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independent 4 by 4 permutation matrices? Solutions to Exercises 40 (a) zero when x D 0
(b) one when x D .1; 1; 1; 1/
(c) three when x D .1; 1; 1; 1/ because all rearrangements of this x are perpendicular to .1; 1; 1; 1/
(d) four when the x ’s are not
equal and don’t add to zero. No x gives dim S D 2. I owe this nice problem to Mike
Artin—the answers are the same in higher dimensions: 0; 1; n 1; n.
The problem is to show that the u’s, v’s, w’s together are independent. We know the
u’s and v’s together are a basis for V , and the u’s and w’s together are a basis for W .
Suppose a combination of u’s, v’s, w’s gives 0. To be proved: All coefﬁcients D zero.
Key idea: In that combination giving 0, the part x from the u’s and v’s is in V . So the
part from the w’s is x . This part is now in V and also in W . But if x is in V \ W it
is a combination of u’s only. Now the combination uses only u’s and v’s (independent
in V !) so all coefﬁcients of u’s and v’s must be zero. Then x D 0 and the coefﬁcients
of the w’s are also zero.
The inputs to an m by n matrix ﬁll Rn . The outputs (column space!) have dimension
r . The nullspace has n r special solutions. The formula becomes r C .n r / D n.
If the left side of dim.V/ C dim.W/ D dim.V \ W/ C dim.V C W/ is greater than n,
then dim.V \ W/ must be greater than zero. So V \ W contains nonzero vectors.
If A2 D zero matrix, this says that each column of A is in the nullspace of A. If the
column space has dimension r , the nullspace has dimension 10 r , and we must have
r 10 r and r 5. 42 The dimension of S spanned by all rearrangements of x is 43 44
45
46 Problem Set 3.6, page 190
1 (a) Row and column space dimensions D 5, nullspace dimension D 4, dim.N .AT //
2 3 4 5 6 D 2 sum D 16 D m C n (b) Column space is R3 ; left nullspace contains only 0.
A: Row space basis D row 1 D .1; 2; 4/; nullspace . 2 ; 1; 0/ and . 4 ; 0; 1/; column
space basis D column1 D .1; 2/; left nullspace . 2 ; 1/. B : Row space basis D
both rows D .1; 2; 4/ and .2; 5; 8/; column space basis D two columns D .1; 2/ and
.2; 5/; nullspace . 4 ; 0; 1/; left nullspace basis is empty because the space contains
only y D 0.
Row space basis D rows of U D .0; 1; 2; 3; 4/ and .0; 0; 0; 1; 2/; column space basis D
pivot columns (of A not U ) D .1; 1; 0/ and .3; 4; 1/; nullspace basis .1; 0; 0; 0; 0/,
.0; 2; 1; 0; 0/, .0; 2; 0; 2 ; 1/; left nullspace .1; 1; 1/ D last row of E 1 !
#
"
10
9
3
(a) 1 0
(b) Impossible: r C.n r / must be 3
(c) Œ 1 1
(d)
3
1
01
(e) Impossible Row space D column space requires m D n. Then m r D n r ;
nullspaces have the same dimension. Section 4.1 will prove N .A/ and N .AT /
orthogonal to the row and column spaces respectively—here those are the same space.
111
2 1 has the
AD
has those rows spanning its row space B D 1
210
same rows spanning its nullspace and BAT D 0.
A: dim 2; 2; 2; 1: Rows .0; 3; 3; 3/ and .0; 1; 0; 1/; columns .3; 0; 1/ and .3; 0; 0/;
nullsp...
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 Spring '12
 Minki
 Mass

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