Determinants and Oriented Volume

The determinant compresses a square matrix to one scalar. That scalar records how a linear map scales volume and whether it reverses orientation. The formulas serve this geometric meaning.

A two-dimensional determinant measures oriented area

For columns $\mathbf a_1=(a,c)^{\mathsf T}$ and $\mathbf a_2=(b,d)^{\mathsf T}$,

$$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc.$$

Its absolute value is the area of their parallelogram. Its sign distinguishes the ordering of the two directions. Exchanging columns reverses orientation and changes the sign; dependent columns collapse the area to zero.

Three properties determine the determinant

View $\det(\mathbf a_1,\ldots,\mathbf a_n)$ as a function of columns. It is linear in each column, changes sign when two columns are exchanged, and satisfies $\det I=1$. These properties imply that repeated columns give zero and that adding a multiple of one column to another changes nothing. They uniquely determine

$$\det A=\sum_{\sigma\in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}.$$

The Leibniz formula establishes the general pattern but is not an efficient method for large matrices.

ProofExistence and uniqueness

Expand each column in the standard basis. Multilinearity expresses any candidate as a sum of values on ordered basis vectors. Repeated basis vectors give zero: swapping equal columns negates the value. Every remaining ordering is a permutation of the standard basis, so its value is its permutation sign times the value at $I$. Normalization therefore forces the displayed formula (replace a permutation by its inverse to obtain the row-index form).

Conversely, the displayed finite sum is linear in each column, because every product contains exactly one entry from each column. Swapping columns pairs its terms through that transposition, which reverses the permutation sign. At $I$, only the identity permutation survives and contributes $1$. Thus the formula exists and has exactly the required properties. Here the permutation sign is $(-1)^{\text{number of inversions}}$; a transposition changes that parity.

Adding $c\mathbf a_j$ to a different column $\mathbf a_i$ adds a determinant with two proportional columns, which is zero. Scaling follows directly from linearity.

Elimination computes determinants

The row operations from elimination and LU have direct effects:

• exchanging rows multiplies the determinant by $-1$;
• multiplying a row by $c$ multiplies it by $c$;
• adding a multiple of another row leaves it unchanged.

After reduction to upper triangular form, multiply the diagonal entries and restore the factors introduced by swaps and scalings. If $PA=LU$ and $L$ has unit diagonal, then

$$\det(P)\det(A)=\prod_i u_{ii}.$$

ProofTranspose, triangular matrices, and row operations

In the permutation sum for $A^{\mathsf T}$, replace $\sigma$ by $\sigma^{-1}$. The factors become those for $A$, and inverse permutations have equal signs. Thus $\det A^{\mathsf T}=\det A$, transferring all column rules to rows.

For an upper triangular matrix a nonzero term requires $i\le\sigma(i)$ for every $i$. Since both sides sum to the same number, all inequalities are equalities. Only the identity term survives, giving the product of the diagonal. Transpose gives the lower triangular case. More generally,

$$\det\begin{pmatrix}C&E\\0&F\end{pmatrix}=\det C\det F.$$

Every surviving permutation maps the bottom row indices to bottom column indices, hence the top indices to top indices. Its product and sign factor into the two block permutations. Summing proves the formula without assuming either diagonal block invertible.

A three-dimensional elimination example

Take

$$A=\begin{pmatrix}1&2&0\\2&5&1\\0&1&3\end{pmatrix}.$$

Apply $R_2\leftarrow R_2-2R_1$, followed by $R_3\leftarrow R_3-R_2$. The resulting upper triangular matrix has diagonal $1,1,2$. Neither operation changes the determinant, so $\det A=2$.

A zero determinant means dimension collapsed

For a square matrix $A$, the following are equivalent:

$$\det A\ne0 \Longleftrightarrow A\text{ is invertible} \Longleftrightarrow \ker A=\{\mathbf0\} \Longleftrightarrow \operatorname{rank}(A)=n.$$

Dependent columns force the determinant to vanish by multilinearity. If elimination finds a pivot in every column, the triangular diagonal is nonzero, so the determinant is nonzero and the matrix is invertible.

Volume factors multiply under composition

Applying $B$ and then $A$ gives $AB$. Volume scales first by $\det B$ and then by $\det A$, hence

$$\det(AB)=\det(A)\det(B).$$

Consequently $\det(A^{-1})=1/\det(A)$, and similar matrices have the same determinant:

$$\det(S^{-1}AS)=\det A.$$

Thus the determinant describes the linear map rather than the chosen coordinate basis.

Derive the product rule from multilinearity

The volume interpretation has an algebraic proof. Fix $A$ and define a function of the columns of $B$:

$$F(\mathbf b_1,\ldots,\mathbf b_n) =\det(A\mathbf b_1,\ldots,A\mathbf b_n).$$

It is alternating and multilinear, and its value on the standard basis is $\det A$. Expand each input in standard coordinates. Repeated indices contribute zero; the remaining terms carry permutation signs. Thus $F(B)=\det(A)\det(B)$. The argument includes singular $A$ and never divides by its determinant.

Cofactor expansion is recursive

Delete row $i$ and column $j$ to form $A_{ij}$, and set $C_{ij}=(-1)^{i+j}\det A_{ij}$. Expansion along row $i$ gives

$$\det A=\sum_{j=1}^n a_{ij}C_{ij}.$$

This is useful for a small sparse matrix. Elimination is preferable for a general dense matrix because recursive expansion repeats many subproblems.

ProofCofactor expansion

Group the Leibniz sum according to the column $j$ selected in row $i$. Move row $i$ to the first position using $i-1$ adjacent swaps and column $j$ to the first position using $j-1$ swaps. Terms using the top-left entry then consist of $a_{ij}$ times a permutation term on the remaining rows and columns, in their original relative order. Undoing the swaps contributes $(-1)^{i+j-2}=(-1)^{i+j}$. Their sum is $a_{ij}C_{ij}$; summing over $j$ proves the expansion. The convention $\det(0\times0)=1$ includes $n=1$. Column expansion follows by transposition.

A determinant does not preserve all geometry

Both $\operatorname{diag}(100,0.01)$ and the identity have determinant $1$, although the former stretches one direction and compresses another. A determinant measures total oriented volume scaling; it does not by itself measure directional amplification or proximity to singularity.

Exercises

ExerciseArea and orientation

Find the oriented area generated by $(2,1)$ and $(1,3)$, then exchange the columns.

Solution

The determinant is $2\cdot3-1\cdot1=5$. Exchanging columns gives $-5$: the area remains $5$ while orientation reverses.

ExerciseCompute without expansion

Matrix $B$ is obtained from $A$ by adding four times row one to row three. Compare their determinants.

Solution

Adding a multiple of another row leaves the determinant unchanged, so $\det B=\det A$.

ExerciseSimilar matrices

Prove that similar matrices have equal determinants.

Solution

$\det(S^{-1}AS)=\det(S)^{-1}\det(A)\det(S)=\det A$.