Section 2C

Dimension

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Problem 2C.1

Show that the subspaces of R2\mathbb {R}^{2} are precisely {0}\left\{ 0\right\}, all lines in R2\mathbb {R}^{2} containing the origin, and R2\mathbb {R}^{2}.

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Problem 2C.2

Show that the subspaces of R3\mathbb {R}^{3} are precisely {0}\left\{ 0\right\}, all lines in R3\mathbb {R}^{3} containing the origin, all planes in R3\mathbb {R}^{3} containing the origin, and R3\mathbb {R}^{3}.

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Problem 2C.3
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(a)

Let U={p∈P4(F):p(6)=0}U = \left\{ p \in \mathcal{P}_{4}(\mathbf{F}): p(6) = 0\right\}. Find a basis of UU.

(b)

Extend the basis in (a) to a basis of P4(F)\mathcal{P}_{4}(\mathbf{F}).

(c)

Find a subspace WW of P4(F)\mathcal{P}_{4}(\mathbf{F}) such that P4(F)=U⊕W\mathcal{P}_{4}(\mathbf{F}) = U \oplus W.

Problem 2C.4
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(a)

Let U={p∈P4(R):p′′(6)=0}U = \left\{ p \in \mathcal{P}_{4}(\mathbb {R}): p^{\prime \prime }(6) = 0\right\}. Find a basis of UU.

(b)

Extend the basis in (a) to a basis of P4(R)\mathcal{P}_{4}(\mathbb {R}).

(c)

Find a subspace WW of P4(R)\mathcal{P}_{4}(\mathbb {R}) such that P4(R)=U⊕W\mathcal{P}_{4}(\mathbb {R}) = U \oplus W.

Problem 2C.5
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(a)

Let U={p∈P4(F):p(2)=p(5)}U = \left\{ p \in \mathcal{P}_{4}(\mathbf{F}): p(2) = p(5)\right\}. Find a basis of UU.

(b)

Extend the basis in (a) to a basis of P4(F)\mathcal{P}_{4}(\mathbf{F}).

(c)

Find a subspace WW of P4(F)\mathcal{P}_{4}(\mathbf{F}) such that P4(F)=U⊕W\mathcal{P}_{4}(\mathbf{F}) = U \oplus W.

Problem 2C.6
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(a)

Let U={p∈P4(F):p(2)=p(5)=p(6)}U = \left\{ p \in \mathcal{P}_{4}(\mathbf{F}): p(2) = p(5) = p(6)\right\}. Find a basis of UU.

(b)

Extend the basis in (a) to a basis of P4(F)\mathcal{P}_{4}(\mathbf{F}).

(c)

Find a subspace WW of P4(F)\mathcal{P}_{4}(\mathbf{F}) such that P4(F)=U⊕W\mathcal{P}_{4}(\mathbf{F}) = U \oplus W.

Problem 2C.7
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(a)

Let U={p∈P4(R):∫−11p=0}U = \left\{ p \in \mathcal{P}_{4}(\mathbb {R}): \int_{-1}^{1} p = 0\right\}. Find a basis of UU.

(b)

Extend the basis in (a) to a basis of P4(R)\mathcal{P}_{4}(\mathbb {R}).

(c)

Find a subspace WW of P4(R)\mathcal{P}_{4}(\mathbb {R}) such that P4(R)=U⊕W\mathcal{P}_{4}(\mathbb {R}) = U \oplus W.

Problem 2C.8

Suppose v1,…,vmv_{1}, \ldots , v_{m} is linearly independent in VV and w∈Vw \in V. Prove that

dim⁡span⁡(v1+w,…,vm+w)≥m−1 \operatorname {dim} \operatorname {span}\left(v_{1}+w, \ldots , v_{m}+w\right) \geq m-1
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Problem 2C.9

Suppose mm is a positive integer and p0,p1,…,pm∈P(F)p_{0}, p_{1}, \ldots , p_{m} \in \mathcal{P}(\mathbf{F}) are such that each pkp_{k} has degree kk. Prove that p0,p1,…,pmp_{0}, p_{1}, \ldots , p_{m} is a basis of Pm(F)\mathcal{P}_{m}(\mathbf{F}).

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Problem 2C.10

Suppose mm is a positive integer. For 0≤k≤m0 \leq k \leq m, let

pk(x)=xk(1−x)m−k. p_{k}(x) = x^{k}(1-x)^{m-k}.

Show that p0,…,pmp_{0}, \ldots , p_{m} is a basis of Pm(F)\mathcal{P}_{m}(\mathbf{F}). The basis in this exercise leads to what are called Bernstein polynomials. You can do a web search to learn how Bernstein polynomials are used to approximate continuous functions on [0,1][0,1].

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Problem 2C.11

Suppose UU and WW are both four-dimensional subspaces of C6\mathbb {C}^{6}. Prove that there exist two vectors in U∩WU \cap W such that neither of these vectors is a scalar multiple of the other.

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Problem 2C.12

Suppose that UU and WW are subspaces of R8\mathbb {R}^{8} such that dim⁡U=3,dim⁡W=5\operatorname {dim} U = 3, \operatorname {dim} W = 5, and U+W=R8U+W = \mathbb {R}^{8}. Prove that R8=U⊕W\mathbb {R}^{8} = U \oplus W.

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Problem 2C.13

Suppose UU and WW are both five-dimensional subspaces of R9\mathbb {R}^{9}. Prove that U∩W≠{0}U \cap W \neq \left\{ 0\right\}.

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Problem 2C.14

Suppose VV is a ten-dimensional vector space and V1,V2,V3V_{1}, V_{2}, V_{3} are subspaces of VV with dim⁡V1=dim⁡V2=dim⁡V3=7\operatorname {dim} V_{1} = \operatorname {dim} V_{2} = \operatorname {dim} V_{3} = 7. Prove that V1∩V2∩V3≠{0}V_{1} \cap V_{2} \cap V_{3} \neq \left\{ 0\right\}.

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Problem 2C.15

Suppose VV is finite-dimensional and V1,V2,V3V_{1}, V_{2}, V_{3} are subspaces of VV with dim⁡V1+dim⁡V2+dim⁡V3>2dim⁡V\operatorname {dim} V_{1}+\operatorname {dim} V_{2}+\operatorname {dim} V_{3} > 2 \operatorname {dim} V. Prove that V1∩V2∩V3≠{0}V_{1} \cap V_{2} \cap V_{3} \neq \left\{ 0\right\}.

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Problem 2C.16

Suppose VV is finite-dimensional and UU is a subspace of VV with U≠VU \neq V. Let n=dim⁡Vn = \operatorname {dim} V and m=dim⁡Um = \operatorname {dim} U. Prove that there exist n−mn-m subspaces of VV, each of dimension n−1n-1, whose intersection equals UU.

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Problem 2C.17

Suppose that V1,…,VmV_{1}, \ldots , V_{m} are finite-dimensional subspaces of VV. Prove that V1+⋯+VmV_{1}+\cdots +V_{m} is finite-dimensional and

dim⁡(V1+⋯+Vm)≤dim⁡V1+⋯+dim⁡Vm. \operatorname {dim}\left(V_{1}+\cdots +V_{m}\right) \leq \operatorname {dim} V_{1}+\cdots +\operatorname {dim} V_{m}.

The inequality above is an equality if and only if V1+⋯+VmV_{1}+\cdots +V_{m} is a direct sum, as will be shown in 3.94.

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Problem 2C.18

Suppose VV is finite-dimensional, with dim⁡V=n≥1\operatorname {dim} V = n \geq 1. Prove that there exist one-dimensional subspaces V1,…,VnV_{1}, \ldots , V_{n} of VV such that

V=V1⊕⋯⊕Vn. V = V_{1} \oplus \cdots \oplus V_{n}.
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Problem 2C.19

Explain why you might guess, motivated by analogy with the formula for the number of elements in the union of three finite sets, that if V1,V2,V3V_{1}, V_{2}, V_{3} are subspaces of a finite-dimensional vector space, then

dim⁡(V1+V2+V3)=dim⁡V1+dim⁡V2+dim⁡V3−dim⁡(V1∩V2)−dim⁡(V1∩V3)−dim⁡(V2∩V3)+dim⁡(V1∩V2∩V3). \begin{aligned} \operatorname {dim}\left(V_{1}+V_{2}+V_{3}\right) = & \operatorname {dim} V_{1}+\operatorname {dim} V_{2}+\operatorname {dim} V_{3} \\ & -\operatorname {dim}\left(V_{1} \cap V_{2}\right)-\operatorname {dim}\left(V_{1} \cap V_{3}\right)-\operatorname {dim}\left(V_{2} \cap V_{3}\right) \\ & +\operatorname {dim}\left(V_{1} \cap V_{2} \cap V_{3}\right). \end{aligned}

Then either prove the formula above or give a counterexample.

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Problem 2C.20

Prove that if V1,V2V_{1}, V_{2}, and V3V_{3} are subspaces of a finite-dimensional vector space, then

dim⁡(V1+V2+V3)=dim⁡V1+dim⁡V2+dim⁡V3−dim⁡(V1∩V2)+dim⁡(V1∩V3)+dim⁡(V2∩V3)3−dim⁡((V1+V2)∩V3)+dim⁡((V1+V3)∩V2)+dim⁡((V2+V3)∩V1)3. \begin{aligned} \operatorname {dim} & \left(V_{1}+V_{2}+V_{3}\right) \\ = & \operatorname {dim} V_{1}+\operatorname {dim} V_{2}+\operatorname {dim} V_{3} \\ & -\frac{\operatorname {dim}\left(V_{1} \cap V_{2}\right)+\operatorname {dim}\left(V_{1} \cap V_{3}\right)+\operatorname {dim}\left(V_{2} \cap V_{3}\right)}{3} \\ & -\frac{\operatorname {dim}\left(\left(V_{1}+V_{2}\right) \cap V_{3}\right)+\operatorname {dim}\left(\left(V_{1}+V_{3}\right) \cap V_{2}\right)+\operatorname {dim}\left(\left(V_{2}+V_{3}\right) \cap V_{1}\right)}{3}. \end{aligned}

The formula above may seem strange because the right side does not look like an integer.

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