Download Math Exam: Midterm, Topic - Linear Algebra and more Exams Linear Algebra in PDF only on Docsity!
Sample Midterm Exam
Math 20F Name: 8/22/08 Section:
Read all of the following information before starting the exam:
• READ EACH OF THE PROBLEMS OF THE EXAM CAREFULLY!
- Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
- A single 8 1/2 × 11 sheet of notes (double sided) is allowed. No calculators are permitted.
- Circle or otherwise indicate your final answers.
- Please keep your written answers clear, concise and to the poin.
- This test has xxx problems and is worth xxx points. It is your responsibility to make sure that you have all of the pages!
- Turn off cellphones, etc.
- Good luck!
∑^5
1. (0 points)
(a) Let A =
, and b =
. Solve Ax = b.
Answer: By row reduction:
x =
(b) Suppose T
and T
. Find the matrix for the
linear transformation T.
Answer: Note that the hypothesis imply T
and T
, so
A =
. (c) Define linear independence. Answer: Vectors v 1 ,... , vn are linearly independent if α 1 v 1 +.. .+αnvn = 0 implies that αi = 0 for all i. (That is, the only linear combination of v 1 ,... , vn giving the 0 vector is the trivial one taking all coefficients to be zero.)
(d) Let A =
. Are the columns of A linearly independent?
Answer: Row reducing A we get that
A ∼
Since not every column is a pivot column, the columns are not linearly independent.
2. (0 points)
(a) Let A =
. Let B =
. Compute AT^ , AB and
BT^ − 3 A.
Answer:
AT^ =
, AB =
, BT^ − 3 A =
(b) Let A =
. Compute A−^1.
Answer:
A−^1 =
(a) If {v 1 , v 2 , v 3 } are linearly dependent, then one of the vectors is a multiple of another.
FALSE: Take v 1 =
, v 2 =
and v 3 =
(b) If Ax = b is consistent for all b, the columns of A span Rm. TRUE: If Ax = b is consistent for all then for any b, b can be written as a linear combination of the columns of A (with coefficients given by x.) This is exactly what it means to span Rm. (c) If AB = AC, then B = C. FALSE: If A is the zero matrix, A times any matrix B is the zero matrix, so this is certainly false. (d) If the transformation T (x) = ABx is onto, then the transformation T ′(x) = Ax is onto TRUE: We need to check that for each b, there is a y such that b = Ay. But, we know that for every b we can write b = ABx for some x. We simply can take y = Bx. This checks the definition of onto. (e) If A and B are n × n, and invertible, then A + B is invertible. FALSE: If A is invertible, so is −A, and A + (−A) is the zero matrix which is not invertible. (f) If A is row equivalent to B, then det(A) = det(B). FALSE: Though det(A) is unchanged by row replacement, swapping rows and multiplying a row by a scalar changes the determinant. Example: ( 1 0 0 1
but the determinants are different. (g) If the columns of A are linearly dependent, then Col(A) is not a vector space. FALSE: Col(A) is the span of the columns of A, and the span of any vectors is a vector space (even if they are linearly dependent.
5. (0 points)
(a) If T : Rn^ → Rm^ is a linear transformation verify that the range of T = {v ∈ Rm^ : T (x) = v for some x ∈ Rn}, is a vector space. (Note the range is not necessarily all of Rm. Rm is the co-domain, not the range.). Answer: It suffices to check that the range is a subspace as it sits inside Rm. Actually this is what I intended to ask. Let R denote the range. If u, v ∈ R then by definition of R, u = T (x) and v = T (y). Note that u + v = T (x) + T (y) = T (x + y)
so u + v ∈ R. If u = T (x) ∈ R then cu = cT (x) = T (cx) ∈ R. Finally, we need to check 0 ∈ R, but T ( 0 ) = 0 so indeed it is. We’ve checked the three conditions so R is a subspace of Rm. (b) Verify that the derivative operator (^) dxd is a linear transformation from the vector space of polynomials of degree at most 3 to the vector space of polynomials of degree at most
- Is this linear transformation 1 − 1? Onto? Answer: Here we use elementary properties of the derivative. Recall a linear transformation satisfies: T (u + v) = T (u) + T (v) and T (cu) = cT (v). In our case, T = (^) dxd and our vectors u and v are polynomials f (x) and g(x). Note:
d dx
(f (x) + g(x)) = d dx
f (x) + d dx
g(x)
so the first property holds, and d dx cf (x) = c d dx f (x)
so the second property also holds, and hence (^) dxd is a linear transformation. d dx 1 =^
d dx 2 = 0, so^
d dx is not 1-1, however^
d dx
f (x)dx = f (x), so (^) dxd is onto.
