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Home , Determinant, Scalar multiplication

Math 217: Multilinearity of Professor Karen Smith (c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. A. Let V −→T V be a linear transformation where V has n.

1. What is meant by the of T ? Why is this well-defined? Solution note: The determinant of T is the determinant of the B- of T , for any B of V . Since all B-matrices of T are similar, and similar matrices have the same determinant, this is well-defined—it doesn’t depend on which basis we pick.

2. Define the of T . Solution note: The rank of T is the dimension of the image.

3. Explain why T is an isomorphism if and only if det T is not zero.

Solution note: T is an isomorphism if and only if [T ]B is invertible (for any choice of basis B), which happens if and only if det T 6= 0.

3 4. Now let V = R and let T be around the axis L (a through the origin) by an 1 0 0  3 angle θ. Find a basis for R in which the matrix of ρ is 0 cosθ −sinθ . Use this to 0 sinθ cosθ compute the determinant of T . Is T othogonal?

Solution note: Let v be any vector spanning L and let u1, u2 be an ⊥ for V = L . Rotation fixes ~v, which means the B-matrix in the basis (v, u1, u2) has 1 first column 0. In the V = L⊥, the rotation is just an ordinary rotation 0 2 like in R . The determinant is 1. The matrix is orthogonal because the columns are orthonormal, or alternatively, because the rotation map preserves the length of every vector.

B. Theorem: The determinant is multilinear in the columns. The determinant is multilinear in the rows. This means that if we fix all but one column of an n × n matrix, the determinant is linear in the remaining column. Ditto for rows.

 1  2 x x 1 2 3 1. Let ~v2 = −1 ,~v3 = 0 . Discuss how the map R → R sending y 7→ det y −1 0 0 1 z z 0 1 illustrates the Theorem in D. Is this map linear? If so, find its matrix.

3 Solution note: Yes, the map is a linear transformation from R to R. This is the content of the theorem! You can (and perhaps should!) also verify this directly by checking it preserves and scalate . Its matrix is 1 × 3 and we find it by seeing where each of the ~ei go. To find this, we compute the three given determinants, replacing the first column by each of ~e1,~e2,~e3. We get A = −1 −1 2 . T T 2. Using the row vectors ~v2 and ~v3 , illustrate what it means that the determinant function on a 3 × 3 matrix is linear in the second row. Solution note: Linearity in row 2 means:

1 −1 0  1 −1 0   1 −1 0  det x y z + det x0 y0 z0 = det x + x0 y + y0 z + z0 2 0 1 2 0 1 2 0 1

x x0 0 3 for all vectors y , y  ∈ R , and z z0

 1 −1 0  1 −1 0 det kx ky kz = k det x y z 2 0 1 2 0 1

for all scalars k.

3. Prove the Theorem in the 2 × 2 case for the first column. That is: Show that the determinant of a 2 × 2 matrix is linear in the first column. Solution note: We need to show a + a0 b a b a0 b ka b a b det = det + det , det = k det . c + c0 d c d c0 d kc d c d

Compute: (a + a0)d − (c + c0)b = ad + a0d − cb −0 cb = (ad − bc) + (a0d −0 cb) and (ka)d − (kc)b = k(ad − bc). QED.

n 4. Suppose ~v1, . . . ,~vn−1 are vectors in R and that

det[~v1 ~v2 . . .~vn−1 ~a] = 5, det[~v1 ~v2 . . .~vn−1 ~b] = 7, det[~v1 ~v2 . . .~vn−1 ~c] = −3.

Find det[~v1 ~v2 . . .~vn−1 (2~a + 4~b − 6~c + 17~v1)]. Solution note: Using multilinearity, we have

det[~v1 ~v2 . . . ~vn−1 2~a] + det[~v1 ~v2 . . . ~vn−1 4~b] + det[~v1 ~v2 . . . ~vn−1 − 6~c] + det[~v1 ~v2 . . . ~vn−1 17~v1)]

=2 det[~v1 ~v2 . . . ~vn−1 ~a] + 4 det[~v1 ~v2 . . . ~vn−1 ~b] − 6 det[~v1 ~v2 . . . ~vn−1 ~c] + 17 det[~v1 ~v2 . . . ~vn−1 ~v1)] =10 + 28 + 18 + 0 = 56.

The final zero is because a matrix with dependent columns always has determinant zero.

5. Prove that the determinant function of an n × n matrix is linear in the last column. [Hint: Use along the last column.] n Solution note: Fix ~v1, . . . ,~vn−1 in R . We need to show

det[~v1 ~v2 . . .~vn−1 ~x + ~y] = det[~v1 ~v2 . . .~vn−1 ~x] + det[~v1 ~v2 . . .~vn−1 ~y],

and similarly for multiplication. Expanding along the last column:

n+1 det[~v1 ~v2 . . .~vn−1 ~x + ~y] = (−1) [(x1 + y1)A1n − (x2 + y2)A2n + · · · ± (xn + yn)Ann] n+1 n+1 = (−1) [x1A1n − x2A2n + · · · ± xnAnn] + (−1) [y1A1n − y2A2n + · · · ± ynAnn]

= det[~v1 ~v2 . . .~vn−1 ~x] + det[~v1 ~v2 . . .~vn−1 ~y].

A similar calculation shows the is preserved too.

6. Use Laplace expansion to explain why the determinant is linear in each column and row.

7. True or False: The determinant map is linear. Justify.

Solution note: FALSE! For example det(2I2) = 4 not 2 det I2.

C. Corollary: Let A be an n × n matrix. Then det(kA) = kn det A for any scalar k. Prove this.

2a + 5p 5q + 2d a −d D. Compute det given that det = 17. p q p −q

2a + 5p 5q + 2d 2a 2d 5p 5q Solution note: det = + by linearity in the p q p q p q first row and the fact that the determinant is zero if there is a relation on the rows. a d This further becomes 2 det , again by linearity in the first row. On the other p q a −d a d hand, using linearity in the second column, = − = −34. So the p −q p q answer is −70.

E. Theorem: The determinant is alternating in the columns. The determinant is alternating in the rows. This means if we interchange two columns, the determinant changes . Ditto for rows.

0 1 2 1. Verify this for swapping the second two columns of 0 −1 0 . Also verify for swapping 1 0 1 the first and third row.

2. Prove the theorem in the case of 2 × 2 matrices.

3. Let A be an n × n matrix. Let B be obtained from A by switching columns i and j. Prove   that det A = − det B. [Hint: Write A = ~v1 ~v2 . . . ~vn . Compute the determinant of   ~v1 ~v2 . . . ~vi + ~vj . . . ~vi + ~vj . . . ~vn , where the ~vi + ~vj is in BOTH the i-th spot, and the j-th spot.] Solution note: (1) Computing the determinant of the matrix in (1) we get 2. We get -2 after doing the intended swaps. (2) We prove only alternating in rows, the case of columns being similar. We need to a b c d check that det = − det . This is easy: ad − cb = −(ac − bd). c d a b   (3) Note that ~v1 ~v2 . . . ~vi + ~vj . . . ~vi + ~vj . . . ~vn has determinant zero be- cause two columns are the same. Now expand out using the multi-linearity: this is the same as

det [~v1 ~v2 . . . ~vi . . . ~vi + ~vj . . . ~vn] + det [~v1 ~v2 . . . ~vj . . . ~vi + ~vj . . . ~vn] . Expanding in the other column and using the fact that the determinant is zero if two columns are the

0 = det [~v1 ~v2 . . . ~vi . . . ~vj . . . ~vn] + det [~v1 ~v2 . . . ~vj . . . ~vi . . . ~vn] .

It is an interesting Theorem that the determinant is the ONLY alternating multilinear function of the columns of an n × n matrix which takes the value 1 on the . More theoretical linear courses (for example, Math 420, which maybe you’ll take someday) usually take this to be the definition of the determinant. We won’t do this in Math 217, however.

2×3 2×3 F. Suppose that we have a linear transformation T : R → R whose matrix in the basis 0 1 2 3 0 0 0 −1 0 0 0 0   1 0 1 0 0 0 (A1,A2,A3,A4,A5,A6) is   . 0 −1 0 0 0 0   0 −1 0 0 0 0 0 −1 0 0 0 0

1. What is the determinant of T ?

2. What is the rank of T ?

3. Find a basis for the and image of T .

4. Suppose that (B1,B2,B3,B4,B5,B6) is another basis, where

B1 = A1 + A2,B2 = A1 − A2,B3 = 3A3,B4 = 4A4,B5 = A5 − A4 − A3 − A2,B6 = A6.

Find the matrix SB→A. Without computing: explain why the change of basis matrix has non-zero determinant.

5. Find the B-matrix of T (do not simplify!) Without computing: What is its determinant? Solution note: The rank is 3, which is less than 6. So the determinant is 0. (3) A basis for the image is represented by a maximal set of linearly independent columns. These can be taken to be the first, second and third columns. So a basis for the image is the set (A3,A1 − A2 − A4 − A5 − A6,A1). A basis for the kernel thus consists of three elements (rank nullity). To find it, we first use row-reduction to solve the system A~x = ~0, and then re-interpret the column vectors as coordinates 2×3 to get a basis in R . Solving the system we get the solutions are spanned by T [3/2 0 −3/2 1 0 0] ,~e5,~e6. Thus a basis for the kernel is (3A1−3A3+2A4,A5,A6). (4) The change of basis matrix is

1 1 0 0 0 0 1 −1 0 0 1 0   0 0 3 0 0 0 SB→A =   . 0 0 0 4 0 0   0 −0 0 0 1 0 0 0 0 0 0 1

Its determinant is non-zero because change of basis matrices are always invertible! The determinant of the B-matrix of T is zero, since it is the same as the determinant of the A-matrix of T .