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MATH-4100-01 09/30/11 Exam #1 (Printed) Name:
Please answer all 5 questions, showing your work in reasonable detail. Closed books; laptops and calculators are not permitted.
1.(15pts). Suppose M is a 4 × 6 matrix with a one-dimensional N (M T^ ) spanned by the vector n = [1, 1 , 2 , 1]T^. (a) What is the maximal number of linearly independent columns in M? (b) What is the condition on the right-hand side vector b for the equation M x = b to have no solutions? Solution: (a) dim N (M T^ )+dim C(M ) = 4. Dimension of N (M T^ ) is one; therefore rankM = dim C(M ) = 3. M must have 3 linearly independent columns. (b) the condition of M x = b being solvable is b ∈ C(M ), which is equivalent to nT^ b = 0. The equation has no solution if nT^ b 6 = 0.
- (25 pts.) (a) For the system Ax = b reduce [A b] to upper triangular form U
[A b] =
3 5 1 b 1 3 3 1 b 2 − 3 − 7 − 1 b 3
(^) → [U c]
(b) What condition(s) on the vector b allow the system Ax = b to be solvable? (c) Factor the 3 × 3 matrix A into LU. (d) Find a basis in the column space C(A). Solution (a)
L 1 =
(^) : [A b] →
3 5 1 b 1 0 − 2 0 b 2 − b 1 0 − 2 0 b 3 + b 1
L 2 =
(^) : [L 1 A L 1 b] →
3 5 1 b 1 0 − 2 0 b 2 − b 1 0 0 0 b 3 − b 2 + 2b 1
(b) The solvability condition: b ∈ C(A) if b 3 − b 2 + 2b 1 = 0. (c)Factorization
A = L− 1 1 L− 21
(d) A has two pivots, therefore C(A) is spanned by its 2 columns which are not proportional, say by [5, 3 , −7]T^ and [1, 1 , −1]T^ ; C(A) is a plane in R^3. (A description of this plane through an equation follows from the solvability condition: b ∈ C(A) if 2b 1 − b 2 + b 3 = 0. This is the equation for C(A) : 2x 1 − x 2 + x 3 = 0.)
3 (15 pts.) True or False? (Explain your answers). (a) If Ax and Ay are linearly independent then x and y are linearly independent. TRUE: If λx + μy = 0 for nonzero coefficients then λAx + μAy = 0, which contradicts independence of x and y
(b)
[
]T
is perpendicular to
[
]T
so the planes x 1 +x 2 +x 3 +x 4 = 0 and x 1 − x 2 − x 3 + x 4 = 0 are orthogonal subspaces. FALSE: There are vectors in the intersection of these two subspaces; they are in the null space of the matrix (^) [ 1 1 1 1 1 − 1 − 1 1
]
which is 2-dimensional. So the subspaces intersect and cannot be orthogonal. A simi- lar explanation: Both planes are null spaces of matrices A =
[
]
[ and^ B^ = 1 − 1 − 1 1
]
. Since the matrices are rank 1, the null subspaces are of dimension 3 and therefore must intersect in the 4-dimensional space. (c) The subspace spanned by
[
]T
and
[
]T
is the orthogonal
complement of the subspace spanned by
[
]T
and
[
]T
FALSE: They do not have enough dimension to cover the whole R^5. The dimensions are both two and the spaces cannot complement each other to R^5.
4 (25 pts.) Suppose that row operations (elimination) factorizes A = LR as follows
A =
(a) Find the null-space of A.
(b) Give the complete solution to Ax =
(^) (this is 2× First column of A).
(c) Suppose a matrix B reduces to the same row echelon form R as for the matrix A. Which
of the four subspaces are sure to be the same for A and B? [C(A) =? C(B), C(AT^ ) =? C(BT^ ),
N (A) ? = N (B), N (AT^ ) ? = N (BT^ )] Solution: (a) N (A) = N (R). There are two free variables x 3 and x 5. N (R) is spanned by x 3 [− 1 , − 2 , 1 , 0 , 0]T^ + x 5 [0, 1 , 0 , − 1 , 1]T^. (b) Find a particular solution by setting the free variables to 0 : xp = [x 1 , x 2 , 0 , x 4 , 0]T^. Ax = x 1 (Column 1) +x 2 (Column 2) + x 4 (Column 4). Since we know that Ax = 2× First column of A, xp = [2, 0 , 0 , 0 , 0]T^ and the complete solution is
[2, 0 , 0 , 0 , 0]T^ + x 3 [− 1 , − 2 , 1 , 0 , 0]T^ + x 5 [0, 1 , 0 , − 1 , 1]T^.
The problem can also be solved by doing forward and backward substitutions in LRx = b. (c) C(AT^ ) = C(BT^ ) = C(RT^ ), the row spaces do not change under row operations and are the same. N (A) = N (B) = N (R).
