MATH 436 HOMEWORK 7–SOLUTIONS
Problem 1. Let V and W be vector spaces of dimension n and m. Let πW : V × W → W be the
projection mapping πW(v,w) = w. Show that there is a basis B1 of V × W and a basis B2 of W
such tha
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MATH 436 HOMEWORK 7–SOLUTIONS
Problem 1. Let V and W be vector spaces of dimension n and m. Let πW : V × W → W be the
projection mapping πW(v,w) = w. Show that there is a basis B1 of V × W and a basis B2 of W
such that the matrix of πW becomes the m × (n + m) matrix
where there are m 1’s.
Solution. Let B2 = {w1,...,wm} be any basis of W. Let v1,...,vn be a basis of V . Then the
list B1 = {(0,w1),..., (0,wm), (v1, 0),..., (vn, 0)} is a basis for V × W since it is the list made up
by writing down a basis for each of the subspaces V 0, W 0 ⊂ V × W such that V × W = V 0 ⊕ W 0.
Now πW acts on this basis by πW(0,wi) = wi and πW(vi, 0) = 0. Therefore the matrix of πW with
respect to the bases B1 and B2 is the required matrix.
Problem 2. For a 2 × 2 matrix B, consider the linear transformation
TB : Mat2×2(F) → Mat2×2(F)
defined by multiplying by B on the right:
TB(A) = AB.
Give Mat2×2(F) the standard basis
1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1
(1) Fix B = 1 3 0 1 . Show that TB is an isomorphism. Write down the matrix of TB−1 with
respect to the standard basis.
(2) Fix B = 1 2 2 4. Show that TB is not an isomorphism.
Solution. (1) The matrix B is invertible, with inverse matrix B−1 = 1 0 1 −3. Then the transformation TB−1 is clearly an inverse of TB, so TB is an isomorphism. To compute the matrix of TB−1,
we can compute the action of TB−1 on the basis. We have
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