MATH347.SP.25 Midterm Solution

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.

Note. Solution briefly states motivation/approach and then concisely carries out required calculations.

1. Find the image of the standard basis vectors 𝒆1,𝒆2,𝒆33 through the linear mapping W:33 defined by the composition W=TS, where:

  1. S is rotation around the x3 axis by θ=π/4;

  2. T is rotation around the x2 axis by θ=π/4.

Solution. Denote by 𝑨,𝑩,𝑪3×3 the standard matrices of mappings S,T,W, respectively, such that

S(𝒖)=𝑨u,T(𝒗)=𝑩𝒗,W(𝒛)=𝑪𝒛.

Since W=TS, 𝑪=𝑩𝑨.

  1. Rotation around the x3 axis by θ=π/4 has standard matrix

    𝑨=[ S(𝒆1) S(𝒆2) S(𝒆3) ]=[ cosθ -sinθ 0 sinθ cosθ 0 0 0 1 ]=[ 1/2 -1/2 0 1/2 1/2 0 0 0 1 ].
  2. Rotation around the x2 axis by θ=π/4 has standard matrix

    𝑩=[ T(𝒆1) T(𝒆2) T(𝒆3) ]=[ cosθ 0 -sinθ 0 1 0 sinθ 0 cosθ ]=[ 1/2 0 -1/2 0 1 0 1/2 0 1/2 ].

    Therefore,

    𝑪=[ 1/2 0 -1/2 0 1 0 1/2 0 1/2 ][ 1/2 -1/2 0 1/2 1/2 0 0 0 1 ]=[ 1/2 -1/2 -1/2 1/2 1/2 0 1/2 -1/2 1/2 ].

    Since W(𝒆j)=𝑪𝒆j, obtain 𝑪=[ 𝒄1 𝒄2 𝒄3 ]=[ W(𝒆1) W(𝒆2) W(𝒆3) ] .

2. Let 𝑪 denote the standard matrix of the linear mapping W defined above. State bases for the four fundamental subspaces of 𝑪.

Solution. Rotation of the orthonormal vectors 𝒆1,𝒆2,𝒆3 by the same angle yields three orthonormal vectors 𝒄1,𝒄2,𝒄3. Verification

𝑪T𝑪=[ 1/2 1/2 1/2 -1/2 1/2 -1/2 -1/2 0 1/2 ][ 1/2 -1/2 -1/2 1/2 1/2 0 1/2 -1/2 1/2 ]=𝑰.

Since columns of 𝑪=[ 𝒄1 𝒄2 𝒄3 ] are orthonormal, they are linearly independent, and rank(𝑪)=3. Any three linearly independent vectors form a basis for both column space and row space, for instance, 𝒆1,𝒆2,𝒆3. The null spaces N(𝑪),N(𝑪T) are of dimension zero, and do not have a basis.

3. Determine an orthonormal basis for C(𝑪) with 𝑪 defined above.

Solution. As noted above, 𝒆1,𝒆2,𝒆3 is an orthonormal basis for C(𝑪). Another basis is 𝒄1,𝒄2,𝒄3.

4. Consider the vector 𝒃=[ 1 1 1 ]T. Find the coordinates of 𝒃 in the basis defined by the column vectors of 𝑪 defined above.

Solution. In the 𝒆1,𝒆2,𝒆3 basis the coordinates of 𝒃 are 1,1,1. In the 𝒄1,𝒄2,𝒄3 basis, the coordinates are

𝒙=𝑪T𝒃=[ 1+12 -1+12 0 ].

5. Compute the inverse of the matrix

𝑨=[ 12 12 12 -12-122 12-122 12 12-122 -12-122 12 ].

Solution. Check if the matrix has orthogonal columns. Computing

𝑨T𝑨=𝑰,

confirming that columns of 𝑨 are orthonormal, hence 𝑨-1=𝑨T.