Deriving the determinant from homogeneity and multiplicativity

Traditionally the determinant is defined using its action on identity and its multilinear and alternating properties, but the latter two are formulated in terms of rows of the matrix, so it is basis dependent. These properties are also rather arbitrary, although it can be motivated by properties of oriented areas spanned by the row vectors. Another common derivation is to simply define it using the Leibniz formula, but then the definition is completely arbitrary.

Instead, in this note we derive it using its action on a multiplication of a matrix. Matrix can be multiplied by either a scalar ($\lambda A$) or by another matrix ($A B$). We define the determinant by its behavior in these two cases, as a function $f$ of a matrix into scalars, and we want this behavior to be distributive. The precise behavior that we assume is $f(A B) = f(A) f(B)$ (homomorphism) and $f(\lambda A) = f(\lambda I) f(A) = \lambda^n f(A)$ (homogeneity), in the second case we really only need to assume how it behaves on the identity matrix multiplied by a scalar: $f(\lambda I) = \lambda^n$. Both definitions are basis independent, just a very natural distributivity in the simplest possible way. We do not assume anything else.

Note: if we assumed just $f(\lambda I) = \lambda$, then for $n\ge2$ there is no such multiplicative function $f$ at all (for $n=1$ it trivially works, since $\lambda^n=\lambda$). The homomorphism property is stronger and forces the $\lambda$-dependence to factor through $\lambda^n$. Specifically, in the steps 1.-3. below we show that homomorphism alone implies $f(\lambda I) = g(\lambda^n)$, where $g(w)$ is a homomorphism $C^* \to C^*$, which can be fully solved with some extra regularity assumptions (continuous or measurable):

Only now we need to make some additional choice other than the original homomorphism of $f$ and the extra regularity assumption. The trivial choice $g\equiv1$ gives the degenerate $f\equiv1$, so the simplest nontrivial choice is the identity $g(w) = w$, which implies $f(\lambda I) = \lambda^n$ (homogeneity); and if we assume just the homogeneity we do not even need to assume continuity. So in this note we simply assume homogeneity in this form, and derive everything else, but it is good to understand that the homomorphism is the strong main assumption, and it heavily restricts the possible forms of the homogeneity assumption, and we choose the simplest possible form given the restrictions, in the above sense.

This note is strictly bottom-up. We assume only that a function $f$ on invertible complex matrices is

• a homomorphism, and

• homogeneous,

and from these two properties alone we derive, step by step, that $f$ must be given by the Leibniz formula. Nothing about the determinant is assumed in advance; each step uses only the steps before it.

Assumptions¶

• $C^* = C\setminus\{0\}$ under multiplication; $M_n(C)$ all complex $n\times n$ matrices; $GL(n,C)$ the invertible ones, a group under multiplication.

• $\operatorname{diag}(x_1,\dots,x_n)$ is the diagonal matrix with the listed entries.

We are given a function $f:GL(n,C)\to C^*$ such that:

(H1) Multiplicativity. $f(AB)=f(A)\,f(B)$ for all $A,B\in GL(n,C)$.

(H2) Homogeneity. $f(\lambda I)=\lambda^{n}$ for all $\lambda\in C^*$.

We use one background fact about $C$:

(R) Roots. Every $w\in C^*$ has an $n$-th root: some $\lambda\in C^*$ with $\lambda^n=w$.

We also use only elementary language about rows, spans, and multilinear functions. We do not assume the determinant, the Leibniz formula, the polar decomposition, or the spectral theorem.

Throughout, $e_1,\dots,e_n$ is the standard basis, and we view a matrix $A$ as its list of rows $r_1,\dots,r_n\in C^n$. The matrix unit $E_{ij}$ is the $n\times n$ matrix with a 1 in row $i$, column $j$, and 0 everywhere else. We write $T_{ij}(c)=I+c\,E_{ij}$ (for $i\neq j$) for the transvection that adds $c$ times row $j$ to row $i$ under left multiplication.

The permutation matrix swapping indicies $i$ and $j$ can be written using transvections as:

It can be shown that $S_{ij} \in GL(n, C)$ and $S^{-1}_{ij} = S_{ji}$.

Remark (the swap via transvections --- might not be needed). Since $\det P_i=-1$ while $\det T_{ij}(c)=1$, the permutation $P_i$ is not itself a product of transvections, but the signed swap

is (this is the $S_{ij}$ identity of (9)/the Appendix on coordinates $\{1,i\}$, where the $2\times2$ block is $\big(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\big)$). One may use $W_{1i}$ in place of $P_i$ in (4): they differ by the diagonal matrix $D=\operatorname{diag}(1,\dots,\underset{i}{-1},\dots,1)$, which commutes with any diagonal matrix, so $W_{1i}\operatorname{diag}(x,1,\dots,1)W_{1i}^{-1}=D\,\big(P_i\operatorname{diag}(x,1,\dots,1)P_i^{-1}\big)D^{-1}=\operatorname{diag}(1,\dots,\underset{i}{x},\dots,1)$, the $D,D^{-1}$ cancelling because the middle factor is diagonal.

Step 1 — $f(I)=1$ and conjugation invariance¶

From $I\cdot I=I$ and (H1): $f(I)=f(I)^2$. Since $f(I)\in C^*$ is nonzero, $f(I)=1$.

For any $P\in GL(n,C)$, using (H1) and the commutativity of $C^*$,

So $f$ is invariant under conjugation.

Step 2 — $f(T_{ij}(c))$ is 1 on every transvection $T_{ij}(c)$¶

Fix $i\neq j$. Two observations.

All nonzero $c$ give conjugate transvections. For a diagonal $D=\operatorname{diag}(d_1,\dots,d_n)$ one computes $D\,T_{ij}(c)\,D^{-1}=T_{ij}\!\big(c\,d_i/d_j\big)$. Choosing $d_i/d_j$ freely, the factor $c\,d_i/d_j$ ranges over all of $C^*$. Hence all $T_{ij}(c)$ with $c\neq 0$ are conjugate, so by (1) the value $f(T_{ij}(c))=:t$ is the same for every $c\neq 0$.

Better: all matrices $T_{ij}(c)$ for all $c\neq 0$ are mutually conjugate. From Step 1 it follows that $f(T_{ij}(c))=t$ is constant for all $c\neq 0$.

The values multiply additively. Since $e_{ij}^2=0$, we have $T_{ij}(c)\,T_{ij}(c')=T_{ij}(c+c')$, so by (H1) the map $c\mapsto f(T_{ij}(c))$ turns addition into multiplication. Taking $c=c'=1$: $t=f(T_{ij}(2))=f(T_{ij}(1))^2=t^2$. As $t\in C^*$, this forces $t=1$.

Therefore

In particular, by (H1), left- or right-multiplying by a transvection does not change $f$ — i.e. adding a multiple of one row (or column) to another leaves $f$ unchanged.

Step 3 — The one-variable function $g$, and $f$ on diagonal matrices¶

Define

$g$ is a homomorphism $C^*\to C^*$: from $\operatorname{diag}(xy,1,\dots,1)=\operatorname{diag}(x,1,\dots,1)\operatorname{diag}(y,1,\dots,1)$ and (H1),

Position does not matter. Let $P_i=S_{1i}$ be the permutation matrix swapping coordinates 1 and $i$ (so $P_i\in GL(n,C)$, $P_i^{-1}=P_i^{T}$). Explicitly $P_i=S_{1i}=I-e_{11}-e_{ii}+e_{1i}+e_{i1}$: the identity with rows (and columns) 1 and $i$ interchanged. This moves $x$ from slot 1 to slot $i$. Then $\operatorname{diag}(1,\dots,\underset{i}{x},\dots,1)=P_i\operatorname{diag}(x,1,\dots,1)P_i^{-1}$, so by (1)

Note: check if the $P^{-1}_i$ should be there or not. We assume it is, and so applying $f$ on the expression above cancels it. Let’s make it explicit.

Note: Conjugating any diagonal matrix by a permutation matrix permutes its diagonal entries (explicitly: $P_\sigma\operatorname{diag}(d_1,\dots,d_n)P_\sigma^{-1}=\operatorname{diag}(d_{\sigma^{-1}(1)},\dots,d_{\sigma^{-1}(n)})$);

Product over the diagonal. Writing $\operatorname{diag}(d_1,\dots,d_n)=\prod_{i=1}^n\operatorname{diag}(1,\dots,\underset{i}{d_i},\dots,1)$ and using (H1), (4), (3),

Layer 1 — the factorization theorem $f=g\circ\det$ (multiplicativity only)¶

Everything so far (Steps 1–3) used only that $f$ is a homomorphism (H1); homogeneity has not yet been touched. Three facts are all we shall use:

• (H1) $f$ is a homomorphism $GL(n,C)\to C^*$;

• (2) $f\big(T_{ij}(c)\big)=1$ for every transvection;

• (5) $f\big(\operatorname{diag}(d_1,\dots,d_n)\big)=g\!\big(\prod_i d_i\big)$, with $g:C^*\to C^*$ the homomorphism of Step 3.

We show these force $f=g\circ\det$, the determinant being supplied — single-valued — by the Leibniz polynomial. We work entirely inside $GL$, never evaluating $f$ on a singular matrix, with no appeal to homogeneity, continuity, exterior algebra, or Zariski density. (The optional Steps 5–7 below develop the classical multilinear/alternating viewpoint of $f$; the present derivation does not use them. The one ingredient missing for $f$ on $GL$ — the alternating property — enters here only as a formal identity about a polynomial, where the singular-matrix obstruction simply does not arise.)

1. Transvections and diagonal matrices generate $GL$. Lemma. Every $A\in GL(n,C)$ factors as

Proof. Row-reduce $A$ by Gaussian elimination. Adding $c$ times row $j$ to row $i$ is left multiplication by $T_{ij}(c)$. Should a pivot vanish, some entry in its column is nonzero (the columns of an invertible matrix are independent), and the row swap that brings it up is the purely algebraic identity (verified in the Appendix)

— again transvections times a diagonal. So a product of transvections and diagonal matrices reduces $A$ to a diagonal matrix. Finally, a diagonal matrix conjugates a transvection to a transvection, $D\,T_{ij}(c)\,D^{-1}=T_{ij}(c\,d_i/d_j)$ (Step 2), so every diagonal factor can be slid to the right past the transvections, collecting into a single $D$ and leaving the form (19):

$\square$

2. A concrete anchor: the Leibniz polynomial. Before drawing any conclusion about the abstract $f$, exhibit one explicit function we can evaluate unambiguously. Let

a polynomial defined for every matrix by this formula — hence single-valued by construction. The following are formal identities in the entries, valid for all $A$ (singular included) and checked directly on the sum:

• (a) $L(I)=1$, and more generally $L\big(\operatorname{diag}(d_1,\dots,d_n)\big)=\prod_i d_i$ — only $\sigma=\mathrm{id}$ survives;

• (b) two equal rows $\Rightarrow L=0$: pairing each $\sigma$ with $\sigma\circ(\text{transposition of the two equal rows})$ cancels the sum term by term (using $\operatorname{char}C=0$);

• (c) $L$ is linear in each row — every summand is.

From (b) and (c), adding a multiple of one row to another leaves $L$ unchanged, $L\big(T_{ij}(c)\,X\big)=L(X)$; taking $X=I$ and using (a) gives in particular $L(T_{ij}(c))=1$. Thus $L$ satisfies $L(T_{ij}(c))=1$ and $L(\operatorname{diag}(d_1,\dots,d_n))=\prod_i d_i$ outright — and, unlike the abstract $f$ on $GL$, with no domain caveat, because $L$ is everywhere-defined and (b) is a genuine polynomial identity even though “two equal rows” is a singular configuration. The obstruction that blocked the alternating property for $f$ has been transferred to $L$, where it evaporates.

Feed $L$ a factorization (19). Peeling the transvections of $E$ off one at a time by $L(T_{ij}(c)X)=L(X)$, then applying (a),

3. $\prod_i d_i$ is well-defined. A priori the product in (20) could depend on the factorization (19) chosen, and the concrete $L$ is exactly what rules this out. If a second factorization $A=E'D'$ gave a different product $\prod_i d_i'$, then (20) applied to it would read $L(A)=\prod_i d_i'$ as well; but $L(A)$ is one unambiguous number, so $\prod_i d_i=\prod_i d_i'$. Hence every factorization of $A$ yields the same product, an honest function of $A$ alone, which we write $p(A):=\prod_i d_i$. (Note we proved independence using the single-valued $L$ — never assuming it beforehand.)

4. Every homomorphism $f$ equals $g\circ\det$. Define $\det:=L$, the everywhere-defined Leibniz polynomial. Let $f$ be any homomorphism — the given $f$ of Steps 1–3 is one such. For any factorization (19), using (H1), then (2) (so $f(E)=1$), then (5) (so $f(D)=g(\prod_i d_i)$),

where $p(A)=\prod_i d_i$ is unambiguous by part 3 and equals $L(A)=\det A$ by (20). This is the factorization theorem:

for some homomorphism $g:C^*\to C^*$ — derived from multiplicativity alone, with no homogeneity, continuity, measurability, or Zariski density. The grounding is fully transparent: the single-valued determinant is the explicit polynomial $L$ (part 2), well-definedness of $p(A)=\det A$ is anchored on $L$ (part 3), and the factorization $f=g\circ\det$ is part 4.

Conversely, every $g\circ\det$ with $g\in\operatorname{Hom}(C^*,C^*)$ is a homomorphism $GL(n,C)\to C^*$, so these are exactly all of them: the determinant is the universal homomorphism, and every other is a “rescaling” $g$ of it. The factor $g$ is genuinely free here — continuity or measurability would narrow it to $g(w)=|w|^s(w/|w|)^k$, and homogeneity (Layer 2) pins it to the identity.

Layer 2 — homogeneity selects $g=\mathrm{id}$, hence $f=\det$¶

Layer 1 leaves exactly one degree of freedom, the homomorphism $g$; homogeneity removes it. Apply the factorization theorem to $\lambda I$, whose determinant is $\det(\lambda I)=\lambda^n$, and compare with (H2):

Given any $w\in C^*$, pick an $n$-th root $\lambda$ with $\lambda^n=w$ (fact (R)); then $g(w)=g(\lambda^n)=\lambda^n=w$. Hence

and the factorization theorem collapses to

So $f$ is the determinant, and the unique homogeneous homomorphism.

This last step is the only place $C$ rather than $R$ is used: it needs every $w\in C^*$ to be an $n$-th power. Over $R^*$ it fails — e.g. $w=-1$ has no real $\lambda$ with $\lambda^2=w$ — so over $R$ with even $n$ the value $g(-1)=\pm1$ stays free, and the choice $g(-1)=+1$ gives a second, sign-blind solution (the permanent-like $|\det|$).

Two consequences of $g=\mathrm{id}$, used in the optional sections below: from (5) and (6),

and, since left-multiplying by $\operatorname{diag}(1,\dots,\underset{i}{\lambda},\dots,1)$ scales row $i$ by $\lambda$, scaling a single row by $\lambda$ multiplies $f$ by $\lambda$:

Optional — the multilinear and alternating viewpoint¶

The two sections that follow (Steps 5–6) develop the classical multilinear and sign properties of $f$, and Step 7 expands them into the Leibniz formula. None of this is used by Layers 1–2 above — it is the traditional textbook route, recorded for context and referenced by the Lie-theoretic correspondence in Part III. A reader after the shortest path may stop here.

Step 5 — $f$ is multilinear in the rows¶

Fix all rows of an invertible matrix except row $i$; call the others $r_1,\dots,\widehat{r_i},\dots,r_n$ (they are linearly independent, since the matrix is invertible). View $f$ as a function of the single varying row $r_i$.

• By Step 2, adding any multiple of another row $r_k$ ($k\neq i$) to $r_i$ leaves $f$ unchanged. Hence $f$ depends on $r_i$ only through its class modulo $V:=\operatorname{span}\{r_k:k\neq i\}$. Because the other rows are independent, $\dim V=n-1$, so the quotient $C^n/V$ is one-dimensional, and the matrix is invertible exactly when the class of $r_i$ is nonzero.

• By (7), $f$ is homogeneous of degree 1 in $r_i$: replacing $r_i$ by $\lambda r_i$ multiplies $f$ by $\lambda$.

A function on the one-dimensional space $C^n/V$ that is homogeneous of degree 1 is linear. Concretely, fix a nonzero class $z_0$ and let $c$ be the value of $f$ there. Every nonzero class is $t\,z_0$ for a unique scalar $t\in C^*$, and degree-1 homogeneity forces the value to be exactly $t\,c$. This rule $t\,z_0\mapsto t\,c$ is linear by the field arithmetic of $C$ (no Hamel-basis pathologies can arise in dimension one). Pulling back, $f$ is an additive (indeed linear) function of $r_i$:

Since $i$ was arbitrary, $f$ is multilinear in the rows. (8)

Step 6 — Swapping two rows changes the sign¶

Work in the plane of coordinates $i,j$ (everything else is left fixed). The factorization

holds (it reduces to the $2\times2$ identity verified in the Appendix), where $S_{ij}$ is the permutation matrix swapping coordinates $i$ and $j$. By (H1), (2) and (5$'$)/(6),

Since left-multiplication by $S_{ij}$ swaps rows $i$ and $j$, (9) and (H1) say: for every invertible $A$, swapping two of its rows multiplies $f$ by -1,

This antisymmetry is a genuine statement about $GL$: a swap sends invertible matrices to invertible matrices. One is tempted to conclude at once that $f=0$ whenever two rows coincide — equate the two rows, swap them, and $f=-f$. But a matrix with two equal rows is singular, where $f$ is not defined, so the alternating property is not available on $GL$; Layer 1 above supplies it in the only place it is needed, as a formal identity about the Leibniz polynomial. Finally, any permutation matrix $P_\sigma$ is a product of swaps, so by (9)

(Here is the only place the choice of branch mattered: (9) used $g(-1)=-1$ from Layer 2. Had $g(-1)=+1$ — the situation that survives over $R$ for even $n$ — we would get the sign-blind “permanent” instead. Homogeneity, through (6), is exactly what selects the alternating sign.)

Step 7 — The same formula via the classical basis expansion (optional)¶

Layers 1–2 above already derived the boxed formula rigorously and elementarily. The familiar textbook route reaches it a different way — by expanding $f$ multilinearly over the standard basis — and is worth recording, with one caveat that Layer 1 was designed to handle.

Multilinearity (8) lets one expand each row $r_i=\sum_{j}A_{ij}\,e_j$:

The surviving terms are those where $(j_1,\dots,j_n)=(\sigma(1),\dots,\sigma(n))$ for a permutation $\sigma$, contributing $f(P_\sigma)=\operatorname{sgn}(\sigma)$ by (11), and one recovers

the Leibniz formula, in agreement with Layers 1–2.

A word on the basis expansion. The intermediate tuples $f(e_{j_1},\dots,e_{j_n})$ with a repeated index correspond to singular matrices, where $f$ was never defined — so the step “these terms vanish” cannot be justified by the abstract $f$ alone (this is the same gap that blocked the alternating property in Step 6). The clean fix is the one used in Layer 1: read every term as a value of the everywhere-defined polynomial $L(A)=\sum_\sigma\operatorname{sgn}(\sigma)\prod_i A_{i,\sigma(i)}$, whose vanishing on repeated indices is the formal identity (b) of Layer 1 (part 2), with no domain caveat. The expansion is then a convenient mnemonic for the rigorous generating-set derivation above — and needs neither continuity nor Zariski density.

Conclusion¶

The argument splits into two independent layers.

Layer 1 — factorization (multiplicativity only). Every homomorphism $f:GL(n,C)\to C^*$ has the form $f(A)=g(\det A)$ for some homomorphism $g:C^*\to C^*$, where $\det$ is the Leibniz polynomial. No homogeneity, continuity, or measurability is used; conversely every $g\circ\det$ is a homomorphism, so these are all of them.

Layer 2 — normalization (adding homogeneity). If in addition $f(\lambda I)=\lambda^n$, then $g=\mathrm{id}$, so $f=\det$: the determinant is the unique homogeneous homomorphism.

Two remarks:

• No regularity, no spectral theory. We never assumed $f$ continuous or measurable, and never used eigenvalues, the polar decomposition, or the determinant itself. Layer 1 came from (H1) alone; Layer 2 added (H2) and one use of (R).

• Where $C$ was essential. Homogeneity is needed only to select $g=\mathrm{id}$, and that is the single step (Layer 2) requiring $C$ rather than $R$: every $w\in C^*$ — in particular $-1=i^2$ — has an $n$-th root, which forces $g=\mathrm{id}$ and hence $g(-1)=-1$, the sign in Step 6. Over $R$ with even $n$ the value -1 has no real $n$-th root, $g(-1)=\pm1$ stays free, and the choice $g(-1)=+1$ yields a second, sign-blind solution (the permanent-like $|\det|$). Over $C$ that ambiguity cannot occur.

Appendix — The swap factorization for $2\times2$¶

The identity used in Step 6 reduces to the $2\times2$ case (all other coordinates are untouched). Write $T_{12}(c)=\begin{pmatrix}1&c\\0&1\end{pmatrix}$, $T_{21}(c)=\begin{pmatrix}1&0\\c&1\end{pmatrix}$. Then

and multiplying on the left by $\operatorname{diag}(1,-1)$ gives

as claimed. All three transvection factors have $f=1$ by (2), so $f(S_{12})=f(\operatorname{diag}(1,-1))=g(-1)=-1$.

Chapter — The Lie-theoretic picture behind $e_{ij}$ and $T_{ij}$¶

Everything in the derivation has a clean interpretation in the language of Lie groups and Lie algebras. This chapter is independent of the main argument — it does not feed back into the proof — but it explains why the elementary steps work, and where they come from. We work over $C$ throughout, so $G=GL(n,C)$ is a complex Lie group of dimension $n^2$, and its Lie algebra is

1. The matrix units $e_{ij}$ are a basis of the Lie algebra¶

The $n^2$ matrix units $\{e_{ij}\}_{1\le i,j\le n}$ form a basis of $\mathfrak{gl}(n,C)$. Their bracket is computed from $e_{ij}e_{kl}=\delta_{jk}\,e_{il}$:

This single rule encodes the entire structure we used.

• The diagonal units $h_i:=e_{ii}$ commute with each other ($[h_i,h_j]=0$) and span the Cartan subalgebra $\mathfrak{h}$ — the diagonal matrices. This is the “torus direction.”

• The off-diagonal units $e_{ij}$ ($i\neq j$) are the root vectors. From (L1),

  $$[h_k,e_{ij}]=(\delta_{ki}-\delta_{kj})\,e_{ij},$$

  (33)

  so $e_{ij}$ is a simultaneous eigenvector for the adjoint action of all diagonal $h_k$. Writing $\varepsilon_i\in\mathfrak{h}^*$ for the functional “read off the $i$-th diagonal entry,” the eigenvalue is the root

  $$\alpha_{ij}=\varepsilon_i-\varepsilon_j,\qquad [H,e_{ij}]=\alpha_{ij}(H)\,e_{ij}\ \ \text{for diagonal }H. \tag{L2}$$

  (34)

Example ($n=2$). Here $\mathfrak{sl}(2,C)$ has the textbook basis

with the famous relations $[h,e]=2e$, $[h,f_-]=-2f_-$, $[e,f_-]=h$. The root of $e_{12}$ is $\varepsilon_1-\varepsilon_2$, evaluating to +2 on $h$; the root of $e_{21}$ is $\varepsilon_2-\varepsilon_1$, evaluating to -2. These are exactly the entries you see in the brackets.

2. The transvections $T_{ij}(c)$ are the exponentials of the root vectors¶

For $i\neq j$ the matrix unit is nilpotent: $e_{ij}^2=0$. Hence the exponential series stops after one term,

So a transvection is literally $\exp$ of a root vector. The map

is a one-parameter subgroup: a Lie-group homomorphism from the additive group $(C,+)$ into $G$. The defining identity used in Step 2,

is exactly the statement $\exp(c\,e_{ij})\exp(c'e_{ij})=\exp((c+c')e_{ij})$, valid because the two exponents commute (they are multiples of the same nilpotent). The image $U_{ij}:=\{T_{ij}(c):c\in C\}\cong(C,+)$ is the root subgroup attached to $\alpha_{ij}$, a unipotent one-dimensional subgroup.

Example. $T_{12}(c)=\exp\!\begin{pmatrix}0&c\\0&0\end{pmatrix}=\begin{pmatrix}1&c\\0&1\end{pmatrix}$, and indeed $\begin{pmatrix}1&c\\0&1\end{pmatrix}\begin{pmatrix}1&c'\\0&1\end{pmatrix}=\begin{pmatrix}1&c+c'\\0&1\end{pmatrix}$.

3. Conjugation by the torus is the adjoint action — and the roots reappear¶

In Step 2 we conjugated a transvection by a diagonal matrix $D=\operatorname{diag}(d_1,\dots,d_n)$ and found

This is the adjoint action $\mathrm{Ad}(D)$ of the group on the subgroup, and at the Lie-algebra level it is $\mathrm{Ad}(D)e_{ij}=\tfrac{d_i}{d_j}\,e_{ij}$. The eigenvalue $d_i/d_j$ is the root $\alpha_{ij}=\varepsilon_i-\varepsilon_j$ exponentiated: if $D=\exp(H)$ with $H=\operatorname{diag}(t_1,\dots,t_n)$, then $d_i/d_j=e^{t_i-t_j}=e^{\alpha_{ij}(H)}$, matching the infinitesimal version (L2) via $\mathrm{Ad}(\exp H)=\exp(\operatorname{ad} H)$. So the fact that “all nonzero-$c$ transvections are conjugate” is the geometric statement that the torus acts on the root line $C\,e_{ij}$ by the nonzero scalars $d_i/d_j$, which sweep out all of $C^*$.

Example ($n=2$). With $D=\operatorname{diag}(d_1,d_2)$,

and choosing $d_1/d_2$ freely rescales $c$ to anything nonzero.

4. Why $f$ must be trivial on the transvections: the derived subgroup¶

Here is the structural punchline. Brackets of root vectors produce more root vectors. From (L1), for instance,

which exponentiates (via the commutator of group elements) to the statement that transvections are commutators in $G$. Concretely, the group commutator of two root subgroups gives a third: