How a Matrix Acts on the Columns
The determinant records how the columns of a square matrix fit together, including the sign produced by their order. When a matrix \(\mathbf{A}\) multiplies another matrix \(\mathbf{B}\), it acts on each column of \(\mathbf{B}\) separately. The multiplicative determinant law says that this action scales the determinant by the fixed factor \(\det\mathbf{A}\). We will prove the law by examining the determinant as a function of the columns being transformed.
Throughout, let \(n\ge 1\), and let \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{n\times n}\). Write the columns of \(\mathbf{B}\) as \(\mathbf{b}_1,\dots,\mathbf{b}_n\), so \(\mathbf{B}=[\mathbf{b}_1\ \cdots\ \mathbf{b}_n]\). The columns of \(\mathbf{A}\mathbf{B}\) are \(\mathbf{A}\mathbf{b}_1,\dots,\mathbf{A}\mathbf{b}_n\): this follows directly from the definition of matrix multiplication.
The determinant is multilinear in its rows, and transposition preserves the determinant. These earlier determinant properties imply multilinearity in the columns: applying row multilinearity to the transpose gives linearity in each column of the original matrix. Also, repeated rows force a zero determinant, and transposition then gives the corresponding fact for repeated columns. We use these properties to establish the behavior of \(F\).
Proof. Fix an argument position \(j\), and hold all other vectors fixed. The map \(\mathbf{b}_j\mapsto\mathbf{A}\mathbf{b}_j\) is linear, and the determinant is linear in its \(j\)th column. Their composition is therefore linear in \(\mathbf{b}_j\). This holds for every \(j\), so \(F\) is multilinear. If \(\mathbf{b}_i=\mathbf{b}_j\) for distinct positions \(i\) and \(j\), then \(\mathbf{A}\mathbf{b}_i=\mathbf{A}\mathbf{b}_j\). The matrix inside the determinant has two equal columns, so its determinant is zero. Hence \(F\) is alternating. \(\square\)
These two properties are restrictive: once the value of an alternating multilinear function is known on the standard basis vectors, all its values are determined. The next result makes that statement precise and supplies the key link to the determinant of \(\mathbf{B}\).
Why the Standard Basis Determines the Function
Proof. First, interchanging any two arguments of \(G\) changes its sign. To justify this from the definition of alternating, place \(\mathbf{u}+\mathbf{v}\) in both of the two selected positions, leaving the other arguments fixed. The value is zero because the two arguments are equal. Expanding by multilinearity gives four terms. The terms using \(\mathbf{u}\) in both positions and \(\mathbf{v}\) in both positions are zero by alternation. Thus the two remaining terms, with \(\mathbf{u},\mathbf{v}\) in opposite orders, sum to zero. They are negatives of each other, as claimed.
Write \(\mathbf{b}_j=\sum_{i=1}^n b_{ij}\mathbf{e}_i\), where \(b_{ij}\) is the entry in row \(i\), column \(j\) of \(\mathbf{B}\). Multilinearity expands \(G(\mathbf{b}_1,\dots,\mathbf{b}_n)\) as a sum over all choices of indices \(i_1,\dots,i_n\), with the term for that choice equal to $$ b_{i_1,1}\cdots b_{i_n,n}\,G(\mathbf{e}_{i_1},\dots,\mathbf{e}_{i_n}). $$ If two of the indices are equal, the corresponding standard basis vectors are equal, so that term is zero by alternation. The remaining choices use each index \(1,\dots,n\) exactly once. They are precisely the permutations of those indices. For such a choice, reordering \(\mathbf{e}_{i_1},\dots,\mathbf{e}_{i_n}\) into \(\mathbf{e}_1,\dots,\mathbf{e}_n\) gives $$ G(\mathbf{e}_{i_1},\dots,\mathbf{e}_{i_n}) =\operatorname{sgn}(\pi)G(\mathbf{e}_1,\dots,\mathbf{e}_n), $$ where \(\pi(j)=i_j\). This sign follows by expressing the reordering as swaps and using the sign change under each swap proved above. Therefore the expansion becomes $$ G(\mathbf{b}_1,\dots,\mathbf{b}_n) =G(\mathbf{e}_1,\dots,\mathbf{e}_n) \sum_{\pi\in S_n}\operatorname{sgn}(\pi)\prod_{j=1}^n b_{\pi(j),j}. $$ The sum is the Leibniz formula for \(\det\mathbf{B}\). This proves the proposition. \(\square\)
The proof also covers \(n=1\): there are no repeated-index terms to discard, and the single term in the sum is the sole identity permutation. No separate assumption about the matrix being invertible has entered.
Worked Examples of the Column Function
Worked Example: Checking the Law for Two Matrices
Take $$ \mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}1&3\\2&-1\end{bmatrix}. $$ The columns of \(\mathbf{B}\) are \(\mathbf{b}_1=(1,2)^{\top}\) and \(\mathbf{b}_2=(3,-1)^{\top}\). Applying \(\mathbf{A}\) to each column gives $$ \mathbf{A}\mathbf{b}_1=\begin{bmatrix}2+2\\1+2\end{bmatrix} =\begin{bmatrix}4\\3\end{bmatrix}, \qquad \mathbf{A}\mathbf{b}_2=\begin{bmatrix}6-1\\3-1\end{bmatrix} =\begin{bmatrix}5\\2\end{bmatrix}. $$ Thus $$ \mathbf{A}\mathbf{B}=\begin{bmatrix}4&5\\3&2\end{bmatrix}, \qquad \det(\mathbf{A}\mathbf{B})=4\cdot2-5\cdot3=8-15=-7. $$ Separately, \(\det\mathbf{A}=2\cdot1-1\cdot1=1\), and \(\det\mathbf{B}=1\cdot(-1)-3\cdot2=-1-6=-7\). Hence \(\det(\mathbf{A}\mathbf{B})=-7=(\det\mathbf{A})(\det\mathbf{B})\).
Check: The product matrix has columns \((4,3)^{\top}\) and \((5,2)^{\top}\), exactly the two computed images. Its determinant and the product of the two original determinants both equal \(-7\).
Worked Example: Equal Columns Make the Transformed Determinant Zero
Let $$ \mathbf{A}=\begin{bmatrix}1&2\\3&5\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}1&2\\1&2\end{bmatrix}. $$ The columns of \(\mathbf{B}\) are \(\mathbf{b}_1=(1,1)^{\top}\) and \(\mathbf{b}_2=(2,2)^{\top}=2\mathbf{b}_1\). Their images are $$ \mathbf{A}\mathbf{b}_1=\begin{bmatrix}1+2\\3+5\end{bmatrix} =\begin{bmatrix}3\\8\end{bmatrix}, \qquad \mathbf{A}\mathbf{b}_2=\begin{bmatrix}2+4\\6+10\end{bmatrix} =\begin{bmatrix}6\\16\end{bmatrix}. $$ The second image is twice the first, and the product is $$ \mathbf{A}\mathbf{B}=\begin{bmatrix}3&6\\8&16\end{bmatrix}. $$ Consequently, \(\det(\mathbf{A}\mathbf{B})=3\cdot16-6\cdot8=48-48=0\). Also \(\det\mathbf{B}=1\cdot2-2\cdot1=0\), while \(\det\mathbf{A}=1\cdot5-2\cdot3=-1\). Thus \((\det\mathbf{A})(\det\mathbf{B})=(-1)\cdot0=0\).
Check: The two columns of \(\mathbf{A}\mathbf{B}\) are \((3,8)^{\top}\) and \(2(3,8)^{\top}=(6,16)^{\top}\), so their determinant is zero, as the alternating property requires.
Worked Example: Expanding in Standard Basis Vectors
Use $$ \mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}1&2\\3&4\end{bmatrix}. $$ Here \(\mathbf{b}_1=\mathbf{e}_1+3\mathbf{e}_2\) and \(\mathbf{b}_2=2\mathbf{e}_1+4\mathbf{e}_2\). For \(F(\mathbf{x},\mathbf{y})=\det(\mathbf{A}\mathbf{x},\mathbf{A}\mathbf{y})\), multilinearity gives $$ F(\mathbf{b}_1,\mathbf{b}_2) =2F(\mathbf{e}_1,\mathbf{e}_1)+4F(\mathbf{e}_1,\mathbf{e}_2) +6F(\mathbf{e}_2,\mathbf{e}_1)+12F(\mathbf{e}_2,\mathbf{e}_2). $$ The repeated-argument terms are zero. Since \(\mathbf{A}\mathbf{e}_1=(2,1)^{\top}\) and \(\mathbf{A}\mathbf{e}_2=(1,1)^{\top}\), $$ F(\mathbf{e}_1,\mathbf{e}_2)=2\cdot1-1\cdot1=1, \qquad F(\mathbf{e}_2,\mathbf{e}_1)=1\cdot1-2\cdot1=-1. $$ It follows that \(F(\mathbf{b}_1,\mathbf{b}_2)=4(1)+6(-1)=-2\). Directly, \(\mathbf{A}\mathbf{B}=\begin{bmatrix}5&8\\4&6\end{bmatrix}\), whose determinant is \(5\cdot6-8\cdot4=30-32=-2\). Also \(\det\mathbf{A}=1\) and \(\det\mathbf{B}=1\cdot4-2\cdot3=-2\), giving the same result.
Check: The columns of the displayed product are \(\mathbf{A}\mathbf{b}_1=(5,4)^{\top}\) and \(\mathbf{A}\mathbf{b}_2=(8,6)^{\top}\); their determinant is \(30-32=-2\), matching the expansion.
The Multiplicative Determinant Law
Proof. Fix \(\mathbf{A}\) and define \(F(\mathbf{b}_1,\dots,\mathbf{b}_n)=\det(\mathbf{A}\mathbf{b}_1,\dots,\mathbf{A}\mathbf{b}_n)\). The first proposition shows that \(F\) is alternating and multilinear. The second proposition therefore gives $$ F(\mathbf{b}_1,\dots,\mathbf{b}_n) =F(\mathbf{e}_1,\dots,\mathbf{e}_n)\det\mathbf{B}. $$ The vector \(\mathbf{A}\mathbf{e}_j\) is the \(j\)th column of \(\mathbf{A}\), so \(F(\mathbf{e}_1,\dots,\mathbf{e}_n)=\det\mathbf{A}\). The columns of \(\mathbf{A}\mathbf{B}\) are \(\mathbf{A}\mathbf{b}_1,\dots,\mathbf{A}\mathbf{b}_n\), so the left side is \(\det(\mathbf{A}\mathbf{B})\). Substitution yields the stated identity. Since the proof placed no restrictions on \(\mathbf{A}\) or \(\mathbf{B}\), it applies whether or not either matrix is invertible. \(\square\)
A Common Sign Pitfall
Swapping two columns does not leave the determinant unchanged: it reverses its sign. For a concrete check, take \(\mathbf{A}=\mathbf{I}_2\) and \(\mathbf{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\). Then \(\mathbf{A}\mathbf{B}=\mathbf{B}\), and $$ \det(\mathbf{A}\mathbf{B})=0\cdot0-1\cdot1=-1, \qquad \det\mathbf{A}\det\mathbf{B}=1\cdot(0\cdot0-1\cdot1)=-1. $$ A wrong calculation would treat the swapped columns \((\mathbf{e}_2,\mathbf{e}_1)\) as having the same determinant as \((\mathbf{e}_1,\mathbf{e}_2)\), and conclude \(+1\). The correction is that alternation makes their values negatives: \(F(\mathbf{e}_2,\mathbf{e}_1)=-F(\mathbf{e}_1,\mathbf{e}_2)\). In the proof, these signs are exactly the permutation signs in the Leibniz formula.
The column-function argument is useful because it proves the law directly from the defining structural properties of determinants. It does not depend on solving a system or dividing by a determinant, so singular matrices are included. It also explains why the determinant of \(\mathbf{B}\) appears: an alternating multilinear function of the columns must be its value on the standard basis multiplied by \(\det\mathbf{B}\). For the transformed columns, that standard-basis value is \(\det\mathbf{A}\).
Check Your Understanding
Use the definitions and proof above to answer each question.
- For fixed \(\mathbf{A}\), explain why \(F(\mathbf{b}_1,\dots,\mathbf{b}_n)=\det(\mathbf{A}\mathbf{b}_1,\dots,\mathbf{A}\mathbf{b}_n)\) is linear in its second argument.
- What happens to \(F\) if two input columns are equal, and which determinant property justifies the result?
- In the basis expansion of an alternating multilinear function, why do choices with a repeated index contribute zero?
- For \(\mathbf{A}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}2&1\\3&4\end{bmatrix}\), compute \(\det(\mathbf{A}\mathbf{B})\) and \(\det\mathbf{A}\det\mathbf{B}\) to check the law.
- Does the multiplicative determinant law require either factor to be invertible? Give the reason from the proof.
