MATH. 108B ANSWER KEY FOR FIRST MIDTERM JANUARY 2021
1. [14 pts] Give complete and accurate definitions (not just notation) for each term below.
(Avoid giving conditions from some theorem, but aim for the basic definition, as stated
when the term was first introduced.)
(a) Let V be a vector space, T ∈ L(V ), and λ ∈ F. The eigenspace of T corresponding
to λ is :
the null space of T − λI.
(b) An inner product on a vector space V is :
a function that sends each ordered pair (u, v) of vectors in V to a scalar 〈u, v〉 in F such
that
〈v, v〉 ≥ 0 for all v ∈ V ;
〈v, v〉 = 0 if and only if v = 0;
〈u+ v, w〉 = 〈u, w〉+ 〈v, w〉 for all u, v, w ∈ V ;
〈λu, v〉 = λ〈u, v〉 for all λ ∈ F and u, v ∈ V ;
〈u, v〉 = 〈v, u〉 for all u, v ∈ V .
2. [14 pts] Carefully and accurately state (but do not prove) each of the results listed
below. Be sure to put in all the hypotheses, and the complete conclusions.
(a) The Cauchy-Schwarz Inequality.
Let V be an inner product space and u, v ∈ V . Then |〈u, v〉| ≤ ‖u‖ ‖v‖.
Equality holds if and only if one of u or v is a scalar multiple of the other.
(b) The theorem about dimensions of orthogonal complements.
If V is a finite dimensional inner product space and U is a subspace of V , then
dimU⊥ = dimV − dimU.
3. [15 pts] No proofs required for this problem. In each case, find the norm of the given
vector u in the given inner product space V .
(a) Let u = (1, 2) and V = R2 with the Euclidean inner product.
‖u‖ = √5
(b) Let u = (1, 2) and V = R2 with the inner product defined by the rule 〈(x, y), (x′, y′)〉 =
3xx′ + 2yy′.
‖u‖ = √11
(c) Let u = x and V = P2(R) with 〈p(x), q(x)〉 =
∫ 1
0
p(x)(q(x)dx.
‖u‖ =
√
1
3
4. [20 pts] Prove the following part of the theorem concerning the direct sum of a subspace
and its orthogonal complement:
If U is a finite dimensional subspace of an inner product space V , then V = U+U⊥.
Proof. Since U is a finite dimensional inner product space itself, it has an ONB, by the
existence theorem for ONBs. Let e1, . . . , em be an ONB for U .
Consider v ∈ V . Define u = 〈v, e1〉e1+ · · ·+ 〈v, em〉em and w = v− u. Then u ∈ U and
v = u+ w. For j = 1, . . . , m, we have
〈w, ej〉 = 〈v, ej〉 −
(〈v, e1〉〈e1, ej〉+ · · ·+ 〈v, em〉〈em, ej〉) = 〈v, ej〉 − 〈v, ej〉 = 0
because e1, . . . , em is orthonormal. It follows that 〈w, x〉 = 0 for any linear combination x
of e1, . . . , em, that is, for any x ∈ U . Thus, w ∈ U⊥.
Therefore v = u+ w ∈ U + U⊥, proving that U + U⊥ = V .
5. [20 pts] Let λ ∈ F, let T be a diagonalizable operator on a finite dimensional vector
space V , and let v1, . . . , vn be a basis for V consisting of eigenvectors of T .
Prove that E(λ, T ) is spanned by some subset of {v1, . . . , vn}.
Proof. Let λj be the eigenvalue corresponding to vj . Then Tvj = λjvj for j = 1, . . . , n.
Set L =
{
j ∈ {1, . . . , n} | λj = λ
}
. Then Tvj = λvj for all j ∈ L, so vj ∈ E(λ, T )
for all j ∈ L. Consequently, span(vj | j ∈ L) ⊆ E(λ, T ). We claim that E(λ, T ) equals
span(vj | j ∈ L).
Consider any v ∈ E(λ, T ). Write v = α1v1 + · · · + αnvn for some αj ∈ F. Then
Tv = α1λ1v1+· · ·+αnλnvn. But also Tv = λv = λα1v1+· · ·+λαnvn. Since v1, . . . , vn is a
basis for V , we must have αjλj = λαj for all j. When λ 6= λj , it follows that αj = 0. Hence,
v =
∑
j∈L αjvj , which lies in span(vj | j ∈ L). Therefore E(λ, T ) = span(vj | j ∈ L).
6. [17 pts] Let V be a real inner product space, i.e., F = R. Let x be a nonzero vector
in V , and define R ∈ L(V ) by the rule Rv = v − 2〈v, x〉‖x‖2 x.
Prove that 〈Rv,Rw〉 = 〈v, w〉 for all v, w ∈ V .
Proof. Let v, w ∈ V . Since V is a real inner product space, 〈w, x〉 = 〈w, x〉 and 〈x, w〉 =
〈w, x〉. Thus, we calculate that
〈Rv,Rw〉 =
〈
v − 2〈v, x〉‖x‖2 x, w −
2〈w, x〉
‖x‖2 x
〉
= 〈v, w〉 −
〈
v,
2〈w, x〉
‖x‖2 x
〉
−
〈
2〈v, x〉
‖x‖2 x, w
〉
+
〈
2〈v, x〉
‖x‖2 x,
2〈w, x〉
‖x‖2 x
〉
= 〈v, w〉 − 2〈w, x〉‖x‖2 〈v, x〉 −
2〈v, x〉
‖x‖2 〈x, w〉+
4〈v, x〉〈w, x〉
‖x‖4 〈x, x〉
= 〈v, w〉 − 2〈w, x〉‖x‖2 〈v, x〉 −
2〈v, x〉
‖x‖2 〈w, x〉+
4〈v, x〉〈w, x〉
‖x‖2
= 〈v, w〉. ()

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