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Solution to Linear Algebra Hoffman & Kunze Chapter 2.6


Exercise 2.6.1

Let α1, α2, , αn be the colunms of A. Then αiFs×1 i. Thus {α1,,αn} are n vectors in Fs×1. But Fs×1 has dimension s<n thus by Theorem 4, page 44, α1,,αn cannot be linearly independent. Thus x1,,xnF such that x1α1++xnαn=0. Thus if
X=[x1xn]then AX=x1α1++xnαn=0.


Exercise 2.6.2

(a) We use the approach of row-reducing the matrix whose rows are given by the αi:
[112130411252][1121031040333][1121001310111][1030011100131][103001110011/13][1003/1301014/130011/13]Let ρ1=(1,0,0,3/13), ρ2=(0,1,0,14/13) and ρ3=(0,0,1,1/13). Thus elements of the subspace spanned by the αi are of the form b1ρ1+b2ρ2+b3ρ3 =(b1, b2, b3, 113(14b23b1b3)).

  • α=(4,5,9,7). We have b1=4, b2=5 and b3=9. Thus if α is in the subspace it must be that
    113(14(5)3(4)9)=?b4 where b4=7. Indeed the left hand side does equal 7, so α is in the subspace.
  • β=(3,1,4,4). We have b1=3, b2=1, b3=4. Thus if β is in the subspace it must be that
    113(143(3)+4)=?b4 where b4=4. But the left hand side equals 9/134 so β is not in the subspace.
  • γ=(1,1,0,1). We have b1=1, b2=1, b3=0. Thus if γ is in the subspace it must be that
    113(143(1)0)=?b4 where b4=1. But the left hand side equals 17/131 so γ is not in the subspace.

(b) Nowhere in the above did we use the fact that the field was R instead of C. The only equations we had to solve are linear equations with real coefficients, which have solutions in R if and only if they have solutions in C. Thus the same results hold: α is in the subspace while β and γ are not.

(c) This suggests the following theorem: Suppose F is a subfield of the field E and α1,,αn are a basis for a subspace of Fn, and αFn. Then α is in the subspace of Fn generated byα1,,αn if and only if α is in the subspace of En generated by α1,,αn.


Exercise 2.6.3

We use the approach of row-reducing the matrix whose rows are given by the αi:
[101234251409][10120411104111][1012011/411/40000].Let ρ1=(1,0,1,2) and ρ2=(0,1,1/4,11/4). Then the arbitrary element of the subspace spanned by α1 and α2 is of the form b1ρ1+b2ρ2 for arbitrary b1,b2R. Expanding we get
b1ρ1+b2ρ2=(b1,  b2,  b1+14b2,  2b1+114b2).Thus the equations that must be satisfied for (x,y,z,w) to be in the subspace are
{z=x+14yw=2x+114y.or equivalently
{x+14yz=02x+114yw=0.


Exercise 2.6.4

We use the approach of row-reducing the augmented matrix:
[10i1001+i1i1010iii001][10i10001ii1i100ii1i01][10i100011+i2i1+i200ii1i01][10i100011+i2i1+i20001+3i21i1i21][10i100011+i2i1+i200012+4i52i513i5][10012i512i53i501012i51+3i52i50012+4i52i513i5]Since the left side transformed into the identity matrix we know that {α1,α2,α3} form a basis for C3. We used the vectors to form the rows of the augmented matrix not the columns, so the matrix on the right is (PT)1 from (2-17). But (PT)1=(P1)T, so the coordinate matrix of (a,b,c) with respect to the basis B={α1,α2,α3} are given by
[(a,b,c)]B=(P1)T[abc]=[12i5a+12i5b+2+4i5c12i5a+1+3i5b+2i5c3i5a+2i5b+13i5c].


Exercise 2.6.5

We row-reduce the matrix whose rows are given by the αi’s.
[10211124202152121352][10211022310110301134][10211011340003900037][10211011340001300037][10214011050001300002][10204011050001300001][10200011000001000001]Let ρ1=(1,0,2,0,0), ρ2=(0,1,1,0,0), ρ3=(0,0,0,1,0) and ρ4=(0,0,0,0,1). Then the general element that is a linear combination of the αi’s is b1ρ1+b2ρ2+b3ρ3+b4ρ4=(b1,  b2,  2b1b2,  b3,  b4).


Exercise 2.6.6

We row-reduce the matrix
[321090171212140616421130][17121003153002103005256][1712100151002103005256][17030001510000100001][17030001500000100000](a) A basis for V is given by the non-zero rows of the reduced matrix
ρ1=(1,7,0,3,0),ρ2=(0,0,1,5,0),ρ3=(0,0,0,0,1).(b) Vectors of V are any of the form b1ρ1+b2ρ2+b3ρ3
=(b1,  7b1,  b2,  3b1+5b2,  b3) for arbitrary b1,b2,b3R.

(c) By the above, the element (x1,x2,x3,x4,x5) in V must be of the form x1ρ1+x3ρ2+x5ρ3. In other words if B={ρ1,ρ2,ρ3} is the basis for V given in part (a), then the coordinate matrix of (x1,x2,x3,x4,x5) is
[(x1,x2,x3,x4,x5)]B=[x1x3x5].


Exercise 2.6.7

To solve the system we row-reduce the augmented matrix [A|Y] resulting in an augmented matrix [R|Z] where R is in reduced echelon form and Z is an m×1 matrix. If the last k rows of R are zero rows then the system has a solution if and only if the last k entries of Z are also zeros. Thus the only non-zero entries in Z are in the non-zero rows of R. These rows are already linearly independent, and they clearly remain independent regardless of the augmented values. Thus if there are solutions then the rank of the augmented matrix is the same as the rank of R. Conversely, if there are non-zero entries in Z in any of the last k rows then the system has no solutions. We want to show that those non-zero rows in the augmented matrix are linearly independent from the non-zero rows of R, so we can conclude that the rank of R is less than the rank of [R|Z]. Let S be the set of rows of [R|Z] that contain all rows where R is non-zero, plus one additional row r where Z is non-zero. Suppose a linear combination of the elements of S equals zero. Since cr=0 r=0, at least one of the elements of S different from r must have a non-zero coefficient. Suppose row rS has non-zero coefficient c in the linear combination. Suppose the leading one in row r is in position i. Then the i-th coordinate of the linear combination is also c, because except for the one in the i-th position, all other entries in the i-th column of [R|Z] are zero. Thus there can be no non-zero coefficients. Thus the set S is linearly independent and |S|=|R|+1. Thus the system has a solution if and only if the rank of R is the same as the rank of [R|Z]. Now A has the same rank as R and [R|Z] has the same rank as [A|Y] since they differ by elementary row operations. Thus the system has a solution if and only if the rank of A is the same as the rank of [A|Y].

From http://greggrant.org

Linearity

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