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My book on linear algebra states the rule $|det(AB)|=|det(A)||det(B)|$ without giving a proof. This annoys me enough to prompt me to think of how I would prove it, and I would like some feedback on my though process.

Proof sketch.

Nonzero determinant:

Assumption: Any matrix with a nonzero determinant can be Gaussian eliminated to a diagonal matrix, without changing the value of its determinant.

If $A,B$ are two square diagonal matrices with the same dimensions, then the matrix AB is a diagonal matrix with each element in the diagonal being the product of the two diagonal elements from $A$ and $B$.

The determinant of a diagonal square matrix is the product of the diagonal elements.
The determinant of $AB$ is thus the product of all diagonal elements of $A$ and $B$.
The determinant of $A$ and $B$ respectively are the product of their diagonal elements and the same as $det(AB)$

Zero determinant:

Assumption: A matrix with a zero determinant can be transformed, without changing the determinant value, to a over triangular matrix who's determinant is the product of its diagonal elements, of which at least one will be zero.

Matrix multiplication of two over triangular matrices $A,B$ will result in $AB$ being an over triangular matrix with its diagonal elements being the product of $A,B$s diagonal element at the same position, of which at least on will be zero and thus $det(AB)=0$.

Erik Eriksson
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  • Did your book already prove $\det(AB)=\det(A)\det(B)$? If yes, $|\det(AB)|=|\det(A)\det(B)|=|\det(A)||\det(B)|$. (For any two real numbers $x,y$ it is true that $|xy|=|x||y|$.) –  Dec 08 '21 at 11:42

1 Answers1

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Although there are rigorous proofs demonstrating the fact: https://math.vanderbilt.edu/sapirmv/msapir/proofdet3.html, How to show that $\det(AB) =\det(A) \det(B)$?, I think about it intuitively to satisfy myself.

A determinant of a matrix corresponds to the factor by which a linear transformation (defined by a matrix) changes in area/volume/n-d volume for an nxn matrix. Therefore a transformation defined by composing j # of matrices with $\det(A_i)=d_i$ scales a unit n-d space before transformation by d₁$\cdot$d₂$\cdot\ $$\cdots$ dⱼ after, and hence $\det(\prod_{i=1}^j (A_i)) = \prod_{i=1}^j \det(A_i)$.

I hope that's intuitively convincing enough.

MathEnthusiast314
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