Question

the last three digit 552

In this HW, the values of a, b and c are the last three digits of your student ID. For example, if your student ID is 201802321 then a = 3, b = 2 and c=1. 1. (5pts) Evaluate the eigenvalues of the following matrix a +5 4 0 0 -1 a+10 0 0 0 0 0 -2 2 + c 2. (7pts) Let -3 1 B= 1 -2 1 3 2 (i) Check if 3 is an eigenvalue of B. (ii) Show that B is Diagonalizable and then compute B" for n = (a + b +3)(c + 2). 3. (4pts) Let A be an nonsingular n by n matrix, prove that N(A) = N(ATA) and then conclude that rank(A) = rank(ATA). 4. (4pts) Find the transition matrix from Sto Tif S = {t2-t+2, –+2+2t - 1, 262 – +2} and T = {t?, t-1, 1} are two given bases for P. 5. (6pts ) Consider the following matrix A A= 1 -1 -2 2 -1 -3 -20 2 0 -1 -3 0 1 1 0 -2 2 1 2 (a) Find Bases for the Null-Space and Row-Space of A. (b) What is the Rank and the Nullity of A and of AT. (C) Check whether (1, 2, 3, 4)' E Col-space(A). 6. (4 Show that the set {(x - y.y+2, 1+2.y - 2); 1,4, 2 € R} is a subspace of R' and find its Basis.

solve only question 4

Answer 1

4. Given S = 7,= 42-4+2, P = =42+24-1, pg = 2824+2] & T= {2, = 4², 2 2 = k1, 2z = 1} . We have to find transition modice from basis & 8 to T. this we write element of s as a linear combination of element of T. p = t²t+2 1 t²-1 (t-1)+1 12-122 + 193 Therefore coordinate vector of g h with respect to basis T is [+] -ld² +2(4-1) + 1 27. P2 = -t²+2t - 2 = -12 + 2q2 + 123 Therefore, [P2] 7 2 P₂ = 2t² rt+2 24-144-1) + 1 = 29, -192t 183

2 - nuo The tramistion matrix from s to I is Ps [M), [no]> [p];] - zers 1 -1 1 2 PS > T 2 -1 Answer 1