MA 1B PRACTICAL - HOMEWORK SET 1 SOLUTIONS
1 (Reading).
2. (40 pts) For each of the following subset of Fn , n ≤ 2, determine whether it is a
subspace of Fn . Explain the reason.
(1){(x1 , x2 , ..., xn )T ∈ Fn |x1 + x2 + · · · + xn = n}.
(2){(x1 , x2 , ..., xn )T ∈ Fn |x1 + x2 + · · · + xn = nxn }.
(3){(x1 , x2 , ..., xn )T ∈ Fn |x21 + x22 + · · · + x2n = 0}. ( You need separate the case
F = R or F = C.)
(4){(x1 , x2 , x3 , x4 , x5 )T ∈ R5 |x1 = 3x2 , x3 = 7x4 , x1 + x2 + x5 = 0}.
Solution. For each part, we write W for the set in question.
1) W is not a subspace since it does not contain the zero vector.
2) Suppose (x1 , . . . , xn ), (y1 , . . . , yn ) ∈ W . Then
(x1 + y1 ) + · · · (xn + yn ) = (x1 + · · · + xn ) + (y1 + · · · + yn ) = nxn + nyn = n(xn + yn ).
Suppose also c ∈ F. Then
cx1 + · · · cxn = c(x1 + · · · + xn ) = cnxn .
Hence W is a subspace.
3) If F = R then W = {(0, . . . , 0)} which is a subspace.
If F = C then (1, i, 0, . . . , 0) ∈ W and (1, −i, 0, . . . , 0) ∈ W but
(1, i, 0, . . . , 0) + (1, −i, 0, . . . , 0) = (1, 0, 0, . . . , 0) ∈
/W
so W is not a subspace.
4) W is defined by a system of linear equations, and hence is a subspace.
3. (20 pts)[Ch. 1, Problem 2.3.]
Solution. Let Ei,j be the matrix with a 1 in the i, j position and zeros elsewhere.
Then
{Ei,j + Ej,i : i, j ∈ {1, . . . , n}, i ≥ j}
is an example of such a basis. The number of elements is n(n + 1)/2.
4. (20 pts) Can the vectors v1 , v2 be extended to a basis of F3 ? If yes, write down
such a basis (there are many possible answers). Justify your answer.
(1)v1 = (1, 0, 2)T and v2 = (2, 0, 4)T ;
(2)v1 = (1, 0, 2)T and v2 = (2, 1, 4)T .
Date: January 9, 2017.
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MA 1B PRACTICAL - HOMEWORK SET 1 SOLUTIONS
Solution. 1) No, because v2 = 2v1 and so {v1 , v2 } is not a linearly independent set.
Any set containing them will not be linearly independent, and hence cannot be a
basis by definition.
2) Yes, {v1 , v2 } are linearly independent because v1 has a zero in the third coordinate while v2 does not so no nonzero multiple of v2 gives v1 . One example of
a third vector to add to make {v1 , v2 } a basis is the cross product v3 = v1 × v2 ,
which will be orthogonal to both v1 and v2 . Please note however, the cross product
is defined only for vector spaces of dimension 3. As a general rule you can check
linear independence of a set {v1 , v2 , v3 } in F3 by computing the determinant of the
matrix [v1 , v2 , v3 ] and finding that it is non-zero. Here, taking v3 = (0, 0, 1) will
also work.
5. (20 pts)[Ch. 1 Problem 7.5.]
Solution. Let Z be the space of 4×4 matrices, V be the subspace of upper triangular
matrices and W be the subspace of symmetric matrices. If M ∈ V ∩ W , then
M = M T means that M is both triangular, and the off-diagonal entries are zero.
Therefore V ∩ W is the set of diagonal matrices, and the largest subspace contained
in V and W . (Recall that the intersection of two subspaces is a subspace).
The smallest subspace containing V and W is V +W , since by exercise 7.2 V +W
is a subspace and any subspace containing V and W must also contain the sums
V + W . Now V ∩ W is of dimension 4, since it is the set of diagonal matrices.
Recall in general for subspaces V, W of any vector space Z, there is a formula
dim(V + W ) = dimV + dimW − dim(V ∩ W )
Consider Z as the set of 16 co-ordinates in F arranged in a square pattern, which
shows Z is dimension 16. Then by counting the number of upper-triangular places,
V is dimension 10. Furthermore, W is dimension 10 because a symmetric matrix
can be written as






0 0 0 0
0 0 1 0
0 1 0 0
 0 0 1 0 
 0 0 0 0 
 1 0 0 0 





M = D + a1 
 0 0 0 0  + a2  1 0 0 0  + a2  0 1 0 0  + · · ·
0 0 0 0
0 0 0 0
0 0 0 0


0 0 0 0
 0 0 0 0 

+ a6 
 0 0 0 1 ,
0 0 1 0
where D is a diagonal matrix. So applying he formula, we obtain dim(V + W ) =
10 + 10 − 4 = 16. Then V + W is a subspace of full dimension 16 in Z, and hence
V + W = Z.