In Artin Algebra 2nd edition page 22, the author proved the uniqueness of determinant by saying that any matrix $A$ can be written in reduced row-echelon form $A'$: $A'=E_1\cdots E_kA$ where $E_i$ are the elementary matrix. Then $A'$ is either $I$ or has a zero row. If $A'=I$, then $\delta(A')=1$. Otherwise, $\delta(A')=0$. In both cases, $\delta(A')$ is determined, and hence by $$\delta(A')=\delta(E_1)\cdots\delta(E_k)\delta(A)$$ $\delta(A)$ is determined uniquely.

However, as he himself pointed out immediately in the following paragraph, the sequence $E_1\cdots E_k$ is not unique. Then why is $\delta(A)$ uniquely determined?

Edit: The author defined determinant as a function $\delta(A)=d\in \mathbb{R}$ satisfying the following 3 conditions:

(i) $\delta(I)=1$

(ii) $\delta$ is linear in the rows of the matrix $A$

(iii) If two adjacent rows of $A$ are equal, then $\delta(A)=0$

He then proved that the above conditions imply some properties that all of us know, e.g.,

(a) Interchanging two rows reverses the sign

(b) If $A$ has a zero row, then $\delta(A)=0$

(c) Multiplying one row by a number and adding it to another row doesn't change the determinant

(d) $\delta(E)=\pm1$ or $c$

(e) $\delta(AB)=\delta(A)\delta(B)$

Then he proved that the function $\delta$ so defined is unique, as shown in the beginning of my post, which I don't understand

  • $\begingroup$ This is still not clear, what is the determinant of $E_i$? $\endgroup$ Jan 29, 2015 at 10:49
  • $\begingroup$ The author has showed that $\delta(E)=\pm 1$ or $c$, and $\delta(EA)=\delta(E)\delta(A)$, which are ok. $\endgroup$
    – velut luna
    Jan 29, 2015 at 10:51
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    $\begingroup$ No, it seems not the most natural definition of determinant. However, he need to show something which correspond to the theorem that saying that the number of transpositions in a factorization of permutaition is always odd or always even. I belive that it have something to do with the number of replacing rows matrices $E_j$ in any decomposition of $A$. $\endgroup$ Jan 29, 2015 at 10:55
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    $\begingroup$ @Kyson The doubt you have pertains to existence of the determinant, not to uniqueness. If the determinant exists, we can compute it by row reduction (that doesn't use the determinant itself), so at most one function with those properties can exist. $\endgroup$
    – egreg
    Jan 29, 2015 at 14:15
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    $\begingroup$ You are misunderstanding Artin's use of the word "uniqueness". What Artin is proving is that if there exist two functions $\delta$ and $\delta'$ satisfying the properties (1.4.7), then $\delta = \delta'$. The question of defining these functions $\delta$ and $\delta'$ is irrelevant here: the assumption is that you are given two such functions, in whatever way. $\delta(A)$ is uniquely determined because $\delta$ is a given function. Finding such a function will be the "existence" part. $\endgroup$ Jan 29, 2015 at 14:33

2 Answers 2


If the determinant $\delta$ exists, you can prove that

  1. $\delta(E)=c$ if $E$ is the elementary matrix corresponding to multiplication of a row by $c$;

  2. $\delta(E)=1$ if $E$ is the elementary matrix corresponding to summing a row to another multiplied by a constant;

  3. $\delta(E)=-1$ if $E$ is the elementary matrix corresponding to switching two rows;

  4. $\delta(A)=0$ if $A$ is not invertible;

  5. $\delta(AB)=\delta(A)\delta(B)$.

If two determinant functions $\delta$ and $\delta'$ exist, then they are both zero on the noninvertible matrices, but, writing an invertible $A$ as $$ A=E_1E_2\dots E_k $$ a product of elementary matrices, we conclude that $$ \delta(A)=\delta(E_1)\delta(E_2)\dots\delta(E_k)=\delta'(A) $$

Note the initial if: we're assuming the existence, not proving it. Also, note that the decomposition into a product of elementary matrices doesn't depend on the determinant; so just use the same decomposition for computing $\delta(A)$ and $\delta'(A)$.

The fact that the decomposition as product of elementary matrices is not unique is indeed a problem, but with respect to the existence of the determinant. In principle you could find two decompositions that produce different values when computed with rules 1–5, but this would simply prove that the determinant doesn't exist.

  • $\begingroup$ Thanks for clearing out the problem I had in common with the one who asked. I found I confused the matter of existence with the matter of uniqueness. $\endgroup$
    – Henry Choi
    Sep 14, 2020 at 12:40

The proof mentioned in the first paragraph of the question is not the proof for uniqueness of the Determinant Function. It says: The determinant of a matrix A, det (A), can be determined from determinant of it's row reduced form det (A'). It just says it can be determined. As you mentioned, the Author also clarifies that since the product of elementary matrices can be written in many ways, uniqueness can't be proved in this way. So, the uniqueness proof is yet incomplete.

When the Author starts off the Determinants topic, he mentions: "The determinants of an nxn matrix is a functions of it's $n^2$ entries. There is one such function for each positive integer n. Unfortunately, there are many formulas for these determinants and all of them are complicated when n is large. Not only are the formulas complicated, but it may not be easy to show that two of them define the same function.

We use the following strategy: We choose one of the formulas, and take it as our definition of the determinant. In that way we are talking about a particular function. We show that our chosen function is the only one having certain special properties. Then, to show that other formula defines the same determinant function, one needs only to check those properties for the other function. This is often not too difficult."

Using the above statements, to prove the uniqueness, the author goes ahead to prove that the chosen standard determinant function (which is the expansion of minors on the first column) satisfies the properties of a Determinant function.

The way this proof is done: Using induction on the size of the matrices. We know that the properties are true for 1x1 matrices. Assuming that the properties are true for determinant of (n-1)x(n-1) matrices, we prove for nxn matrix. This completes the uniqueness proof.


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