UPSC Civil Services Main Mathematics Linear Algebra

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1 UPSC Civil Services Main 99 - Mathematics Linear Algebra Sunder Lal Retired Professor of Mathematics Panjab Universit Chandigarh June 4, 7 Question (a Let U and V be vector spaces over a field K and let V be of finite dimension. Let T : V U be a linear transformation, prove that dim V = dim T(V + dim nullit T. Solution. See question 3(a, ear 998. Question (b Let S = {(x,, z x + + z =, x,, z R}. Prove that S is a subspace of R 3. Find a basis of S. Solution. S because (,, S. If (x,, z, (x,, z S then α (x,, z + α (x,, z S because (α x + α x + (α + α + (α z + α z = α (x + + z + α (x + + z =. Thus S is a subspace of R 3. Clearl (,,, (,, S and are linearl independent. Thus dim S. However (,, S, so S R 3. Thus dim S = and {(,,, (,, } is a basis for S. Question (c Which of the following are linear transformations?. T : R R defined b T(x = (x, x.. T : R R 3 defined b T(x, = (x,, x. 3. T : R R 3 defined b T(x, = (x +,, x. 4. T : R R defined b T(x = (,. Solution.

2 . Thus T is a linear transformation. T(αx + β = (αx + β, αx β = (αx, αx + (β, β = αt(x + βt(. T((, = T(, = (4,, T(, = (,, Thus T is not a linear transformation. 3. T(α(x, + β(x + = T(αx + βx, α + β Thus T is a linear transformation. = (αx + βx + α + β, α + β, αx + βx = α(x +,, x + β(x +,, x = αt(x, + βt(x, 4. T((, = T(, = (, T(, Thus T is not a linear transformation. Question (a Let T : M, M,3 be a linear transformation defined b (with the usual notation ( ( ( ( 3 6 T =, T = 4 5 Find T. Solution. T ( ( ( = x + ( 3 = (x ( 6 = ( x + 4 x 3x 3 4x 4 x 5x 3

3 Question (b For what values of η do the following equations x + + z = x + + 4z = η x z = η have a solution? Solve them in each case. Solution. Since the determinant of the coefficient matrix ( 4 is, the sstem has to 4 be consistent to be solvable. Clearl x z = 3(x + + 4z (x + + z. Thus for the sstem to be consistent we must have η = 3η, or η =,. If η =, then x + + z =, x + + 4z = so + 3z =, or = 3z, x = + z. Thus the space of solutions is {( + z, 3z, z z R}. Note that the rank of the coefficient matrix is, and consequentl the space of solutions is one dimensional. If η =, then x + + z =, x + + 4z =, so + 3z = or = 3z, hence x = z. Consequentl, the space of solutions is {(z, 3z, z z R}. Question (c Prove that a necessar and sufficient condition of a real quadratic form x Ax to be positive definite is that the leading principal minors of A are all positive. Solution. Let all the principal minors be positive. We have to prove that the quadratic form is positive definite. We prove the result b induction. If n =, then a x > a >. Suppose as induction hpothesis the result is true for n = m. Let S = ( B B B be a matrix of a quadratic form in m + variables, where B is m m, B k is m and k is a single element. Since all principle minors of B are leading principal minors of S, and are hence positive, the induction hpothesis gives that B is positive definite. This means that there exists a non-singular m m matrix P such that P BP = I m (We shall prove this presentl. Let C be an m-rowed column to be determined soon. Then ( ( ( ( P B B P C P C = BP P BC + P B k C B P + B P C BC + C B + B C + k B Let C be so chosen that BC + B =, or C = B B. Then ( ( ( ( P B B P C P C = BP k B C + k B Taking determinants, we get P S P = B C + k, because P BP = I m, and B C + k is a single element. Since S ( >, it( follows that B C + k >, so let B C + k = α. Then P C Q Im SQ = I m+ with Q = α. Thus the quadratic forms of S and I m+ take the same values. Hence S is positive definite, so the condition is sufficient. 3

4 The condition is necessar - Since x Ax is positive definite, there is a non-singular matrix P such that P AP = I A P = A >. Let r < n. Let x r+ =... = x n =, then we obtain a quadratic form in r variables which is positive definite. Clearl the determinant of this quadratic form is the r r principal minor of A which shows the result. Proof of the result used: Let A be positive definite, then there exists a non-singular P such that P AP = I. We will prove this b induction. If n =, then the form corresponding to A is a x and a >, so that P = ( a. Take a a... P =. (n (n then a a 3... a n P AP = a 3. (n (n a n Repeating this process, we get a non-singular Q such that a... Q AQ =. (n (n Given the (n (n matrix on the lower right, we get b induction P s.t. P ((n (n matrixp is diagonal. Thus P, P, P AP = [α,..., α n ] sa. Take R = diagonal[ α,..., α n ], then R P APR = I n. Question 3(a State the Cale-Hamilton theorem and use it to find the inverse of ( 4 3. Solution. Let A be an n n matrix. If λi A = λ n + a λ n a n = is the characteristic equation of A, then the Cale-Hamilton theorem sas that A n + a A n a n I = i.e. a matrix satisfies its characteristic equation. The characteristic equation of A = ( 4 3 is λ 4 3 λ = λ 5λ + = 4

5 B the Cale-Hamilton theorem, A 5A + I =, so A(A 5I = I, thus A = (A 5I. Thus A = [( ( ] ( 5 3 = Question 3(b Transform the following into diagonal form x + x, 8x 4x + 5 and give the transformation emploed. Solution. Let A = ( (, B = 8 5 Let = A λb = 8λ + λ + λ 5λ = 5λ + 4λ 4λ 4λ Thus 36λ 9λ =, so λ = 9± 8+44 =,. 7 3 Let (x, x be the vector such that (A λb ( x x = with λ =. Thus 5x x 3 = x = x. We take x = ( so that (A λbx = with λ =. Similarl, if (x 3, x is the vector such that (A λb ( x x = with λ =, then 5x 3 + 5x 6 =, so x + x =. We take x = (. Now x Ax = ( ( ( = ( ( = 3 x Ax = ( ( ( = ( ( = 3 x Ax = ( ( ( = ( ( = If P = (x x, then P AP = ( 3 3 Similarl Thus P BP = ( 9 36, thus x + x 3X 3Y b P = (. x Bx = ( ( ( 8 5 = ( ( 6 3 = 9 x Bx = ( ( ( 8 = ( ( 5 = 36 x Bx = ( ( ( 8 5 = ( ( =, so 8x 4x + 5 is transformed to 9X + 36Y b ( ( X Y = P x Question 3(c Prove that the characteristic roots of a Hermitian matrix are all real, and the characteristic roots of a skew Hermitian matrix are all zero or pure imaginar. Solution. For Hermitian matrices, see question (c, ear 995. If H is skew-hermitian, then ih is Hermitian, because (ih = ih = ih = ih as H = H. Thus the eigenvalues of ih are real. Therefore the eigenvalues of H are ix where x R. So the must be (if x = or pure imaginar. 5