MTH 102: Linear Algebra Department of Mathematics and Statistics Indian Institute of Technology - Kanpur Problem Set 5 Problems marked (T) are for discussions in Tutorial sessions. 4 ? 1. Let S = fe1 + e4; −e1 + 3e2 − e3g ⊂ R . Find S . ? Solution: (e1 +e4) is the set of all vectors that are orthogonal to e1 +e4. That is, the set of 1 0 0 1 0 all xT = (x ; : : : ; x ) such that x +x = 0. So S? is the solution space of . 1 4 1 4 −1 3 −1 0 0 Apply GJE and get it. Otherwise apply GS with fe1 + e4; −e1 + 3e2 − e3; e1; e2; e3; e4g. Linear span of the last two vectors of the orthonormal basis is S?. 2 2. Show that there are infinitely many orthonormal bases of R . cos θ − sin θ Solution: Columns of , for 0 ≤ θ < 2π, form bases of 2. Idea is that take sin θ cos θ R fe1; e2g and then counter-clockwise rotate the set by an angle θ. 3. (T) What is the projection of v = e1+2e2−3e3 on H := f(x1; x2; x3; x4): x1+2x2+4x4 = 0g? 8203 2 23 2 439 <> 0 −1 0 => Solution: Basis for H: 6 7; 6 7; 6 7 . 415 4 05 4 05 > > : 0 0 −1 ; 8 203 2 23 2 439 <> 0 −1 8 => Orthonormalize: w = 6 7; w = p1 6 7; w = p1 6 7 : 1 415 2 5 4 05 3 105 4 05 > > : 0 0 −5 ; 2 03 2 43 2 163 0 8 32 The projection is hv; w iw + hv; w iw + hv; w iw = 6 7 + 0w + 20 6 7 = 1 6 7. 1 1 2 2 3 3 4−35 2 105 4 05 21 4−635 0 −5 −20 213 2 Alternately: Let x be the projection. Then v − x is parallel to 6 7, the normal vector of H. 405 4 As ub is the unit vector in the direction of the vector u, we get 213 2 13 213 2 163 2 2 2 32 v − x = hv; v − xiv − x = 5 6 7. So x = 6 7 − 5 6 7 = 1 6 7. \ \ 21 405 4−35 21 405 21 4−635 4 0 4 −20 n T 4. Let V be a subspace of R . Then show that dim V = n − 1 if and only if V = fx : a x = 0g for some a 6= 0. T T Solution: Let V = fx : a x = 0g = N (A); where A = a : Since a 6= 0; we see that rank(A) = 1 and hence dim V = n − 1 (use dim(N (A)) + dim(col space(A)) = n). 2 Conversely, suppose that dim V = n − 1. Get an orthonormal basis fu1;:::; un−1g of V and n T extend it to an orthonormal basis fu1;:::; ung of R . Then V = fx : un x = 0g. 5. (T) Does there exist a real matrix A, for which, the Row space and column space are same but the null-space and left null-space are different? Solution: Not possible. Use the fundamental theorem of linear algebra which states that ? N (A) = col space(AT ) and N (AT ) = (col space(A))? : That is, same row and column spaces require us to have a square matrix. This further implies that the dimension of null-spaces have to be same. Now, null-space and left-null-space are orthogonal to row and column spaces, respectively (which are same in this case). Hence, the Null-spaces are also same. 6. (T) Consider two real systems, say Ax = b and Cy = d. If the two systems have the same nonempty solution set, then, is it necessary that row space(A) = row space(C)? Solution: Yes. Observe that they have to be systems with the same number of variables. So, the two matrices A and C have the same number of columns. If they have the unique solution then N (A) = f0g = N (C). If it has infinite number of solutions then let Sh be the solution set of the corresponding homogeneous system Ax = 0 and Cy = 0. Thus, N (A) = N (C). So, by fundamental theorem of linear algebra, col space(AT ) = col space(CT ). That is, row space(A) = row space(C). 7. Show that the system of equations Ax = b given below x1 + 2x2 + 2x3 = 5 2x1 + 2x2 + 3x3 = 5 3x1 + 4x2 + 5x3 = 9 has no solution by finding y 2 N (AT ) such that yT b 6= 0. Solution: Note that if the system has a solution x0 then, we get Ax0 = b. Thus, for any y 2 N (AT ), we have T T T T T T y b = y (Ax0) = (y A)x0 = (A y) x0 = 0 b = 0: (1) 2−13 253 T But, it is easy to check that 4−15 is in N (A ) and −1 −1 1 455 = −1. A contradiction 1 9 to Equation (1). Thus, the given system has no solution. 8. (T) Suppose A is an n by n real invertible matrix. Describe the subspace of the row space of A which is orthogonal to the first column of A−1. Solution: Let A[:; j] (respectively, A[i; :]) denote the j-th column (respectively, the i-th row) −1 −1 of A. Then, AA = In implies hA[i; :];A [:; 1]i = 0 for 2 ≤ i ≤ n. So, the row subspace of A which is orthogonal to the first column of A−1 equals LS(A[2; :];A[3; :];:::;A[n; :]). 3 9. (T) Let An×n be any matrix. Then, the following statements are equivalent. (i) A is unitary. n (ii) For any orthonormal basis fu1;:::; ung of C , the set fAu1;:::;Aung is also an or- thonormal basis. ∗ Solution: (i) ) (ii): Suppose A is unitary. Then hAui;Auji = hui;A Auji = hui; uji. It n follows that fAu1;:::;Aung is orthonormal, hence a basis of C . (ii) ) (i): Suppose (ii) is satisfied by A. Consider the standard basis fe1;:::; eng. By hypothesis fAe1;:::;Aeng is an orthonormal basis. That is the columns of A form an or- thonormal basis, that is, A∗A = I. 10. Let V be an inner product space and S be a nonempty subset of V. Show that (i) S ⊂ (S?)?. ? ? (ii) If V is finite dimensional and S is a subspace then (S ) = S. ? ? (iii) If S ⊂ T ⊂ V; then S ⊃ T . (iv) If S is a subspace then S \ S? = f0g. Solution: (i) x 2 S ) hw; xi = 0,for all w 2 S? ) x ? S? ) x 2 (S?)?. (ii) If S = f0g; V we have nothing to show. So let S 6= f0g; V. Take a basis of S, apply GS to get an orthonormal basis fu1;:::; ukg of S. Extend that to an orthonormal basis ? fu1;:::; uk; w1;:::; wmg of V. It is easy to show that wi 2 S . ? ? P P ? ? Now let x 2 (S ) ⊂ V. Thus x = αiui + βjwj, for some αi; βj 2 C. As x 2 (S ) , ? we have hx; yi = 0, for all y 2 S . In particular hx; wji = 0,for all j. Thus βj = 0; for all j. P Thus x = αiui 2 S. (iii) Obvious. (iv) Let x 2 S \ S?. Then x ? S. In particular hx; xi = 0. Thus x = 0. P 2 11. Let A1; ··· ;Ak be k real symmetric matrices of order n such that Ai = 0. Show that each Ai = 0. n Solution: For each x 2 R we have T X 2 X T 2 X T T X 2 0 = x Ai x = x Ai x = x Ai Aix = kAixk : Hence, Aix = 0 for each i and for each x. In particular, Aie1 = 0;Aie2 = 0;:::;Aien = 0 ) Ai = 0. 12. Let V be a normed linear space and x; y 2 V. Is it true that kxk − kyk ≤ kx − yk? 13. (T) Polar Identity: The following identity holds in an inner product space. • Complex IPS : 4hx; yi = kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2. 4 • Real IPS : 4hx; yi = kx + yk2 − kx − yk2 Solution: We see that kx + yk2 = hx; xi + hx; yi + hy; xi + hy; yi; kx − yk2 = hx; xi − hx; yi − hy; xi + hy; yi ikx + iyk2 = ihx; xi + ihx; iyi + ihiy; xi + hiy; iyi and ikx − iyk2 = ihx; xi − ihx; iyi − ihiy; xi + hiy; iyi: Hence kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2 = 2hx; yi + 2hy; xi + 2ihx; iyi + 2ihiy; xi = 2hx; yi + 2hy; xi − 2i2hx; yi + 2i2hy; xi = 4hx; yi: 14. Just for knowledge, will NOT be asked Let k · k be a norm on V. Then k · k is induced by some inner product if and only if k · k satisfies the parallelogram law: kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2: Solution: See the appendix in my notes. 15. Show that an orthonormal set in an inner product space is linearly independent. n P Solution: Let S be an orthonormal set and suppose that αixi = 0; for some xi 2 S. i=1 n P Then αi = hxi; αjxji = hxi; 0i = 0; for each i: Thus, S is linearly independent. j=1 16. Let A be unitarily equivalent to B (that is A = U ∗BU for some unitary matrix U). Then P 2 P 2 jaijj = jbijj . ij ij Solution: We have X 2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ X 2 jaijj = tr(A A) = tr(U B UU BU) = tr(U B BU) = tr(B BU U) = tr(B B) = jbijj : ij ij 17. For the following questions, find a projection matrix P that projects b onto the column space of A, that is, P b 2 col(A) and b − P b is orthogonal to col(A).