In this section we give a geometric interpretation of determinants, in terms of volumes. This will shed light on the reason behind three of the four defining properties of the determinant. It is also a crucial ingredient in the change-of-variables formula in multivariable calculus.
The determinant computes the volume of the following kind of geometric object.
The parallelepiped determined by vectors in is the subset
In other words, a parallelepiped is the set of all linear combinations of vectors with coefficients in We can draw parallelepipeds using the parallelogram law for vector addition.
The parallelepiped determined by the standard coordinate vectors is the unit -dimensional cube.
When a parallelepiped is just a paralellogram in Note that the edges come in parallel pairs.
When a parallelepiped is a kind of a skewed cube. Note that the faces come in parallel pairs.
When does a parallelepiped have zero volume? This can happen only if the parallelepiped is flat, i.e., it is squashed into a lower dimension.
This means exactly that is linearly dependent, which by this corollary in Section 5.1 means that the matrix with rows has determinant zero. To summarize:
The parallelepiped defined by has zero volume if and only if the matrix with rows has zero determinant.
The key observation above is only the beginning of the story: the volume of a parallelepiped is always a determinant.
Let be vectors in let be the parallelepiped determined by these vectors, and let be the matrix with rows Then the absolute value of the determinant of is the volume of
Since the four defining properties characterize the determinant, they also characterize the absolute value of the determinant. Explicitly, is a function on square matrices which satisfies these properties:
The absolute value of the determinant is the only such function: indeed, by this recipe in Section 5.1, if you do some number of row operations on to obtain a matrix in row echelon form, then
For a square matrix we abuse notation and let denote the volume of the parallelepiped determined by the rows of Then we can regard as a function from the set of square matrices to the real numbers. We will show that also satisfies the above four properties.
Since is the only function satisfying these properties, we have
This completes the proof.
Since by the transpose property, the absolute value of is also equal to the volume of the parallelepiped determined by the columns of as well.
A matrix is just a number In this case, the parallelepiped determined by its one row is just the interval (or if ). The “volume” of a region in is just its length, so it is clear in this case that
When is a matrix, its rows determine a parallelogram in The “volume” of a region in is its area, so we obtain a formula for the area of a parallelogram: it is the determinant of the matrix whose rows are the vectors forming two adjacent sides of the parallelogram.
It is perhaps surprising that it is possible to compute the area of a parallelogram without trigonometry. It is a fun geometry problem to prove this formula by hand. [Hint: first think about the case when the first row of lies on the -axis.]
You might be wondering: if the absolute value of the determinant is a volume, what is the geometric meaning of the determinant without the absolute value? The next remark explains that we can think of the determinant as a signed volume. If you have taken an integral calculus course, you probably computed negative areas under curves; the idea here is similar.
Let be an matrix with columns and let be the associated matrix transformation Then and so takes the unit cube to the parallelepiped determined by
Since the unit cube has volume and its image has volume the transformation scaled the volume of the cube by a factor of To rephrase:
If is an matrix with corresponding matrix transformation and if is the unit cube in then the volume of is
The notation means the image of the region under the transformation In set builder notation, this is the subset
In fact, scales the volume of any region in by the same factor, even for curvy regions.
Let be an matrix, and let be the associated matrix transformation If is any region in then
Let be the unit cube, let be the columns of and let be the parallelepiped determined by these vectors, so and For we let be the cube with side lengths i.e., the parallelepiped determined by the vectors and we define similarly. By the second defining property, takes to The volume of is (we scaled each of the standard vectors by a factor of ) and the volume of is (for the same reason), so we have shown that scales the volume of by
By the first defining property, the image of a translate of is a translate of
Since a translation does not change volumes, this proves that scales the volume of a translate of by
At this point, we need to use techniques from multivariable calculus, so we only give an idea of the rest of the proof. Any region can be approximated by a collection of very small cubes of the form The image is then approximated by the image of this collection of cubes, which is a collection of very small parallelepipeds of the form
The volume of is closely approximated by the sum of the volumes of the cubes; in fact, as goes to zero, the limit of this sum is precisely Likewise, the volume of is equal to the sum of the volumes of the parallelepipeds, take in the limit as The key point is that the volume of each cube is scaled by Therefore, the sum of the volumes of the parallelepipeds is times the sum of the volumes of the cubes. This proves that
