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Section5.3Determinants and Volumes

Objectives
  1. Understand the relationship between the determinant of a matrix and the volume of a parallelepiped.
  2. Learn to use determinants to compute volumes of parallelograms and triangles.
  3. Learn to use determinants to compute the volume of some curvy shapes like ellipses.
  4. Pictures: parallelepiped, the image of a curvy shape under a linear transformation.
  5. Theorem: determinants and volumes.
  6. Vocabulary word: parallelepiped.

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.

Subsection5.3.1Parallelograms and Paralellepipeds

The determinant computes the volume of the following kind of geometric object.

Definition

The paralellepiped determined by n vectors v 1 , v 2 ,..., v n in R n is the subset

P = C a 1 x 1 + a 2 x 2 + ··· + a n x n EE 0 a 1 , a 2 ,..., a n 1 D .

In other words, a parallelepiped is the set of all linear combinations of n vectors with coefficients in [ 0,1 ] . We can draw parallelepipeds using the parallelogram law for vector addition.

Example(The unit cube)

The parallelepiped determined by the standard coordinate vectors e 1 , e 2 ,..., e n is the unit n -dimensional cube.

e 2 e 1 e 1 e 2 e 3
Example(Parallelograms)

When n = 2, a paralellepiped is just a paralellogram in R 2 . Note that the edges come in parallel pairs.

v 2 v 1 P
Example

When n = 3, a parallelepiped is a kind of a skewed cube. Note that the faces come in parallel pairs.

v 1 v 2 v 3 P

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.

P v 1 v 2 v 1 v 2 v 3 P

This means exactly that { v 1 , v 2 ,..., v n } is linearly dependent, which by this corollary in Section 5.1 means that the matrix with rows v 1 , v 2 ,..., v n has determinant zero. To summarize:

Key Observation

The parallelepiped defined by v 1 , v 2 ,..., v n has zero volume if and only if the matrix with rows v 1 , v 2 ,..., v n has zero determinant.

Subsection5.3.2Determinants and Volumes

The key observation above is only the beginning of the story: the volume of a parallelepiped is always a determinant.

Proof

Since the four defining properties characterize the determinant, they also characterize the absolute value of the determinant. Explicitly, | det | is a function on square matrices which satisfies these properties:

  1. Doing a row replacement on A does not change | det ( A ) | .
  2. Scaling a row of A by a scalar c multiplies | det ( A ) | by | c | .
  3. Swapping two rows of a matrix does not change | det ( A ) | .
  4. The determinant of the identity matrix I n is equal to 1.

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 A to obtain a matrix B in row echelon form, then

| det ( A ) | = EEEE (productofthediagonalentriesof B ) (productofscalingfactorsused) EEEE .

For a square matrix A , we abuse notation and let vol ( A ) denote the volume of the paralellepiped determined by the rows of A . Then we can regard vol as a function from the set of square matrices to the real numbers. We will show that vol also satisfies the above four properties.

  1. For simplicity, we consider a row replacement of the form R n = R n + cR i . The volume of a paralellepiped is the volume of its base, times its height: here the “base” is the paralellepiped determined by v 1 , v 2 ,..., v n 1 , and the “height” is the perpendicular distance of v n from the base.
    base v 2 v 1 height base height v 1 v 2 v 3
    Translating v n by a multiple of v i moves v n in a direction parallel to the base. This changes neither the base nor the height! Thus, vol ( A ) is unchanged by row replacements.
    base v 2 v 1 height −−−−→ base v 2 v 2 .5 v 1 v 1 height base height v 1 v 2 v 3 −−−−→ base height v 1 v 2 v 3 + .5 v 1 v 3
  2. For simplicity, we consider a row scale of the form R n = cR n . This scales the length of v n by a factor of | c | , which also scales the perpendicular distance of v n from the base by a factor of | c | . Thus, vol ( A ) is scaled by | c | .
    base v 2 v 1 height −−−−→ base 3 4 v 2 v 1 3 4 height base height v 1 v 2 v 3 −−−−→ base 4 3 height v 1 v 2 4 3 v 3
  3. Swapping two rows of A just reorders the vectors v 1 , v 2 ,..., v n , hence has no effect on the parallelepiped determined by those vectors. Thus, vol ( A ) is unchanged by row swaps.
    v 2 v 1 −−−−→ v 1 v 2 v 1 v 3 v 2 −−−−→ v 1 v 2 v 3
  4. The rows of the identity matrix I n are the standard coordinate vectors e 1 , e 2 ,..., e n . The associated paralellepiped is the unit cube, which has volume 1. Thus, vol ( I n )= 1.

Since | det | is the only function satisfying these properties, we have

vol ( P )= vol ( A )= | det ( A ) | .

This completes the proof.

Since det ( A )= det ( A T ) by the transpose property, the absolute value of det ( A ) is also equal to the volume of the paralellepiped determined by the columns of A as well.

Example(Length)

A 1 × 1 matrix A is just a number A a B . In this case, the parallelepiped P determined by its one row is just the interval [ 0, a ] (or [ a ,0 ] if a < 0 ). The “volume” of a region in R 1 = R is just its length, so it is clear in this case that vol ( P )= | a | .

vol ( P )= | a | 0 a
Example(Area)

When A is a 2 × 2 matrix, its rows determine a parallelogram in R 2 . The “volume” of a region in R 2 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.

H ab I H cd I area = EEEE det H abcd IEEEE = | ad bc |

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 A lies on the x -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.

Subsection5.3.3Volumes of Regions

Let A be an n × n matrix with columns v 1 , v 2 ,..., v n , and let T : R n R n be the associated matrix transformation T ( x )= Ax . Then T ( e 1 )= v 1 and T ( e 2 )= v 2 , so T takes the unit cube C to the parallelepiped P determined by v 1 , v 2 ,..., v n :

C e 2 e 1 T F || v 1 v 2 || G v 2 v 1 P

Since the unit cube has volume 1 and its image has volume | det ( A ) | , the transformation T scaled the volume of the cube by a factor of | det ( A ) | . To rephrase:

If A is an n × n matrix with corresponding matrix transformation T : R n R n , and if C is the unit cube in R n , then the volume of T ( C ) is | det ( A ) | .

The notation T ( S ) means the image of the region S under the transformation T . In set builder notation, this is the subset

T ( S )= C T ( x ) | x in S D .

In fact, T scales the volume of any region in R n by the same factor, even for curvy regions.

Proof

Let C be the unit cube, let v 1 , v 2 ,..., v n be the columns of A , and let P be the paralellepiped determined by these vectors, so T ( C )= P and vol ( P )= | det ( A ) | . For A> 0 we let A C be the cube with side lengths A , i.e., the paralellepiped determined by the vectors A e 1 , A e 2 ,..., A e n , and we define A P similarly. By the second defining property, T takes A C to A P . The volume of A C is A n (we scaled each of the n standard vectors by a factor of A ) and the volume of A P is A n | det ( A ) | (for the same reason), so we have shown that T scales the volume of A C by | det ( A ) | .

A e 2 A e 1 A C vol ( A C )= A n T F || v 1 v 2 || G A v 2 A v 1 A P vol ( A P )= A n | det ( A ) |

By the first defining property, the image of a translate of A C is a translate of A P :

T ( x + A C )= T ( x )+ A T ( C )= T ( x )+ A P .

Since a translation does not change volumes, this proves that T scales the volume of a translate of A C by | det ( A ) | .

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 S can be approximated by a collection of very small cubes of the form x + A C . The image T ( S ) is then approximated by the image of this collection of cubes, which is a collection of very small paralellepipeds of the form T ( x )+ A P .

x + A C S T T ( x )+ A P T ( S )

The volume of S is closely approximated by the sum of the volumes of the cubes; in fact, as A goes to zero, the limit of this sum is precisely vol ( S ) . Likewise, the volume of T ( S ) is equal to the sum of the volumes of the paralellepipeds, take in the limit as A 0. The key point is that the volume of each cube is scaled by | det ( A ) | . Therefore, the sum of the volumes of the paralellepipeds is | det ( A ) | times the sum of the volumes of the cubes. This proves that vol ( T ( S ))= | det ( A ) | vol ( S ) .