PART II: Do any FIVE of the following SIX problem sets and show all work (12 poi
ID: 3117401 • Letter: P
Question
PART II: Do any FIVE of the following SIX problem sets and show all work (12 points for each problem). 6. Show that the polynomials 1+1, 2-1, 2-1 form a basis for the vector space P of polynomials with real coefficients with degree at most 2. 7. Consider the bases B = {vi, va) and B' = {vi, , where (a) Find the transition matrix from B to B. (b) Knowing that the B coordinate vector of w is (wlB (1,2), find (wlgy 8. Find bases for the row space and column space of the matrix 1 452 A- 2 1 3 0 -1 3 2 2 9. Consider the transformation T : R2 R2 given by (a) Show that T is a linear transformation. (b) Show that T is invertible and find T(,y). 10. Consider the linear transformations T. R3 R3 and Ta : R3 R3 given by Tj(z, y, z) (4x,-2x + y,-z-Sy) = (a) Find the standard matrix of T, oT (b) Give a formula for (Ti oT)(, y,z). 11. Consider the matrix A given by 1 0 0 A 0 1 1. 0 1 1 Find a matrix P that diagonalizes the matrix A and provide the corresponding diagonal matrix.Explanation / Answer
6. Let A =
1
1
0
0
0
2
1
-1
-1
It may be observed that the entries in the columns of A are the coefficients of x2,x and the scalar multiples of 1 in the given polyniomials. To determine whether the given polyniomials form a linearly independent set, we will reduce A to its RREF. The RREF of A is I3. It implies that the given vectors are linearly independent and span P2 and hence form a basis for P2 .
7. Let A =
1
-1
2
4
3
-1
2
-1
The RREF of A is
1
0
0
-5/2
0
1
-2
-13/2
Hence the transition matrix from B to B’ is M =
0
-5/2
-2
-13/2
If [w]B = (1,2)T, then [w]B’ =M(1,2)T = (-5,-15)T.
8. For determining the bases for Row(A) and Col(A), we will reduce A to its RREF which is
1
0
1
-2/7
0
1
1
4/7
0
0
0
0
Thus, {(1,0,1,-2/7),(0,1,1,4/7) } is a basis for Row A and { (1,0,0)T,(0,1,0)T} is a basis for Col(A).
9. (a). Let X = (x,y) and Y = (p,q) be 2 arbitary vectors in R2 and let k be an arbitrary scalar. Then T(X+Y) = T(x+p,y+q) = (x+p+2y+2q, x+p-y-q) and T(X)+T(Y) = (x+2y,x-y)+ (p+2q,p-q) =(x+p+2y+2q, x+p-y-q) so that T(X+Y) = T(X)+T(Y). Thus T preserves vector addition. Also, T(kX) = T(kx,ky) = (kx+2ky,kx-ky)=k(x+2y,x-y). Hence T preserves scalar multipliucation. Therefore, T is a linear transformation.
(b).We have T(x,y) = (x+2y,x-y). Then T(1,0) = (1,1) and T(0,1) = (2,-1). Hence the standard matrix of T is A =
1
2
1
-1
Since det(A) = -30, hence A is invertible. Therefore T is invertible. Also T-1(x,y) = A-1(x,y) , where A =
1/3
2/3
1/3
-1/3
Then T-1(x,y) = (x/3+2y/3,x/3-y/3).
11. We have A =
1
0
0
0
1
1
0
1
1
The eigenvalues of A are solutions to its characteristic equation det(A-I3)=0 or, 3-32+2=0 or,(-1)(-2) = 0. Thus, the eigenvalues of A are 0,1 and 2. The eigenvectors of A associated with these eigenvalues are (0,1,-1),(1,0,0) and (0,1,1) respectively. Then A = PDP-1 , if P
0
1
0
1
0
1
-1
0
1
and D =
0
0
0
0
1
0
0
0
2
1
1
0
0
0
2
1
-1
-1
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