Math 310 Hour Exam 2 Instructor: Phil Grizzard Fri 3 November 2006
(Solutions)
1.(20pts) As commonly done in class, we represent an arbitrary vector in R2by x= (x1, x2)T.
Prove whether or not each of the following is a linear transformation from R2to R3.
(a) L(x) = (x1,0,0)T.
L(αx+y) = L((αx1+y1, αx2+y2)T)=(αx1+y1,0,0)T= (αx1,0,0)T+ (y1,0,0)T=
α(x1,0,0)T+ (y1,0,0)T=αL(x) + L(y). Thus Lis linear.
(b) L(x) = (x1,0,1)T.
Lis not linear. For example, if α6= 1, then L(αx)=(αx1,0,1) 6=αL(x). L also fails
summation.
2.(15pts) Find the distance from the point p= (3,3,3) to the plane x−y+ 3z= 0.
The plane goes through the origin, and its normal vector (obtained from the coefficients of
the equation) is N= (1,−1,3)T. We let vbe the vector from the origin to our point p, so
v= (3,3,3)T. The distance dfrom vto the plane is the absolute value of the scalar projection of
vonto N. So d=|vTN|
||N|| =9
√11 .
3.(10pts) Using the basis {1,1 + x, 1 + x2}for the space of polynomials of degree at most 2, give
the coordinates of the “vector” 1 + x+x2.
We are trying to find the coordinates (a, b, c)Tsuch that 1+x+x2=a·1+ b·(1 +x)+ c·(1+x2).
We can see immediately that we need c= 1 and b= 1, and so we have
1 + x+x2=−1·1+1·(1 + x)+1·(1 + x2).
Thus the coordinates of this vector are (−1,1,1)Tin the given basis.
4.(20pts) In polynomial space, are the “vectors” 1 + x, 1 −x, and 2 + x2linearly independent?
Write your solution clearly and precisely.
By definition of linear independence, these vectors are linearly independent ⇔given the equa-
tion c1(1 + x) + c2(1 −x) + c3(2 + x2) = 0, the only solution is c1=c2=c3= 0. The given
equation distributes out to c1+c1x+c2−c2x+ 2c3−c3x2= 0. Collecting like terms tells us that
c1+c2+ 2c3= 0,c1−c2= 0, and c= 0. Substituting c= 0 into c1+c2+ 2c3= 0 gives c1+c2= 0.
Combining this with c1−c2= 0 shows that c1=c2=c3= 0. Thus these vectors are linearly
independent.
5.(15pts)
(a) Find the dimension of the subspace Vof R3, where Vis spanned by the vectors:
u1=
2
0
1
,u2=
1
2
3
,u3=
1
4
5
.
We make matrix Aout of these vectors, so A=
211
024
135
. Taking the determinant of this
gives det A=−26= 0. Hence these vectors are linearly independent, and since there are 3 of them,
they form a basis for the entire space R3, and so V=R3. Thus the dimension of Vis 3.
Alternatively, we could row reduce A, which gives the identity matrix. Vis the same as the column
space of this matrix, which has dimension 3.
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