Group  ^{List No.}^{Last Name}^{Name}  ^{Student No.}^{Department}  ^{Section}^{Signature} ^{:}^{:}  ^{:}^{:}^{:}^{:}^{:}^{:}^{:}^{:}^{:}^{:}^{:}^{Code}^{Acad. Year}^{Semester}^{Coordinator}^{Date}^{Time}^{Duration}  M E T U Department of Mathematics^{BASIC LINEAR ALGEBRA}^{Midterm II}^{Math 260}^{2007-2008}^{Fall}^{S.F, S.O, A.S.}^{November 28 2007}^{17:40}^{90 minutes}  ^{4 QUESTIONS ON 4 PAGES}^{TOTAL 100 POINTS}^{1}  ^{2}  ^{3}  ^{4}

Show your work ! Partial credits will not be given for correct answers if they are not justified.

Question 1 (12+6+6=24 points)

The product of two (3 × 3)-matrices A and B is known to be AB = [ \begin{matrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{matrix} ], and the adjoint matrix Adj(A) has

determinant 4.

a) Find the determinants det(A^{−1}) = det A = det B =

Solution:

(det A)(det B) = det [ \begin{matrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{matrix} ] = 24 and Adj(A) = (det A)A^{−1}

Since A is 3 × 3-matrix, det(Adj(A)) = (det A)^{3} det(A^{−1}) = \frac{(det A)^{3}}{det A} = (det A)^{2} = 4

There are two possibilities: either det A = 2, det B = 12, det A^{−1} = \frac{1}{2}, or det A = −2, det B = −12, det A^{−1} = −\frac{1}{2}.

b) Can we conclude that the row [ 1 2 3 ] is a linear combination of the rows of matrix B ? (Present complete and detailed arguments !)

Solution: Yes. The rows of matrix B form a basis of ℝ^{3}, because det B ≠ 0 (a theorem). So, any row and in particular [ 1 2 3 ] is a linear combination of the rows of matrix B.

c) Can we conclude that matrices A and B are row-equivalent ? (Present complete and detailed arguments !)

Solution: Yes. A and B are both row equivalent to the unit matrix, because det A ≠ 0, det B ≠ 0 (a theorem). Thus, A and B are row equivalent to each other.

Question 2 (10+10=20 points)

Consider the subspace V ⊂ ℝ^{2×3} formed by (2 × 3)-matrices A such that A [ \begin{matrix}1 \\ 2 \\ 0 \end{matrix} ] = [ \begin{matrix}0 \\ 0 \end{matrix} ].

a) Find a basis of V and determine the dimension of V. Solution:

[ \begin{matrix}a & b & c \\ d & e & f \end{matrix} ] [ \begin{matrix}1 \\ 2 \\ 0 \end{matrix} ] = [ \begin{matrix}a + 2b \\ d + 2e \end{matrix} ] = [ \begin{matrix}0 \\ 0 \end{matrix} ] \begin{matrix}a + 2b = 0 \\ d + 2e = 0 \end{matrix}

Four free variables b, c, e, f, so dim V = 4.

Fundamental solutions (basis of V): [ \begin{matrix}−2 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} ], [ \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ], [ \begin{matrix}0 & 0 & 0 \\ −2 & 1 & 0 \end{matrix} ], [ \begin{matrix}0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} ].

b) Extend your basis of V to a basis of ℝ^{2×3} by choosing additional vectors from the standard basis of ℝ^{2×3}. Solution:

[ \begin{matrix}−2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & −2 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ] → [ \begin{matrix}0 & 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} ] → [ \begin{matrix}1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 0 \end{matrix} ]

So, additional vectors-matrices (corresponding to 5th and 8th columns) are [ \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ] and [ \begin{matrix}0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} ].

Question 3 (8+8+8+8=32 points)

Consider the vector space of polynomials, P_{2} = {ax^{2} + bx + c|a, b, c ∈ ℝ}, and its subset V formed by those p(x) ∈ P_{2} which satisfy the condition p(1) + p′(2) = 0.

a) Show that V is a subspace of P_{2}.

Solution \begin{matrix}p_{1}(1) + p′_{1}(2) = 0 \\ p_{2}(1) + p′_{2}(2) = 0 \end{matrix} ⇒ (p_{1} + p_{2})(1) + (p_{1} + p_{2})′(2) = 0, so V is closed with respect to addition

p(1) + p′(2) = 0 ⇒ cp(1) + cp′(2) = c(p(1) + p′(2)) = 0, so V is closed with respect to scalar multiplication. Therefore V is a subspace of P_{2}.

b) Show that B = {p_{1}, p_{2}} and C = {q_{1}, q_{2}} are two bases of V, where p_{1} = x^{2} − 2x − 1, p_{2} = x^{2} − x − 3, q_{1} = x − 2, q_{2} = x^{2} − 3x + 1. Solution if p(x) = ax^{2} + bx + c, then p′(x) = 2ax + b and p(1) + p′(2) = (a + b + c) + (4a + b) = 5a + 2b + c = 0. The solution space V is two dimensional (one equation with three unknowns a, b, c). This equation is satisfied for p_{1}, p_{2}, q_{1}, q_{2}:

\begin{matrix}5 − 4 − 1 = 0 \\ 5 − 2 − 3 = 0 \\ 0 + 2 − 2 = 0 \\ 5 − 6 + 1 = 0 \end{matrix} , so p_{1}, p_{2}, q_{1}, q_{2} ∈ V.

p_{1}, p_{2} are not proportional and thus linearly independent. Then they form a basis of V (because V is two-dimensional). Similarly, q_{1}, q_{2} form a basis.

c) Determine the coordinates of p_{1}, p_{2} with respect to basis C.

Solution

x^{2} − 2x − 1 = c_{1}(x − 2) + c_{2}(x^{2} − 3x + 1) ⇒ c_{2} = 1, c_{1} = 1,

coordinates [p_{1}]_{C} = [ \begin{matrix}1 \\ 1 \end{matrix} ]

x^{2} − x − 3 = c_{1}(x − 2) + c_{2}(x^{2} − 3x + 1) ⇒ c_{2} = 1, c_{1} = 2,

coordinates [p_{2}]_{C} = [ \begin{matrix}2 \\ 1 \end{matrix} ]

d) Find the both transition matrices: P_{B→C} (from B to C) and P_{C→B} (from C to B). Write down how the coordinate columns [v]_{B} and [v]_{C} are related to P_{B→C} and P_{C→B}.

Solution

P_{C→B} = [p_{1}]_{C}[p_{2}]_{C} = [ \begin{matrix}1 & 2 \\ 1 & 1 \end{matrix} ]

P_{B→C} = P_{C→B}^{−1} = [ \begin{matrix}1 & 2 \\ 1 & 1 \end{matrix} ]^{−1} = [ \begin{matrix}1 & −2 \\ −1 & 1 \end{matrix} ]

The relations: [v]_{B} = (P_{C→B})[v]_{C}, [v]_{C} = (P_{B→C})[v]_{B}.

Question 4 (8+8+8=24 points)

a) Show that (p|q) = p(1)q(1)+p(2)q(2)+p(3)q(3) is an inner product in the space of polynomials P_{2} = {ax^{2} +bx+c : a, b, c ∈ ℝ}.

Solution: Positivity: (p|p) = (p(1))^{2} + (p(2))^{2} + (p(3))^{2} ≥ 0. If (p|p) = 0, then p(1) = p(2) = p(3) = 0, but polynomial p cannot have 3 roots (because its degree ≤ 2), except the case p = 0 (constant zero polynomial).

Symmetry: (p|q) = (q|p) (obvious).

Linearity: (c_{1}p_{1} + c_{2}p_{2}|q) = (c_{1}p_{1}(1) + c_{2}p_{2}(1))q(1) + (c_{1}p_{1}(2) + c_{2}p_{2}(2))q(1) + (c_{1}p_{1}(3) + c_{2}p_{2}(3))q(1) = c_{1}(p_{1}|q) + c_{2}(p_{2}|q).

b) Determine the norms ||1|| and ||x|| with respect to this inner product.

Solution: ||p|| = \sqrt{(p|p)} = \sqrt{(p(1))^{2} + (p(2))^{2} + (p(3))^{2}} ||1|| = \sqrt{1 + 1 + 1} = \sqrt{3} ||x|| = \sqrt{1 + 4 + 9} = \sqrt{14}

c) Determine cos(ϕ), for angle ϕ between the vectors-polynomials 1 and x, with respect to the above inner product.

Solution: (1|x) = 1 · 1 + 1 · 2 + 1 · 3 = 6

cos(ϕ) = \frac{(1|x)}{||1||·||x||} = \frac{6}{\sqrt{3}\sqrt{14}} = \frac{6}{\sqrt{42}}