MATH347.SP.01 Midterm Practice Examination

Instructions. Answer the following questions. Provide a motivation of your approach and the reasoning underlying successive steps in your solution. Write neatly and avoid erasures. Use scratch paper to sketch out your answer for yourself, and then transcribe your solution to the examination you turn in for grading. Illegible answers are not awarded any credit. Presentation of calculations without mention of the motivation and reasoning are not awarded any credit. Each complete, correct solution to an examination question is awarded 3 course grade points. Your primary goal should be to demonstrate understanding of course topics and skill in precise mathematical formulation and solution procedures.

1. Consider a vector

𝒃 \in ℝ^{3 \times 1}

with components

b_{1} = 1

b_{2} = 2

b_{3} = - 1

in the canonical basis

Let

x_{1}, x_{2}, x_{3}

denote the components of

𝒃

with respect to the basis

Compute

𝒙 = {[\begin{array}{lll} x_{1} & x_{2} & x_{3} \end{array}]}^{T} \in ℝ^{3 \times 1}

. (3 points)

Solution. Problem asks for solution of system $𝒃 = 𝑨 𝒙$

𝒃 = [\begin{array}{l} 1 \\ 2 \\ - 1 \end{array}] = 𝑨 𝒙 = [\begin{array}{lll} 1 & 0 & 2 \\ - 1 & 1 & 1 \\ 2 & 2 & 1 \end{array}] [\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}] .

Form bordered matrix and bring to upper trinagular form (Gauss elimination)

[\begin{array}{llll} 1 & 0 & 2 & 1 \\ - 1 & 1 & 1 & 2 \\ 2 & 2 & 1 & - 1 \end{array}] \sim [\begin{array}{llll} 1 & 0 & 2 & 1 \\ 0 & 1 & 3 & 3 \\ 0 & 2 & - 3 & - 3 \end{array}] \sim [\begin{array}{llll} 1 & 0 & 2 & 1 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & - 9 & - 9 \end{array}]

By back substitution, find $x_{3} = 1$ , $x_{2} = 0$ , $x_{1} = - 1$

Solution. a) Denote $𝑨 = [\begin{array}{ll} 𝒂_{1} & 𝒂_{2} \end{array}]$ . Since $𝒂_{1}, 𝒂_{2}$ are linearly independent, a basis for $C (𝑨)$ is ${𝒂_{1}, 𝒂_{2}}$ and $rank (𝑨) = 2 = \dim C (𝑨) = \dim C (𝑨^{T})$ . A basis for the row space is

{[\begin{array}{l} 1 \\ 2 \end{array}], [\begin{array}{l} 2 \\ 1 \end{array}]} .

The null space is $N (𝑨) = {𝟎}$ , with $\dim N (𝑨) = 0$ , empty basis set. For $N (𝑨^{T})$ consider the system

𝑨^{T} 𝒚 = 𝟎 \Rightarrow [\begin{array}{llll} 1 & 2 & 3 & 0 \\ 2 & 1 & 1 & 0 \end{array}] \sim [\begin{array}{llll} 1 & 2 & 3 & 0 \\ 0 & - 3 & - 5 & 0 \end{array}]

𝒚 = [\begin{array}{l} (- 3 + \frac{10}{3}) y_{3} \\ - \frac{5}{3} y_{3} \\ y_{3} \end{array}] = \frac{1}{3} [\begin{array}{l} 1 \\ - 5 \\ 3 \end{array}] y_{3} .

{𝒘} = {[\begin{array}{l} 1 \\ - 5 \\ 3 \end{array}]}

b) To conpute the projection $𝒚$ , subtract from $𝒃$ its component along $𝒘$ determined above (FTLA states $𝒃 = 𝒚 + 𝒛$ , with $𝒚 \in C (𝑨)$ , $𝒛 \in N (𝑨^{T})$ )

Let 𝒗 = \frac{𝒘}{|| 𝒘 ||} = \frac{1}{\sqrt{35}} [\begin{array}{l} 1 \\ - 5 \\ 3 \end{array}], 𝒛 = (𝒗^{T} 𝒃) 𝒗 = - \frac{1}{35} [\begin{array}{l} 1 \\ - 5 \\ 3 \end{array}] \Rightarrow 𝒚 = [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] + \frac{1}{35} [\begin{array}{l} 1 \\ - 5 \\ 3 \end{array}] = \frac{1}{35} [\begin{array}{l} 36 \\ 30 \\ 38 \end{array}] .

𝒒_{1} = \frac{𝒂_{1}}{|| 𝒂_{1} ||} = \frac{1}{\sqrt{14}} [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] .

𝒒_{2} = 𝒂_{2} - (𝒒_{1}^{T} 𝒂_{2}) 𝒒_{1} = [\begin{array}{l} 2 \\ 1 \\ 1 \end{array}] - \frac{1}{14} (7) [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] = [\begin{array}{l} 3 / 2 \\ 0 \\ - 1 / 2 \end{array}], || 𝒒_{2} || = \frac{2}{\sqrt{10}} [\begin{array}{l} 3 \\ 0 \\ - 1 \end{array}]

𝑨 = [\begin{array}{ll} 1 & 2 \\ 2 & 1 \\ 3 & 1 \end{array}] = [\begin{array}{ll} 1 / \sqrt{14} & 3 / \sqrt{10} \\ 2 / \sqrt{14} & 0 \\ 3 / \sqrt{14} & - 1 / \sqrt{10} \end{array}] [\begin{array}{ll} \sqrt{14} & 7 / \sqrt{14} \\ 0 & \sqrt{5 / 2} \end{array}]