Math 416, Spring 2010 The algebra of determinants March 16, 2010 THE ALGEBRA OF DETERMINANTS. 1. Determinants

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1 THE ALGEBRA OF DETERMINANTS 1. Determinants We have already defined the determinant of a 2 2 matrix: det = ad bc. We ve also seen that it s handy for determining when a matrix is invertible, and when it is, for actually calculating that inverse. Our goal for today is to define determinants for a general square matrix, and see if some of these properties come along for the ride. We ll start with the following Definition 1.1. The ijth minor of an n n matrix A, written A ij, is the n 1 n 1 matrix one gets upon deleting the ith row and jth column of A., then If A is the matrix A 22 = and A 11 = We can now give a definition of the determinant. Definition 1.2. If A is an n n matrix, then deta = 1 1+j a 1j deta 1j. Notice that our definition is recursive: finding the determinant of an n n matrix requires us to compute the determinant of many n 1 n 1 matrices, which each require us to compute the determinant of many n 2 n 2 matrices, etc. Here s an example of a determinant calculation: det = 1 det 3 det det 1 1 = = = 34. We won t be able to prove this in class today, but in fact one has the following acs@math.uiuc.edu acs/w10/math416 Page 1 of 7

2 Theorem 1.1. The determinant can be computed as deta = 1 i+j a ij deta ij for a fixed i this is expanding along the row i. The determinant can also be computed as for a fixed j this is expanding along the row j. deta = 1 i+j a ij deta ij This theorem is awfully handy in computing determinants, because it lets us choose a row that simplifies calculations as much as possible. Find the determinant of A = Solution. The definition of the determinant says we should expand along the first row, but since the first column has lots of zeroes I m going to compute the determinant by expanding along it. deta = 1 det det + 0 det 0 det I haven t bothered to write down the other three matrices since their determinants will not count: they have a coefficient of 0 in front! Now to compute the determinant of the residual 3 3 matrix I ll again choose to expand along the first column: it has lots of zeros which make calculations easy. det = 1 det det + 0 det = 11 0 = 1. Putting all our calculations together we have = 1 det = = 1. This example shows that the smartest way to calculate determinants is to expand along a row or column which is sparse i.e., which has lots of zeros. It also is indicative of another result which is very handy: Theorem 1.2. If A is a lower or upper triangular matrix with diagonal entries a 11,, a nn then n deta = a ii. acs@math.uiuc.edu acs/w10/math416 Page 2 of 7

3 One final note is that in practice it can be hard when expanding along a given row or column to remember whether one should add or subtract the determinant of a given minor i.e., it s sometimes hard to remember whether the coefficient 1 i+j will be 1 or 1. For this, it can be helpful to write down a checkerboard that keeps track of which minors have a coefficient 1 and which have a coefficient 1. Just start by putting + in the top left hand corner and then alternate. For instance, the checkerboard for 4 4 matrices is just Determinants by Row Reduction The problem with the technique we ve described above is that it s the wrong way to compute determinants. I say this because for a matrix A that isn t full of lots of convenient 0 s which we ve already seen can make determinant calculations fairly easy, the algorithm we ve provided takes about n! many multiplications to run. The number n! grows extremely rapidly as n grows, and that makes this algorithm extrememly computationally expensive, even for moderately sized n. Suppose that A is a matrix. To compute deta using the algorithm above, we d need to perform about 50! operations. The faster compute today runs at about 1.75 petaflops i.e., about operations per second, and so computing deta would require about = seconds. Since there are about 10 7 seconds in a year, this means the calculation will take a whopping years to finish! For a frame of reference, the accepted age of the universe is around years. This means that if the fastest computer in the world today were around at the time of the big bang, and if we asked it to compute deta using the algorithm above, then it would only be about % through with the calculation today. Clearly, this is a junky way to compute determinants. A very reasonable question to ask, then, is whether there is a way to compute determinants which is more effecient. We ll talk about one such answer right now. The method we will tell about involves row reduction, which is pretty exciting pedagogically because row reduction has played such a huge role in this class all term long. In fact, with the exception of the Gram-Schmidt algorithm, I think all of our results have relied on being able to compute the reduced row echelon form of a matrix. The result we ll describe is motivated by the following Consider what happens to the determinant of a 2 2 matrix after an elementary row operation has been performed on it. Solution. Let A = and we ll consider the determinant of the matrices which result from performing an elementary row operation on A. We first consider the row operation of a scalar multiple of one row into another row. In this case we have a + kc b + kd det = a + kcd cb + kd = ad + kcd cb ckd = ad bc = deta., acs@math.uiuc.edu acs/w10/math416 Page 3 of 7

4 For the next row operation scaling a row by a nonzero constant we have det = akd bkc = kad bc = k deta. kc kd Finally we notice the effect of swapping a row: det = cb da = ad bc = deta. In fact these identities on 2 2 matrices carry over to arbitrary square matrices. Theorem 2.1. Suppose that we reach a matrix B by performing elementary row operations on a matrix A. The number of row swaps in these operations is s and the number of row scalings is r, with constants k 1,, k r. Then detb = 1 s r k i deta. This theorem let s us use elementary row operations to transform a matrix into a convenient form for computing the determinant reduced row echelon form which is always upper triangular, for instance. As long as we remember the operations we took to get to a convenient matrix form, we can calculate the determinant of the initial matrix. Suppose that rrefa = I n and that the move A into reduced row echelon form we had to swap 7 rows and scale rows by the constants k 1 = 2, k 2 = 1 2, k 3 = 11, and k 4 = 2. Then deta = deti n = The previous theorem is not only computationally convenient: it is also theoretically quite useful. In fact, it proves the fact about determinants we care about most: Theorem 2.2. A matrix A has deta = 0 if and only if A is not invertible. Proof. A matrix is not invertible only if rrefa has a last row which is the vector 0. Hence detrrefa = 0 and since 0 = detrrefa = k deta with k is some nonzero constant, we have deta = 0. If deta = 0, on the other hand, then detrrefa = k deta = 0. But a square matrix in reduced row echelon form is upper triangular, and so the determinant is the product of the diagonal entries. This can be 0 only if the there is a 0 entry on the diagonal of rrefa, which implies that the last diagonal entry of rrefa = 0. But this means that the last row of rrefa = 0, and hence A is not invertible. There s one last comment about using row operations to find the determinant of a matrix. Calculating the determinant using row operations is generally speaking much quicker than calculating the determinant by expanding along a row or column. However when one is attempting to find the determinant of a 2 2, a 3 3 or a 4 4 matrix, it is usually more convenient to just expand along a row or column. acs@math.uiuc.edu acs/w10/math416 Page 4 of 7

5 3. Algebraic Properties of the Determinant There are several algebraic properties of the determinant which will be useful. Theorem 3.1. Suppose that A and B are two n n matrices. Then detab = deta detb. Solution. First we ll assume that deta = 0. This means that A is not invertible, and hence ima R n. But since imab ima R n we have imab R n, and so AB is not invertible. Therefore we have detab = 0, and so detab = deta detb as desired no matter what detb is. Now assume that deta 0. Suppose that to move A into reduced row echelon form rrefa = I n since deta 0 implies A is invertible we require s row swaps and scalar multiplications k 1,, k r. Then we have r 1 = deti n = 1 s 1 k i deta = deta = 1 s r k i. But notice that if one applies the same row operations to the matrix AB we will wind up at matrix B performing these row operations is like multiplying by A 1 on the left. This means that detb = 1 s r k i detab = detab = 1 1 s r k i detb = deta detb. Corollary 3.2. For an invertible matrix A, deta 1 = deta 1. Proof. We know that AA 1 = I n, so that Solving for deta 1 gives the desired result. 1 = deti n = detaa 1 = deta deta 1. There is another handy fact to know about determinants, though we won t prove it in class today. Theorem 3.3. For a square matrix A, deta = deta T. Proof. We ll prove this by induction. The base case is when A is a 1 1 matrix, in which case there s nothing to to since A = A T in this case. So suppose that A is n n, and that we know the result for all square matrices that are smaller than n n. To compute deta, the definition says we should expand minors along the first row of A. This gives deta = 1 1+j a 1j deta 1j. But A 1j is an n 1 n 1 matrix, and hence by induction we know that deta 1j = deta T 1j. Note that A T 1j is just AT j1. Since a 1j is the entry in the jth row and 1st column of A T, this means that the formula 1 1+j a 1j deta 1j = 1 1+j a 1j deta T 1j = 1 1+j a 1j deta T j1 is simply the expansion of A T along the 1st column of A T, which we know to be deta T. acs@math.uiuc.edu acs/w10/math416 Page 5 of 7

6 Finally, we have the following Theorem 3.4. Suppose that v 1,, v j 1, v j+1,, v n are vectors in R n. Then the function D j x = det v 1 v j 1 x v j+1 v n is linear. That is to say, for any x, y R n and any c R, we have D j x + y = D j x + D j y and D j c x = cd j x. 4. Determinants and geometry Having thoroughly discussed the algebra of the determinant, it s now time to go on and talk about how determinant manifests itself on the geometric side of linear algebra. We ll start by talking about a highdimensional analogue of parallelograms. Definition 4.1. For vectors v 1,, v m R n, the m-parallelepiped defined by the v i is the set of all vectors which can be written c 1v1 + + c mvm for scalars 0 c i 1. Notice that if we had put no restriction on the scalars then we would have the span of v 1,, v m ; since we insist 0 c i 1, however, we have only a subset of Span v 1,, v m. Let v 1 = e 1 and v 2 = e 2 R 2. Then the 2-parallelepiped is the unit square. v 1 v 2 e2 e1 Figure 1. Two 2-parallelepipeds in R 2 Definition 4.2. The n-parallelepiped defined by e 1,, e n R n is called the unit n-cube. Let v 1 and v 2 be as shown on the right of Figure 1. Then the shaded region is the 2-parallelepiped defined by v 1 and v 2. acs@math.uiuc.edu acs/w10/math416 Page 6 of 7

7 When the vectors v 1,, v m are not linearly independent then the m-parallelepiped they define is degenerate in the sense that they live in an n 1 dimensional space. Suppose that v 1 = v 2 as shown in Figure 3. Then the 2-parallelepiped is the line segment shown on the right. v 1 = v 2 P Figure 2. Two linearly dependent vectors and the degenerate 2-parallelepiped P they define We can calculate the area of a 2-parallelepiped by computing the length of its base and the length of its height: the area is just the product of these two numbers. Since the length of the base of the parallelepiped can be taken to be v 1, its height is just v 2. This generalizes into higher dimensions as well. v 2 v 2 v 1 Figure 3. Calculating the area of a 2-parallelepiped: area = v 1 v 2 Definition 4.3. The m-volume V v 1,, v m of an m-parallelepiped defined by v 1,, v m is defined recursively by V v 1 = v 1 and V v 1,, v m = V v 1,, v m 1 v m. You can think of V v 1,, v m 1 as the m 1 volume of the base of the parallelepiped, and then v m is its height. If you don t like recursive definitions, it isn t too hard to see that this definition is equivalent to another. Theorem 4.1. The m-volume can be calculated as V v 1,, v m = v 1 v 2 vm. acs@math.uiuc.edu acs/w10/math416 Page 7 of 7

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