Someone please help answer these Determine whether {(x1, x2 } x 2/3 = x 2/3 } fo

Question

Someone please help answer these

Determine whether {(x1, x2 } x 2/3 = x 2/3 } forms a subspace of R2. Determine whether the following set is spanning set of R2 Determine whether the vectors cos2 (x/2), cos(x), 1 linearly independent in C([-pi, pi] Let Show that x2, x2, and x3 are linearly independent. What is the dimension of Span (x1, x2, x3)? Give a geometric description of Span (x1, x2, x3) Find a basis and compute the dimension of (i) row space, (ii) the column space, and (iii) the null space of the following matrix. Determine the dimension of the subspace of R3 spanned by the given vectors. Determine whether b is in column space of A, and state whether the system Ax = b is constant Determine whether the following are linear transformations: L(x) = (1 + x1, x2)T, from R3 into R2. L(x) = (x3, x1 + x2)T, from R3 into R2. L(p(x)) = xp' (x) from P2 to P3. Find the Kernel the linear transformation L : R3 rightarrow R2 defined by L((x1, x2, x3)T) = (x1 + 2x2, x2 + 2x3)T and image of the subspace S spanned by e1 and e2.

Explanation / Answer

1.

(a) No,since (1,1) and (1,-1) are both in the given set but their sum (2,0) is not.

(b) Yes,since (1/3)(b+a)(1;2) + (1/3)(-2a+b)(-1;1) = (a;b), every vector (a,b) in R^2 is in the span{(1;2), (-1;1)}

2.

(a) 4x1+(-2)x2+(-1)x3 = 0.

Hence x1, x2 and x3 are dependent. (since there is a triple (a,b,c) not all zero, here (4,-2,-1) so that ax1+bx2+cx3 = 0)

(b)Neither is a linear multiple of other.Hence x1,x2 are linearly independent.

(c)x1,x2 are independent and x3 is in the span(x1,x2) as x3 = 4x1-2x2.Hence dimension of span(x1,x2,x3) = dimension of span(x1,x2) = 2.

(d)It is a plane formed by vectors x1,x2.

3.

(a)

(i)if we call rows as r1,r2,r3 (1st row,2nd row and 3rd row resply), then r1 and r2 are independent as none is scalar multiple of other.Also r3 = 3r2 - 2r1.

Hence basis of row space = r1,r2 and thereby dimension = 2.

(ii)lets call the columns as c1,c2,c3,c4 (1st,2nd,3rd,4th columns resply)

Then c1 and c2 are independent as none is scalar multiple of another.

Also c3 = (7/3)c1 - (5/3)c2; c4 = (1/3)c1 + (1/3) c2; c5 = (5/3)c1 - (4/3)c2.

Thus c3,c4,c5 are in the span(c1,c2).

Hence basis of the column space = c1,c2 and thereby its dimension = 2.

(iii)We have (7/3; -5/3; -1; 0; 0) , (1/3; 1/3; 0; -1; 0) , (5/3; -4/3; 0; 0; -1) as elements of null space by the observation in part ii.

Also these three are independent as x (7/3; -5/3; -1; 0; 0) + y(1/3; 1/3; 0; -1; 0)+z(5/3; -4/3; 0; 0; -1) = (0; 0; 0; 0; 0) iff x=0=y=z by looking at thrid,fourth and fifth coordinates on both sides.

Finally by rank nullity theorem dimension of null space = number of columns of the matrix - rank of the matrix

= 5 - 2 (rank of matrix = dimension of row space = dimension of column space)

= 3.

Hence (7/3; -5/3; -1; 0; 0) , (1/3; 1/3; 0; -1; 0) , (5/3; -4/3; 0; 0; -1) is a basis of null space as these are independent and also null spacew has dimension 3.

(b)

The given vectors are independent since x(1; 1; 1)+y(1; 2; 3)+z(2; 3; 1) = (0; 0; 0)

=> x+y+2z = 0 ; x+2y+3z = 0; x+3y+z = 0

=> y+z = 0,(first equation - second equation); y-2z = 0(second equation - third equation); x+y+2z = 0 (first equation)

=> y = z = 0 (y+z = 0 = y-2z => z = -2z => z = 0,which in turn imply y = 0), x = 0 (x+y+2z = 0 together with y=0=z).

Hence dimesnion of subspace spanned by these vectors = 3 (as the vectors are independent).

(c) column space of A = { x(1;2)+y(2;4) : x,y are reals } (when i put comma it is separating elements in the same row, when i put semi colon it is separating columns)

= {(x+2y; 2x+4y) : x,y are reals}

b=(2;3) is in the column space of A iff (2;3) = (x+2y; 2x+4y) for some reals x,y

iff x+2y = 2 and 2x+4y = 4 for some x,y

but this cant happen as 2x+4y = 3 not equal to 4 = 2(x+2y) = 2x+4y.

thus b is not in the column space of A.

This also means the given system is inconsistent.

5.

(a)

(i) No since (0;0) is not going to (0;0) (For a linear transformation origin has to be mapped to origin).

(ii) Yes,L(x) = Ax, where A = (0,0,1; 1,1,0) , a 2 by 3 matrix with rows (0, 0, 1) and (1,1,0).

Hence L is a linear transformation.

(iii)L(p(x)) = xp'(x).

L(p(x)+q(x)) = L((p+q)(x)) = x(p+q)'(x) = x[p'(x)+q'(x)] (linearity of differentiation)

= xp'(x)+xq'(x)

= L(p(x)) + L(q(x))

Also if k is real, then

L(kp(x))= L((kp)(x)) = x(kp)'(x) = x.[kp'(x)] (linearity of differentiation)

= k.xp'(x) = k.L(p(x)).

Hence L is linear.

(b)

Kernel = {(x1;x2;x3) : x1+2x2 = 0 and x2+2x3 = 0}

= {(4t;-2t;t) : t is real } (Just assign value t to x3 and get x2 and x1 in terms of t)

= {t (4;-2;1) : t is real }

Thus kernel is the subspace spanned by (4; -2; 1)

image of e1 = (1; 0; 0) is (1; 0)

image of e2 = (0; 1; 0) is (2; 1)

Thus span of images of e1 and e2 = span { (1; 0) , (2; 1) }

= R^2

This is because (a; b) = (a-2b) (1; 0) + b (2;1) for all (a;b) in R^2.

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