a field $\mathbb{F}$): (a) $f([a_1, \dots, a_k + \lambda a_p, \dots, a_n]) = f([a_1, \dots, a_n])$ for any $p \neq k$ and any $\lambda \in \mathbb{F}$; (b) $f([a_1, \dots, \mu a_k, \dots, a_n]) = \mu f([a_1, \dots, a_n])$ for any $\mu \in \mathbb{F}$. Swapping two columns can be achieved by (a) and (b). The two properties seem to be more known than the multilinear and alternating properties are. $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 10:40
  • $\begingroup$ Also, those conditions tie in directly with notions of (signed) area/volume. $\endgroup$
    – copper.hat
    Jun 17, 2023 at 15:18
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Let $\mathbb{A} = \mathbb{R} $ or $\mathbb{C}$. Consider the $n$-fold anti-symmetric tensor product $(\mathbb{A}^n)^{\otimes_A n}$. Note that this space is one dimensional. The vector $e = \otimes^n_A( e_1, e_2, \dots, e_n)$ is a basis of $(\mathbb{A}^n)^{\otimes_A n}$, where $e_i$ is the $i$-th natural basis vector. Note that by definition the determinant $\det$ is just the map obtained by defining a linear map by $T(e) = 1$ and setting $\det = T \circ \otimes^n_A$.

Since $f$ is an alternating $n$-linear form there exist a unique linear map $\tilde{f}:(\mathbb{A}^n)^{\otimes_A n} \to \mathbb{A}$ so that $f = \tilde{f} \circ \otimes^n_A$ (universal property). Because $(\mathbb{A}^n)^{\otimes_A n}$ is one dimensional $\tilde{f}$ is uniquely determined by its value $\tilde{f}(e) = f(I)$. This means that $\tilde{f} = f(I) T$ and so $f = f(I) \det$.

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  • $\begingroup$ The proof is good, but it is not for those who are just beginners of (elementary) linear algebra. It is for those who take a higher linear algebra course. $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 6:22
  • $\begingroup$ @Juliamisto If the students have studied multi-linear forms it is easy to introduces the tensor product. In my opinion concepts like tensor products, dual space, quotient space, projections, adjoint operators, abstract vector spaces etc. should be introduced as early as possible. They remove many of the tedious and non illuminating aspects of the usual "we prove everything by explicit matrix/vector calculation in $\mathbb{R}^n$" linear algebra. In my opinion the usual beginner linear algebra leaves the students puzzled and dazed since it robs them of any ability to see the big picture. $\endgroup$
    – jd27
    Jun 17, 2023 at 6:48
  • $\begingroup$ You are right. An "elementary/beginner linear algebra" course is... actually a "matrix algebra" course. I have always wanted to teach a "true linear algebra course", but I cannot, since most of my students... do not study math "mathematically". I do not want to teach determinants in the first place, but I still have to. $\endgroup$
    – Juliamisto
    Jun 17, 2023 at 6:59
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Here is a solution by a friend of mine; it is not mine.


Instead of $f(A)$, consider $g(A) = f(A) - f(I) \det {(A)}$ (for any $A \in \mathbb{A}^{n \times n}$). It is easy to show that $g$ has the alternating property and the multilinear property (in which it has been assumed that $\det$ is alternating and multilinear), so according to the attempt in the description of the question, $$ \begin{aligned} g(A) = \sum_{\substack{ 1 \leq i_1, i_2, \dots, i_n \leq n \\ i_1, i_2, \dots, i_n\,\text{are distinct} }} {[A]_{i_1,1} [A]_{i_2,2} \dots [A]_{i_n,n}\, g([e_{i_1}, e_{i_2}, \dots, e_{i_n}])}. \end{aligned} $$

The antisymmetric property can be proved by the alternating property and the multilinear property:

Suppose that $g$: $\mathbb{A}^{n \times n} \to \mathbb{A}$ is alternating and multilinear. Suppose that $B \in \mathbb{A}^{n \times n}$ is the square matrix obtained from $A \in \mathbb{A}^{n \times n}$ by interchanging the location of two columns. Then $g(B) = -g(A)$.

Proof. Suppose that $1 \leq p < q \leq n$. Then $$\begin{aligned} & g([\dots, \overset{\text{column}\, p}{a_p}, \dots, \overset{\text{column}\, q}{a_q}, \dots]) \\ = {} & g([\dots, a_p, \dots, a_q, \dots]) + 0 \\ = {} & g([\dots, a_p, \dots, a_q, \dots]) + g([\dots, a_q, \dots, a_q, \dots]) \\ = {} & g([\dots, a_p + a_q, \dots, a_q, \dots]) \\ = {} & g([\dots, a_p + a_q, \dots, (a_p + a_q) - a_p, \dots]) \\ = {} & g([\dots, a_p + a_q, \dots, a_p + a_q, \dots]) - g([\dots, a_p + a_q, \dots, a_p, \dots]) \\ = {} & 0 - g([\dots, a_p + a_q, \dots, a_p, \dots]) \\ = {} & 0 - g([\dots, a_p, \dots, a_p, \dots]) - g([\dots, a_q, \dots, a_p, \dots]) \\ = {} & {-g([\dots, a_q, \dots, a_p, \dots])}. \end{aligned}$$

Interchanging the locations of the columns of $[e_{i_1}, e_{i_2}, \dots, e_{i_n}]$ yields $$ g([e_{i_1}, e_{i_2}, \dots, e_{i_n}]) = \pm g(I) = \pm (f(I) - f(I) \det {(I)}) = \pm 0 = 0, $$ in which it has been assumed that the determinant of the identity matrix is unity (a proof of which is left as an exercise). Hence $g(A) = 0$, which shows that $f(A) = f(I) \det {(A)}$.


The trick is that it does not matter how many times the sign before zero changes.

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