Math 110 Answers for Homework 8

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1 Math 110 Answers for Homework 8 1. Determine if the linear transformations described by the following matrices are invertible. If not, explain why, and if so, nd the matrix of the inverse transformation. (a) (e) fl (b) (f) f f fl (c) (d) Solution. In order for a matrix to be invertible, it must be square (say n ˆn) and have rank n, i.e., if we put the matrix in RREF, we should get the n ˆn identity matrix I n. In class we ve seen that this is exactly the condition so that the linear transformation described by the matrix is invertible as a function. (a) This matrix is pretty clearly invertible: we could either row reduce it, or even, thinking geometrically, see that this transformation stretches the x-axes by a factor of 4 and the y-axis by a factor of 3. Its inverse is, from either point of view, the matrix (b) This matrix can not be invertible; it s not even square. (c) This matrix is invertible. Row reducing: we see that the inverse is 9 7 RREF ù (d) This is matrix is not invertible, it has rank 1. If we row reduce, we get 3 6 RREF 1 2 ù

2 (e) This matrix isn t invertible. Its RREF is and so it has rank 2. In fact, there was no need to even compute the RREF. The original matrix has a row of zeros on the bottom, and so the RREF will also have a bottom row which is all zeros, and so the matrix can t have rank 3. (f) This matrix is invertible. Row reducing, we get f f fl RREF ù so that the inverse matrix is f f fl. f f fl, 2. Suppose that A is the matrix A (a) Find the inverse of A. (b) Explain why, for any values of a, b, and c, the equations always have a unique solution. 5x ` 2y ` 4z a 2x ` 3y ` z b 5x ` 6y ` 3z c (c) Find this unique solution (in terms of a, b, and c). 2

3 Solution. (a) We can check that A is invertible, and nd the inverse at the same time, by row-reducing: fl RREF ù and so the inverse of A is the matrix B (b) If T is the linear map T : R 3 ÝÑ R 3 corresponding to the matrix A, solving the system of equations is the same as nding those vectors px,y,zq with T px,y,zq pa, b, cq. Since the transformation T is invertible (its matrix A is invertible), we know that there is a unique solution px,y,zq for each pa,b,cq in R 3. Alternatively, since the RREF of A is the identity matrix I 3, the usual argument with the row reduced form shows us that there is a unique solution. This is of course really the same argument as the one above. (c) The denition of the inverse transformation T 1 is that it undoes what T does, so that for any vector pa,b,cq, T 1 pa,b,cq is exactly the vector px,y,zq such that T px,y,zq pa,b,cq. Since we already know the matrix B for T 1, we can use this to compute px,y,zq in terms of pa,b,cq: x y z fl fl a b c fl fl 3a `18b 10c a 5b `3c 3a 20b `11c 3. Suppose that T 1 : R n ÝÑ R m and T 2 : R m ÝÑ R p are linear transformations. (a) If T 1 and T 2 are inective, prove that T 2 T 1 is inective. (b) If T 1 and T 2 are surective, prove that T 2 T 1 is surective. (c) If T 1 and T 2 are invertible, prove that T 2 T 1 is invertible. 3

4 Solution. (a) A linear transformation T is inective if v 1 v 2 implies that T p v 1 q T p v 2 q. In other words, if different input vectors are always sent to different output vectors. We want to show that T T 2 T 1 is inective, using the fact that both T 1 and T 2 are inective. Suppose that v 1 and v 2 P R n, and v 1 v 2. Since T 1 is inective, T 1 p v 1 q T 1 p v 2 q. But since T 2 is inective, this means that T 2 pt 1 p v 1 qq T 2 pt 1 p v 2 qq, i.e,. that pt 2 T 1 qp v 1 q pt 2 T 1 qp v 2 q. That is, if v 1 v 2 then pt 2 T 1 qp v 1 q pt 2 T 1 qp v 2 q, so T 2 T 1 is inective. Alternate Solution We can solve the problem using the equivalent characterization of inectivity: T p v 1 q T p v 2 q implies that v 1 v 2. Let s start with v 1 and v 2 in R n, and suppose that pt 2 T 1 qp v 1 q pt 2 T 1 qp v 2 q. To show that T 2 T 1 is inective we have to show that this implies that v 1 v 2. But if T 2 pt 1 p v 1 qq T 2 pt 1 p v 2 qq, then, since T 2 is inective, this can only happen if T 1 p v 1 q T 1 p v 2 q. (They re the things that we re feeding into T 2 ). But since T 1 is inective, this can only happen if v 1 v 2 which is exactly what we wanted to show. So, since we ve shown that for any v 1 and v 2, T 2 pt 1 p v 1 qq T 2 pt 1 p v 2 qq implies that v 1 v 2, this means that T 2 T 1 is inective. (b) Saying that a linear transformation T : R a ÝÑ R b is surective is the same thing as saying that for any w in R b, there is at least one v in R a with T p vq w. We re given that both T 1 and T 2 are surective, and we want to show that T 2 T 1 is too. So suppose that w is any vector in R p. We need to show that there is a vector v in R n with T 2 pt 1 p vqq w. Since T 2 is surective, we know that there is some vector u in R m with T 2 p uq w. Since T 1 is surective, we also know that there is some vector v in R n with T 1 p vq u. But then T 2 pt 1 p vqq T 2 p uq w, so T 2 T 1 is surective too. (c) Saying that a linear transformation T is invertible is the same as saying that T is inective and surective. We re given that both T 1 and T 2 are invertible, and we want to show that T 2 T 1 is. From part (a), T 2 T 1 is inective, since both T 1 and T 2 are. From part (b), T 2 T 1 is surective since both T 1 and T 2 are. Therefore, T 2 T 1 is both inective and surective, and so is invertible. Remark. Nothing about these arguments depended on T 1 and T 2 being linear transformations. The arguments really prove that the composition of two inective functions are inective, that the composition of two surective functions are surective, and that the composition of two biective functions are biective, without the functions needing to be linear. 4

5 4. For each of the following subsets W of R 3, either show that they are subspaces, or show why they aren t subspaces by explaining which of the three conditions don t hold. (a) W tpx,y,zq P R 3 x 2 `y 2 z 2 u. (b) W tpx,y,zq P R 3 px,y,zq is orthogonal to p3,1, 2qu. (c) W tpx,y,zq P R 3 x `y `z ě 0u. Solution. (a) The set in(a) isnotasubspace ofr 3, althoughit passes two ofthetests. Certainly 0 p0,0,0q P W since 0 2` This set also passes test (iii) w px,y,zq P W and c is any scalar then c w pcx,cy,czq P W, since if w P W then x 2 `y 2 z 2 and, multiplying by c 2, this gives pcxq 2 ` pcyq 2 pczq 2, which is the test for c w pcx,cy,czq to be in W. However the set does not pass test (iii). If w 1, w 2 P W then this does not imply that w 1 ` w 2 P W. A simple example is to take w 1 p1,0,1q, w 2 p0,1,1q which are both in W, but then w 1 ` w 2 p1,1,2q is not in W since 1 2 ` The set W is actually a cone in R 3 ; here is a sketch at the side. Hopefully by looking at the picture you can see that 0 P W ( 0 is the vertex of the cone), and that W is closed under scalar multiplication but not closed under addition. (b) The set W in part (b) is a subspace. x 2 ` y 2 z 2 ù For a vector px,y,zq to be orthogonal to p3,1, 2qmeans that px,y,zq p3,1, 2q 3x `y 2z 0. Let s go through the tests to be a subspace : (i) 0 p0,0,0q P W since p3,1, 2q 0 3p0q `1p0q 2p0q 0. (ii) If w 1 px 1,y 1,z 1 q and w 2 px 2,y 2,z 2 q P W then w 1 ` w 2 px 1 ` x 2,y 1 ` y 2,z 1 `z 1 q P W since 3px 1 `x 2 q ` py 1 `y 2 q 2pz 1 `z 2 q p3x 1 `y 1 2z 1 q ` p3x 2 `y 2 2z 2 q 0 `0 0. In the second last equality we have used the fact that so we know that 3x 1 `y 1 2z 1 0 and 3x 2 `y 2 2z 2 0, since w 1 and w 2 are in W. (iii) If w px,y,zq P W then for any scalar c, c w pcx,cy,czq P W since 3cx `cy 2cz cp3x `y zq cp0q 0. In the second last equality we have used the fact that 3x`y 2z 0 since w P W. Therefore W is a subspace of R 3, since W passes all the tests. 5

6 (c) The set W in part (c) is not a subspace of R 3. It fails the scalar multiplication test for multiplication by negative numbers, but passes all the other tests. Certainly 0 p0,0,0q P W since 0 `0`0 0 ě 0. If w 1 px 1,y 1,z 1 q and w 2 px 2,y 2,z 2 q are in W (so x 1 `y 1 `z 1 ě 0 and x 2 `y 2 `z 2 ě 0) then w 1 ` w 2 px 1 `x 2,y 1 ` y 2,z 1`z 2 q P W since px 1`x 2 q`py 1`y 2 q`pz 1`z 2 q px 1`y 1`z 1 q`px 2`y 2`z 2 q ě 0 (each of the pieces we re adding is ě 0). However, as stated above, the set fails to be closed under scalar multipliction when multiplying by a negative number. For example, w p1,1,1q P W (since 1 ` 1 ` 1 3 ě 0, but w p 1, 1, 1q is not in W. Therefore W is not a subspace. In this case the set W is a half space it s the plane x `y `z 0 and everything in R 3 to one side of this plane. Thinking about this picture, you should be able to see that W contains 0, is closed under addition and scalar multiplication by positive numbers, but not scalar multiplication by negative numbers. 6