Math 108B - Home Work # 1 Solutions
1. For T to have the matrix
1 0
0 1
0 0
with respect to a basis {u1 , u2 } of R2 and a basis {v1 , v2 , v3 } for R3 , means simply
that T u1 = v1 and T u2 = v2 . Hence {u1 , u2 } can remain the standard basis, and then
v1 = (1, 2, 0) and v2 = (−1, 2, 3) will be the columns of the given matrix for T . Since
v1 and v2 are linearly independent, we can complete them to a basis. To do this we
just need to find a third vector of R3 that is not a linear combination of v1 and v2 . For
instance, v3 = e3 = (0, 0, 1) works.
2. We must multiply the given matrix on the right by the change of basis matrix C whose
columns are the coordinates of the new basis w1 , w2 in the old basis {v1 , v2 }, and we
must multiply it on the left by the change of basis matrix C −1 whose columns are the
coordinates of v1 , v2 in the new basis {w1 , w2 }. To find C, note that w1 = (1, 2) =
3
2
(1, 1) + 12 (−1, 1) = 23 v1 + 12 v2 and w2 = (0, 1) = ((1, 1) + (−1, 1))/2 = 12 v1 + 12 v2 . Hence
3/2 1/2
C= .
1/2 1/2
To get C −1 , note v1 = (1, 1) = (1, 2) − (0, 1) = w1 − w2 and v2 = (−1, 1) = −(1, 2) +
3(0, 1) = −w1 + 3w2 . Hence
−1 1 −1
C = ,
−1 3
and the matrix for T in the new basis is
−1 1 −1 4 −1 3/2 1/2
C AC =
−1 3 2 4 1/2 1/2
13/2 3/2
=
−11/2 3/2
3. Let T : V → W be a linear transformation, and let {v1 , . . . , vn } be a basis for V . Show
that T is invertible if and only if {T v1 , . . . , T vn } is a basis for W .
Solution. ⇐: Suppose {T v1 , . . . , T vn } is a basis for W , and write wi = T vi for each
i. Then we can define a linear transformation L : W → V by
L(c1 w1 + · · · + cn wn ) = c1 v1 + · · · cn vn , ∀ c1 , . . . , cn ∈ F.
1
L is well-defined since the vectors w1 , . . . , wn are linearly independent, and since these
vectors span W , L is defined for all vectors in W . Clearly LT vi = Lwi = vi and
T Lwi = T vi = wi for every i. Using linearity of T and L it follows that LT v = v and
T Lw = w for all v ∈ V and all w ∈ W . Thus L is the inverse of T and T is invertible.
⇒: Let L = T −1 . We first show that T v1 , . . . , T vn span W . Let w ∈ W and write
Lw = c1 v1 + · · · + cn vn in V . Applying T , we get
w = T Lw = c1 T v1 + · · · + cn T vn .
To show that T v1 , . . . , T vn are linearly independent, suppose that c1 T v1 +· · ·+cn T vn =
0 for scalars ci . Applying L, we get
0 = L(0) = c1 LT v1 + · · · + cn LT vn = c1 v1 + · · · cn vn .
Since v1 , . . . , vn are linearly independent, we must have ci = 0 for all i.
4. The trace of an n × n matrix A is defined as the sum of all the entries on the main
diagonal of A. That is,
Xn
tr(A) = Aii ,
i=1
th
where Aij denotes the entry of A in the i row and j th column.
(a) Show that for any two n × n matrices A and B, tr(AB) = tr(BA).
(b) Use (a) to show that if X and Y are similar matrices then tr(X) = tr(Y ).
Solution. (a)
n
X n X
X n n X
X n n
X
tr(AB) = (AB)ii = Aij Bji = Bji Aij = (BA)jj = tr(BA).
i=1 i=1 j=1 j=1 i=1 j=1
(b) If X and Y are similar matrices, then X = C −1 Y C for some invertible matrix C.
Thus
tr(X) = tr(C −1 (Y C)) = tr((Y C)C −1 ) = tr(Y ).
5. Let V be an inner-product space, and let W be a subspace of V . Define the orthogonal
complement of W by
W ⊥ = {v ∈ V | hv, wi = 0 ∀w ∈ W }.
Show that W ⊥ is a subspace of V .
2
Solution. Clearly, 0 ∈ W ⊥ since h0, wi = 0 for any w ∈ W . If v ∈ W ⊥ and a ∈ F ,
then av ∈ W ⊥ since hav, wi = ahv, wi = 0 for any w ∈ W . Finally, if u, v ∈ W ⊥ , then
u + v ∈ W ⊥ since hu + v, wi = hu, wi + hv, wi = 0 + 0 = 0 for any w ∈ W . Thus W ⊥
is a subspace of V .
