Triangular Maps and Products of Derivative Factors

Why Triangular Maps Simplify Jacobians

A general three-variable coordinate change can produce a Jacobian matrix with nine partial derivatives and a determinant that requires several terms. But many useful maps have a dependency pattern: the first output depends only on the first input, the second output depends only on the first two inputs, and the third output may depend on all three. The resulting Jacobian is triangular, so its determinant is determined by just three entries.

This structure occurs naturally in nested coordinate descriptions. For example, one can define $x$ from $u$, then define $y$ from $u$ and $v$, and finally define $z$ from $u$, $v$, and $w$. The later outputs may depend on earlier inputs, but the earlier outputs do not depend on later inputs. That one-way dependency is what creates the triangular matrix.

Triangular Coordinate Maps

Definition: A three-variable map $(u,v,w)\mapsto (x,y,z)$ is lower triangular in this ordering if it has the form $x=f(u)$, $y=g(u,v)$, and $z=h(u,v,w)$, where the functions have the partial derivatives used below. Its Jacobian matrix has the form $$ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{pmatrix} \frac{\partial x}{\partial u} & 0 & 0\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & 0\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{pmatrix}. $$

The zeros are consequences of the dependencies: since $x$ does not depend on $v$ or $w$, its partial derivatives with respect to those variables are zero. Since $y$ does not depend on $w$, its partial derivative with respect to $w$ is zero. The entries on the diagonal are the derivatives of each output with respect to the corresponding input, holding the other inputs fixed.

There is a reversed version as well. A map with $x=f(u,v,w)$, $y=g(v,w)$, and $z=h(w)$ has an upper-triangular Jacobian: its entries below the diagonal are zero. The determinant formula is the same for either orientation of triangularity.

Theorem: The determinant of a square triangular matrix is the product of its diagonal entries. In particular, for a lower-triangular three-variable map, $$ \det\!\left(\frac{\partial(x,y,z)}{\partial(u,v,w)}\right) = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} \frac{\partial z}{\partial w}. $$ For an upper-triangular map, the determinant is likewise the product of the three diagonal entries.

Why the Product Formula Holds

For a three-by-three lower-triangular Jacobian, write the entries as $a,b,c,d,e,f$. Its determinant expansion immediately shows why only the diagonal product remains:

\begin{vmatrix} a&0&0\\ b&c&0\\ d&e&f \end{vmatrix} = a\begin{vmatrix}c&0\\e&f\end{vmatrix} = a(cf) = acf.

The same fact holds for triangular matrices of any square size. Here is a justification using the determinant formula as a sum over permutations. In a lower-triangular matrix, an entry in row $i$, column $j$ can be nonzero only when $j\leq i$. A term in the determinant corresponds to a permutation $\sigma$ and selects one entry from row $i$, column $\sigma(i)$, for each row. For such a term to be nonzero, it must have $\sigma(i)\leq i$ for every $i$. But the sum of the values $\sigma(i)$ equals $1+2+\cdots+n$, which also equals the sum of the row indices $i$. If every $\sigma(i)\leq i$, equality of these sums is possible only if $\sigma(i)=i$ for every $i$. Thus every determinant term except the diagonal one is zero, and the remaining term is the product of the diagonal entries. For an upper-triangular matrix the same reasoning gives $\sigma(i)\geq i$, again forcing the identity permutation.

Notice what the theorem does and does not say. The off-diagonal entries may be nonzero and may be large; triangularity does not mean that the matrix is diagonal. Those entries simply do not contribute additional terms to its determinant.

Worked Examples

Worked Example: A Nested Map with Mixed Dependencies

Consider the map $x=2u+1$, $y=3u+4v-2$, and $z=uv+5w+1$. The first output depends only on $u$, the second on $u,v$, and the third on $u,v,w$, so the Jacobian is lower triangular.

Differentiate each output with respect to each input. For the last row, for example, $\partial z/\partial u=v$, $\partial z/\partial v=u$, and $\partial z/\partial w=5$. Therefore, $$ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{pmatrix} 2&0&0\\ 3&4&0\\ v&u&5 \end{pmatrix}. $$

The diagonal product gives $$ \det\!\left(\frac{\partial(x,y,z)}{\partial(u,v,w)}\right) =2\cdot4\cdot5=40. $$

Check by expanding along the first row: the determinant is $2(4\cdot5-0\cdot u)=2\cdot20=40$. At $(u,v,w)=(1,2,0)$, the outputs are $(x,y,z)=(3,9,3)$, and the Jacobian becomes $\begin{pmatrix}2&0&0\\3&4&0\\2&1&5\end{pmatrix}$. Its determinant is still $2(4\cdot5-0\cdot1)=40$, as the formula predicts.

Worked Example: Diagonal Factors That Depend on Earlier Inputs

Let $x=u^2$, $y=u^2v$, and $z=u^2v^3w$, on the region $u>0,\ v>0$. Each later output depends on the earlier inputs as well as its own input. Compute the diagonal derivatives carefully: $\partial x/\partial u=2u$, $\partial y/\partial v=u^2$ with $u$ held fixed, and $\partial z/\partial w=u^2v^3$ with $u,v$ held fixed. Thus $$ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{pmatrix} 2u&0&0\\ 2uv&u^2&0\\ 2uv^3w&3u^2v^2w&u^2v^3 \end{pmatrix}. $$

The determinant is the product of the diagonal entries: $$ (2u)(u^2)(u^2v^3)=2u^5v^3. $$

At $(u,v,w)=(2,3,1)$, the outputs are $x=4$, $y=12$, and $z=108$. The Jacobian matrix is $\begin{pmatrix}4&0&0\\12&4&0\\108&108&108\end{pmatrix}$. Its determinant by first-row expansion is $4(4\cdot108-0\cdot108)=4\cdot432=1728$. Substituting into the formula also gives $2(2^5)(3^3)=2\cdot32\cdot27=1728$. The nonzero diagonal entries on the stated region make the determinant nonzero there.

Worked Example: An Upper-Triangular Map

Consider $x=u+v^2+2w$, $y=3v+w^2$, and $z=4w$. Here $x$ depends on all three inputs, $y$ depends on $v,w$, and $z$ depends only on $w$. The Jacobian is upper triangular: $$ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{pmatrix} 1&2v&2\\ 0&3&2w\\ 0&0&4 \end{pmatrix}. $$

The diagonal product is $1\cdot3\cdot4=12$. To check, evaluate at $(u,v,w)=(1,2,1)$. The outputs are $(x,y,z)=(7,7,4)$, and the matrix is $\begin{pmatrix}1&4&2\\0&3&2\\0&0&4\end{pmatrix}$. Expanding along the last row gives $4\begin{vmatrix}1&4\\0&3\end{vmatrix}=4(3)=12$. This agrees with the diagonal product.

Reading the Diagonal in a Nested Map

For a lower-triangular map, the factors are not all the derivatives that appear in the formulas. They are specifically the partial derivatives that pair each output with its corresponding input in the chosen order:

x=f(u),\qquad y=g(u,v),\qquad z=h(u,v,w) \quad\Longrightarrow\quad \det J=f'(u)\,\frac{\partial g}{\partial v}\,\frac{\partial h}{\partial w}.

In $\partial g/\partial v$, $u$ is held fixed. In $\partial h/\partial w$, both $u$ and $v$ are held fixed. A factor such as $\partial g/\partial u$ belongs below the diagonal; it records a real dependency, but is not one of the diagonal factors.

If a diagonal derivative vanishes at a point, the determinant vanishes there. If every diagonal derivative is nonzero, their product is nonzero, so the Jacobian matrix is invertible at that point. This is a pointwise conclusion: nonzero determinant at a point does not, by itself, prove that the map is one-to-one on an entire region.

Common Pitfall: Multiplying Every Nonzero Entry

A triangular matrix is not generally diagonal. Multiplying all of its nonzero entries is not a determinant rule. For the matrix from the upper-triangular example at $(1,2,1)$, multiplying every nonzero entry would give $1\cdot4\cdot2\cdot3\cdot2\cdot4=192$, which is not its determinant.

The correction is to use only the diagonal entries, in their matrix order: $1\cdot3\cdot4=12$. The determinant expansion confirms this value. Off-diagonal entries affect the matrix and its action on vectors, but in a triangular determinant they do not create extra surviving terms.

Using the Determinant in a Coordinate Change

The determinant retains its sign, which records orientation. In the change-of-variables formula for volume integrals, the volume factor is the absolute value of the Jacobian determinant. Thus, for a triangular map, the factor used for volume is the absolute value of the product of its diagonal derivatives. Taking the absolute value does not change how the determinant is computed; it is applied after the signed product has been found.

A reliable calculation follows a short sequence: identify the order of the inputs and outputs, determine which dependencies force zeros, compute the diagonal partial derivatives with the other inputs held fixed, and multiply those entries. If the determinant is needed for volume, take its absolute value at the end. When a map is triangular, this avoids unnecessary expansion while preserving the exact determinant.

Check Your Understanding

Use the dependency pattern and diagonal entries to answer the following questions.

For $x=f(u)$, $y=g(u,v)$, and $z=h(u,v,w)$, which three partial derivatives form the determinant product?
Why is $\partial y/\partial u$ not a factor in the diagonal product, even though $y$ may depend on $u$?
What zero entries appear in the Jacobian of an upper-triangular map with $x=f(u,v,w)$, $y=g(v,w)$, and $z=h(w)$?
If one diagonal derivative of a triangular Jacobian is zero, what does that imply about its determinant at that point?
For volume in a change of variables, when should the absolute value be applied to the determinant product?