I'm solving the following problem:

Let $\Delta : M_n(\mathbb{F}) \to \mathbb{F}$ satisfy the following properties:

  1. $\Delta(AB) = \Delta(A)\Delta(B), \quad\forall A, B \in M_n(\mathbb{F})$
  2. $\Delta(\lambda I) = \lambda^n, \quad\forall \lambda \in \mathbb{F}$

Prove that $\Delta = \det$.

$M_n(\mathbb{F})$ is the space of $n\times n$ matrices with entries from the field $\mathbb{F}$, and $I \in M_n(\mathbb{F})$ is the identity matrix. $\mathbb{F}$ can pretty much be assumed to be $\mathbb{R}$ or $\mathbb{C}$, since it was the general setting where the problem was given.

$\DeclareMathOperator{\sgn}{sgn}$ The definition of the determinant I'm using is $\det A = \sum_{p\in S_n} \sgn p\cdot a_{1p(1)}a_{2p(2)}\cdots a_{np(n)}$, for $A = (a_{ij})$. I'm also aware of certain characterizations, such as:

Let $f : M_n(\mathbb{F}) \to \mathbb{F}$ be a multilinear and alternating function in the columns of the matrix such that $f(I) = 1$. Then $f = \det$.

My attempt:

My idea was to prove that $\Delta = \det$ on all elementary matrices and then use the fact that every invertible matrix can be written as a product of elementary matrices. I would separately prove that $\Delta$ of all singular matrices is $0$.

Some results I obtained along the way (using the notation for elementary matrices from the linked Wikipedia article):

  • By plugging in $\lambda = 1$ in $(2)$ we get $\Delta(I) = 1$.
    $\lambda = 0$ gives $\Delta(0) = 0$.
  • $\Delta$ is homogenous of degree $n$: By pluggin in $\lambda = 1$ in $(2)$ we get $\Delta(I) = 1$. Then for any matrix $A$ we have $\Delta(\lambda A) = \Delta(\lambda I\cdot A) = \Delta(\lambda I)\Delta(A) = \lambda^n \Delta(A)$.
  • Let $A$ be invertible. Then $1 = \Delta(I) = \Delta(AA^{-1}) = \Delta(A)\Delta(A^{-1})$. Thus, $\Delta(A) \ne 0$.
  • ${T_{i,j}}^2 = I$ so $|\Delta(T_{i,j})| = 1$
  • For $\lambda \notin \{0, 1\}$ we have $$\Delta(D_1(\lambda))\,\Delta(D_2(\lambda))\cdots \Delta(D_n(\lambda))) = \Delta(D_1(\lambda)\,D_2(\lambda)\cdots D_n(\lambda)) = \Delta(\lambda I) = \lambda^n$$ so there surely exists $i_0 \in \{1, \ldots, n\}$ such that $|\Delta(D_{i_0}(\lambda))|\ne 1$.
  • $\Delta$ of a matrix with a zero row is $0$.

    Let $A$ be such that its $i$-th row is $0$. Then there exists a sequence of row transpositions $T_1, T_2, \ldots, T_k$ of the type $T_{r,s}$ such that $A$ is transformed to a matrix with its $i_0$-th row is a zero row, where $|\Delta(D_{i_0}(\lambda))|\ne 1$. Now, since multiplying that row with $\lambda \notin \{0, 1\}$ does not change the matrix, by doing the row transpositions again in reverse order we return to the matrix $A$. Then we have $$\Delta(A) = \Delta(T_1\ldots T_{n-1}T_nD_{i_0}(\lambda)T_n\ldots T_2T_1A) = \Delta(T_1)^2\Delta(T_2)^2\cdots\Delta(T_n)^2\Delta(D_{i_0}(\lambda))\Delta(A)$$ By taking the absolute value of both sides, since $|\Delta(T_{r,s})| = 1$ and $|\Delta(D_{i_0}(\lambda))|\ne 1$, we obtain $\Delta(A) = 0$.
  • $\Delta$ of a singular matrix is $0$.

    Let $A$ be singular. Then $A$ can be transformed by row and column operations to a matrix $$E_r = \begin{bmatrix} 1 & & & & \\ & \ddots & & & \\ & & 1 & & & \\ & & & 0 & & \\ & & & & \ddots \\ & & & & & 0 \\ \end{bmatrix}$$ with $r$ nonzero rows, where $r$ is the rank of $A$. Since $\Delta$ of elementary matrices is nonzero because they are invertible, and $\Delta(E_r) = 0$ because there is always a zero row in it, we can conclude that $\Delta(A) = 0$.
  • $L_{i,j}(\lambda)$ can be diagonalized to $I$, and since $\Delta$ is obviously a similarity invariant we have $\Delta(L_{i,j}(\lambda)) = 1$.

This is where I'm stuck, we still have to prove $\Delta(T_{i,j}) = -1$ and $\Delta(D_i(\lambda)) = \lambda$.

I have had some further progress if we assume $\mathbb{F} = \mathbb{R}$:

  • From $|\Delta(T_{i,j})| = 1$ we can conclude $\Delta(T_{i,j})$ = $\pm 1$.
  • Notice that $D_i(\lambda) = T_{i,j}D_j(\lambda)$. Applying $\Delta$ we conclude that $\Delta(D_i(\lambda))$ and $\Delta(D_j(\lambda))$ can differ only by sign. Since we know $\Delta(D_1(\lambda))\,\Delta(D_2(\lambda))\cdots \Delta(D_n(\lambda))) = \lambda^n$, we get $|\Delta(D_i(\lambda))| = |\lambda|$.
  • If we further assume $\lambda > 0$, we have $\Delta(D_i(\lambda)) = \Delta(D_i(\sqrt{\lambda})D_i(\sqrt{\lambda})) = \Delta\left(D_i(\sqrt{\lambda})\right)^2 > 0$. Finally we obtain $\Delta(D_i(\lambda)) = \lambda$.
  • 1
    $\begingroup$ Note that $1 = \Delta(T_{i,j}^2) = \Delta(T_{i,j})^2$, so $\Delta(T_{i,j}) \in \{\pm 1\}$ independent of $\Bbb F$. $\endgroup$ Sep 4, 2017 at 10:26
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    $\begingroup$ The statement isn't true. E.g. when $n=2$ and $\mathbb F=\mathbb R$, $\Delta=|\det|$ will also satisfy all requirements of the problem statement. $\endgroup$
    – user1551
    Sep 4, 2017 at 12:57
  • $\begingroup$ @user1551 You are right. This is strange as I found this problem in a textbook. Does the statement perhaps hold for odd $n \in \mathbb{N}$? $\endgroup$ Sep 4, 2017 at 13:14
  • $\begingroup$ I don't think so. Let $n=3,\ \mathbb F=GF(7),\ S=\{1,2,4\}$ and $T=\{3,5,6\}$. Define $f(0)=0,\ f=1$ on $S$ and $f=6$ on $T$. Then $\Delta(A)=f(\det(A))$ will satisfy the two given conditions, because (a) $S^2=T^2=S,\ 1^2=6^2=1,\ ST=T$ and $(1)(6)=6$, and (b) $S^3=\{1\}$ and $T^3=\{6\}$. $\endgroup$
    – user1551
    Sep 4, 2017 at 15:12
  • $\begingroup$ The matrices $T_{ij}$ can be written as products of $L$'s and $D$'s. $\endgroup$
    – orangeskid
    Sep 5, 2017 at 10:46

1 Answer 1


The general solution to your problem is $f(A) = \phi(\det A)$, where $\phi \colon F \to F$, $\phi(0)=0$, $\phi\colon F \to F$ multiplicative and $\phi(t)^n = t^n$ for all $t \in F$. So, $f(t) = \omega(t) \cdot t$, where $\omega \colon F^{\times} \to \mu_n$ is a morphism of groups from $F^{\times}$ to the group of $n$-th roots of $1$ lying in $F$.

Sketch of the proof:

  1. If $A$, $B$ are conjugate then $f(A) = f(B)$.

  2. If $A = L_{ij}(m)$ then $f(A) = 1$. Indeed, all $L_{ij}(m)$, with $m\ne 0$ are conjugate, and moreover, $L_{ij}(m)^2 = L_{ij}(2m)$ ( distinguish two cases: $\operatorname{char} F \ne 2$ and $\operatorname{char} F =2 $ )

  3. $f(D_i(\lambda)) = f(D_j(\lambda))$ ( conjugate matrices). Moreover, $\prod_{i=1}^n D_i(\lambda) = \lambda \cdot I_n$.

  4. Define $\phi(t) =f(D_i(t))$. Notice that it's multiplicative and $\phi(t)^n = t^n$.

  5. Use that every matrix is a product of matrices of type $D_i$ and $L_{ij}$

$\bf{Added 1:}$ One should probably give examples of such maps $\phi$, like noticed, if $n$ is even, and $F=\mathbb{R}$, $\phi(t) = |t|$ works. However, if say $F= \mathbb{C}$ then $\phi(t) = t$ for all $t$. Indeed, $\phi(t^n) = t^n$ and every element in $\mathbb{C}$ is an $n$-th power.

$\bf{Added 2:}$ Note that $D_i(a) L_{ij}(\lambda) D_i(a)^{-1} = L_{ij}( a \lambda)$, hence the conjugation of all the $L_{ij}(\lambda)$ where $\lambda \ne 0$. Moreover, $L_{ij}(\lambda)^2 = L_{ij}(2 \lambda)$. If $2 \ne 0$ this implies $f(L_{ij}(\lambda))= 1$. But if $2=0$ in the field $F$ then we get $f(L_{ij}(\lambda) = \pm 1 = 1$ again.

(Note that $L_{ij}(\lambda)$ is not conjugate to the indentity if $\lambda \ne 0$.)

You can do some $D$ and $L$ operations on a matrix $T_{ij}$ and get the identity.

  • $\begingroup$ Ok, I have been able to prove $(1)-(4)$, however $(5)$ eludes me: let $A$ be the product of $D_i(\lambda_i)$ and $L_{jk}$ in some order. Then $f(A) = \prod_{i} f(D_i(\lambda_i))\prod_{j,k}f(L_{jk}) = \prod_{i} f(D_i(\lambda_i))= \prod_{i}\phi(\lambda_i) = \phi\left(\prod_{i} \lambda_i\right)$. Is $\prod_{i} \lambda_i = \det A$? Take $J$, the Jordan canonical form of $A$ and $\lambda_i$ the eigenvalues of A. I think that $J$ can be obtained only with $D_i(\lambda_i)$ and $L_{ij}$. Then it would follow $f = \phi \circ \det$. $\endgroup$ Sep 5, 2017 at 11:40
  • $\begingroup$ How would you prove that $\phi(t)$ has to be of the form $\omega(t)\cdot t$ (assuming that is what you meant)? Also, I proved $(2)$ using the fact that $L_{ij}$ is diagonalizable to $I$, and then applying $f$ and using $(1)$. Could you explain your proof of $(2)$? I would prove that $L_{ij}$ are all conjugate because they are conjugate to $I$. (Is there another way? I used the same reasoning for $(3)$). If $\mathrm{char}\,\mathbb{F} \ne 2$ then from $x^2 = x$ and $x\ne0$ follows $x = 1$, right? How would you conclude this if $\mathrm{char}\,\mathbb{F} = 2$? Is $x^2 = 0$? $\endgroup$ Sep 5, 2017 at 11:40
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    $\begingroup$ @mechanodroid; I've added some details. $\endgroup$
    – orangeskid
    Sep 5, 2017 at 15:18
  • $\begingroup$ Ok, thaks a lot for the explanations. Sorry to bother you again, why do you say that $L_{ij}$ is not conjugate to $I$? $L_{ij}$ diagonalizes to $I$ because its characteristic polynomial is $(x - 1)^n$. Also, regarding the field with characteristic $2$, we have $L_{ij}(2\lambda) = L_{ij}(0)$ which is singular, so $f(L_{ij}(\lambda))^2 = 0$. How does this imply $f(L_{ij}(\lambda)) = \pm1$? $\endgroup$ Sep 6, 2017 at 10:16
  • $\begingroup$ $L_{ij}(0) = Id$, and $L_{ij}(\lambda)$ is not diagonalisable for $\lambda \ne 0$. $\endgroup$
    – orangeskid
    Sep 6, 2017 at 14:05

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