MATH347.SP.25 Midterm Examination

Instructions. Answer the following questions. Provide concise motivation of your approach. Illegible answers are not awarded any credit. Presentation of calculations without mention of the motivation and reasoning are not awarded any credit. Each correct question answer is awarded 2 course points.

Notation:

\cos^{2} (θ) = {[\cos (θ)]}^{2}

\sin^{2} (θ) = {[\sin (θ)]}^{2}

𝑹_{θ} = [\begin{array}{ll} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}]

describes rotation by angle $θ$ in $ℝ^{2}$ . Rotation by angle $3 θ$ is obtained by repeated application

𝑹_{θ}^{3} = 𝑹_{3 θ} \Rightarrow [\begin{array}{ll} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] [\begin{array}{ll} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] [\begin{array}{ll} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] = [\begin{array}{ll} \cos 3 θ & - \sin 3 θ \\ \sin 3 θ & \cos 3 θ \end{array}]

Carrying out the multiplications gives

𝑹_{θ}^{3} = [\begin{array}{ll} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array}] [\begin{array}{ll} \cos^{2} θ - \sin^{2} θ & - 2 \sin θ \cos θ \\ 2 \sin θ \cos θ & \cos^{2} θ - \sin θ \end{array}] = [\begin{array}{ll} \cos^{3} θ - \cos^{} θ \sin^{2} θ - 2 \sin^{2} θ \cos θ & - \\ - & - \end{array}]

and equality of the 1,1 component gives

\cos (3 θ) = \cos^{3} θ - 3 \cos θ \sin^{2} θ,

2. Determine the standard matrix

𝑷

of the orthogonal projection of a vector

𝒗 \in ℝ^{3}

onto the line

x_{1} = x_{2} = x_{3}

Solution. The unit vector along the line $x_{1} = x_{2} = x_{3}$ is

𝒒 = \frac{1}{\sqrt{3}} [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}],

and the projection matrix along this direction is

𝑷 = 𝒒 𝒒^{T} = \frac{1}{3} [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] [\begin{array}{lll} 1 & 1 & 1 \end{array}] = \frac{1}{3} [\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}] .

3. Determine bases for the fundamental subspaces of the matrix

𝑷

defined above.

Solution. For $𝒖 \in ℝ^{3}$ , $𝒗 = 𝑷 𝒖 = (𝒒 𝒒^{T}) 𝒖 = (𝒒^{T} 𝒖) 𝒒$ is in the direction of $𝒒$ hence a basis for $C (𝑷)$ is ${𝒒}$ and $r = rank (𝑷) = 1$ . The left null space contains vectors orthogonal to $𝒒$ , for example

𝒚 = [\begin{array}{l} 1 \\ - 1 \\ 0 \end{array}], 𝒛 = [\begin{array}{l} 1 \\ 0 \\ - 1 \end{array}],

that verify $𝒚^{T} 𝒒 = 0$ , $𝒛^{T} 𝒒 = 0$ , and linearly independent, hence ${𝒚, 𝒛}$ is a basis. Note that $𝑷^{T} = 𝑷$ such that ${𝒒}$ is a basis for the row space $C (𝑷^{T})$ , and ${𝒚, 𝒛}$ is a basis for $N (𝑷)$ .

4. Find the inverse of the standard matrix

𝑴 \in ℝ^{3 \times 3}

of the linear mapping,

L = F \circ G

with

Solution. Consider $𝒗, 𝒘, 𝒛 \in ℝ^{3}$ and diagram in Fig. 1, with $𝒘$ the projection of $𝒗$ onto the direction of $𝒖$ . Let

𝒒 = \frac{𝒖}{|| 𝒖 ||} = \frac{1}{\sqrt{2}} [\begin{array}{l} 1 \\ 1 \\ 0 \end{array}] .

𝑷 = 𝒒 𝒒^{T} = \frac{1}{2} [\begin{array}{l} 1 \\ 1 \\ 0 \end{array}] [\begin{array}{lll} 1 & 1 & 0 \end{array}] = \frac{1}{2} [\begin{array}{lll} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{array}],

leading to $𝒘 = 𝑷 𝒗$ . The vector $𝒘$ is also the sum

𝒘 = 𝒗 + 𝒛 .

The reflection of $𝒗$ across $𝒖$ is reached by traveling from endpoint of $𝒗$ by $2 𝒛$

𝒚 = 𝒘 + 2 𝒛 = 𝒘 + 2 (𝒘 - 𝒗) = 2 𝒘 - 𝒗 = 2 𝑷 𝒗 - 𝒗 = (2 𝑷 - 𝑰) 𝒗 .

From the above deduce the standard matrix for reflection across $𝒖$ is

𝑨 = 2 𝑷 - 𝑰 = [\begin{array}{lll} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{array}] - [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - 1 \end{array}] .

𝑩 = [\begin{array}{lll} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{array}] .

The matrix for the composite transformation is

𝑴 = 𝑨 𝑩,

𝑴^{- 1} = 𝑩^{- 1} 𝑨^{- 1} .

𝑩^{- 1} = [\begin{array}{lll} 1 / λ_{1} & 0 & 0 \\ 0 & 1 / λ_{2} & 0 \\ 0 & 0 & 1 / λ_{3} \end{array}],

assuming non-zero $λ_{1}, λ_{2}, λ_{3}$ . Note that reflection of $𝒚$ across $𝒖$ gives the original vector $𝒗$ . Stated in matrix terms

𝑨 𝑨 = 𝑰,

𝑴 = 𝑩^{- 1} (2 𝑷 - 𝑰) = [\begin{array}{lll} 1 / λ_{1} & 0 & 0 \\ 0 & 1 / λ_{2} & 0 \\ 0 & 0 & 1 / λ_{3} \end{array}] (2 \cdot \frac{1}{2} [\begin{array}{lll} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{array}] - [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]) \Rightarrow

𝑴^{- 1} = [\begin{array}{lll} 1 / λ_{1} & 0 & 0 \\ 0 & 1 / λ_{2} & 0 \\ 0 & 0 & 1 / λ_{3} \end{array}] [\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - 1 \end{array}] = [\begin{array}{lll} 0 & 1 / λ_{1} & 0 \\ 1 / λ_{2} & 0 & 0 \\ 0 & 0 & - 1 / λ_{3} \end{array}] .

Solution. The stage 1 operation is

𝑳_{1} 𝑨 = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} 1 & 2 & 3 \\ - 1 & 2 & 2 \\ 0 & - 8 & - 4 \end{array}] = [\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & - 8 & - 4 \end{array}] .

𝑳_{2} (𝑳_{1} 𝑨) = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}] [\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & - 8 & - 4 \end{array}] = [\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}] = 𝑼 .

𝑨 = 𝑳_{1}^{- 1} 𝑳_{2}^{- 1} 𝑼 = 𝑳 𝑼,

𝑳 = 𝑳_{1}^{- 1} 𝑳_{2}^{- 1} = [\begin{array}{lll} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 2 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & - 2 & 1 \end{array}] .