Homework 3 Solutions

Section 1.5

Problem 1

a. If $S$ is a linearly dependent set, then each vector in $S$ is a linear combination of other vectors in $S$. False: one possible counterexample is $\{v, 0\}$ with $v\neq 0$. The vectors $\{v,0\}$ are dependent, but $v$ is not a linear combination of $0$.

b. Any set containing the zero vector is linearly dependent. True.

d. Subsets of linearly dependent sets are linearly dependent. False. Same counterexample as in (a): $\{v, 0\}$ is dependent, but $\{v\}$ is independent.

e. Subsets of linearly independent sets are linearly independent. True.

f. If $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = 0$ and $x_1$, $x_2$ , … , $x_n$ are linearly independent, then all the scalars $a_i$ are zero. True. This is the definition of linear independence.

Problem 2a

If $M=\begin{pmatrix}1 &-3 \\ -2 &4 \end{pmatrix}$, $N=\begin{pmatrix} -2 & 6 \\4 & -8\end{pmatrix}$ then $\{M, N\}$ is dependent, because $N=-2M$, i.e. $2M+N=0$.

Problem 3

Suppose

$a\begin{pmatrix} 1 & 1 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 0 & 0 \end{pmatrix} + c\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 1 \end{pmatrix} + d\begin{pmatrix} 1 & 0 \\ 1 & 0 \\ 1 & 0 \end{pmatrix} + e\begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 &0 \\ 0 & 0 \end{pmatrix}$

Then

$\begin{pmatrix} a+d & a+e \\ b+d & b+e \\ c+d & c+e \end{pmatrix} = \begin{pmatrix} 0&0\\0&0\\0&0 \end{pmatrix}\implies \begin{aligned} a+d &=0 & a+e&=0 \\ b+d &=0 & b+e&=0 \\ c+d &=0 & c+e&=0 \end{aligned}$

Solving this you find $d=e$, $a=b=c$, and $a=-d$. One solution is $a=b=c=1$, and $d=e=-1,$ i.e. we have

$\begin{pmatrix} 1 & 1 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 1 & 0 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 &0 \\ 0 & 0 \end{pmatrix}$

which implies that the matrices are linearly dependent.

Problem 7

The set of diagonal matrices is

$V = \left\{\begin{pmatrix} a_1 & 0 \\ 0 & a_2 \end{pmatrix} \Big| \; a_1, a_2\in\mathbb{F}\right\}$

A basis for this vector space is given by $\{M_1, M_2\}$ where

$M_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad M_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

There are many other bases.

Problem 12: write a detailed proof.

We have to prove:

Theorem 1.6. Let $V$ be a vector space, and let $S_1 \subset S_2 \subset V$. If $S_1$ is linearly dependent, then $S_2$ is linearly dependent.

Proof. Since $S_1$ is linearly dependent, there are vectors $v_1, \dots, v_k\in S_1$ and numbers $a_1, \dots, a_k\in\mathbb{F}$, at least one of which is not zero, and for which $a_1v_1+\cdots+a_kv_k=0$. We are given that $S_1\subset S_2$, so each of the vectors $v_i$ also belongs to $S_2$. Therefore we have a set of vectors $v_1, \dots, v_k\in S_2$ and the same set of numbers $a_1, \dots, a_k$, one of which is not zero, with $a_1v_1+\cdots+a_kv_k=0$. This implies that $S_2$ is dependent. ////

Extra problem

If $u,v,w∈V$ are given, and if $p=u+v$, $q=u+w$, $r=v+w$, then express $u,v,w$ in terms of $p,q,r$. Solution:

$\begin{alignedat}{4} &u&+v&&=p \\ &u&&+w&=q \\ &&v&+w&=r \end{alignedat}\implies \begin{alignedat}{4} &u&+v&&&=p \\ &&-v&+w&&=-p+q \\ &&v&+w&&=r \end{alignedat}\implies \begin{alignedat}{4} &u&+v&&&=p \\ &&-v&&+w&=-p+q \\ &&&&2w&=-p+q+r \end{alignedat}\implies$

$\begin{alignedat}{4} &u&+v&&&=p \\ &&-v&&+w&=-p+q \\ &&&&w&=-\tfrac12p+\tfrac12q+\tfrac12r \end{alignedat}\implies \begin{alignedat}{4} &u&+v&&&=p \\ &&-v&&\;\;\;&=-\tfrac12p+\tfrac12q -\tfrac12r\\ &&&&w&=-\tfrac12p+\tfrac12q+\tfrac12r \end{alignedat}\implies \begin{alignedat}{4} &u&=\tfrac12p+\tfrac12q-\tfrac12r \\ &v&=\tfrac12p-\tfrac12q+\tfrac12r \\ &w&=-\tfrac12p+\tfrac12q+\tfrac12r \end{alignedat}$

Problem 13b

We are given that $\{u,v,w\}$ are linearly independent. To see if $\{u+v, u+w, v+w\}$ is independent, assume that for $a,b,c\in\mathbb{F}$ one has

$a(u+v)+b(u+w)+c(v+w)=0.$

Then

$(a+b)u+(a+c)v+(b+c)w=0.$

Independence of $\{u,v,w\}$ implies

$a+b=a+c=b+c=0.$

Solving these equations leads to $a=b=c=0$. So the only linear combination of $\{u+v, u+w, v+w\}$ that adds up to zero is the trivial linear combination. Therefore $\{u+v, u+w, v+w\}$ is linearly independent.

Section 1.6

Problem 1

b. Every vector space that is generated by a finite set has a basis. True.

c. Every vector space has a finite basis. False. Example: $\mathcal{P}(\R)$ does not have a finite basis.

d. A vector space cannot have more than one basis. False.

e. If a vector space has a finite basis, then the number of vectors in every basis is the same. True. This follows from the dimension theorem. The number of vectors in a basis is called the dimension of the vector space.

f. The dimension of $\mathcal{P}_n(\mathbb{F})$ is $n$. False. $\{1, x, x^2, \dots, x^n\}$ is a basis for $\mathcal{P}_n(\mathbb{F})$, so the dimension is $n+1$.

g. The dimension of $\mathcal{M}_{m×n} (\mathbb{F})$ is $m + n$. False. The dimension of $\mathcal{M}_{m\times n}(\mathbb{F})$ is $mn$.

Problem 4

Do the polynomials $x^3 −2x^2 +1$, $4x^2 −x+3$, and $3x−2$ generate $\mathcal P_3 (\R)$?

Answer: no. This follows from the dimension theorem: If the three vectors $x^3 −2x^2 +1$, $4x^2 −x+3$, and $3x−2$ spanned $\mathcal P_n(\R)$, then the four vectors $\{1, x, x^2, x^3\}$ would be linear combinations of $x^3 −2x^2 +1$, $4x^2 −x+3$, and $3x−2$. By the dimension theorem $\{1, x, x^2, x^3\}$ would be linearly dependent. Since $\{1, x, x^2, x^3\}$ is independent we have a contradiction.

Problem 5

Is $\{p=(1, 4, −6), q=(1, 5, 8), r=(2, 1, 1), s=(0, 1, 0)\}$ a linearly independent subset of $\R^3$? Answer: no. The four given vectors $\{p,q,r,s\}$ are linear combinations of the three standard basis vectors $e_1=\left(\begin{smallmatrix} 1 \\ 0\\ 0\end{smallmatrix}\right)$, $e_2=\left(\begin{smallmatrix} 0 \\ 1\\ 0\end{smallmatrix}\right)$, $e_3=\left(\begin{smallmatrix} 0 \\ 0\\ 1\end{smallmatrix}\right)$. By the dimension theorem $\{p,q,r,s\}$ is linearly dependent.

Problem 7

Given:

$u_1 = \left(\begin{smallmatrix} 2 \\ -3 \\1 \end{smallmatrix}\right),\quad u_2 = \left(\begin{smallmatrix} 1 \\ 4 \\ -2 \end{smallmatrix}\right),\quad u_3 = \left(\begin{smallmatrix} -8 \\ 12 \\-4 \end{smallmatrix}\right),\quad u_4 = \left(\begin{smallmatrix} 1 \\ 37 \\ -17 \end{smallmatrix}\right),\quad u_5 = \left(\begin{smallmatrix} 3 \\ -5 \\8 \end{smallmatrix}\right)$

We have to choose three vectors from this collection and verify that they are independent.
Note that $u_3=-4u_1$, so $\{u_1, u_2, u_3\}$ won't work. Let's try $\{u_1, u_2, u_4\}$. Suppose $au_1+bu_2+cu_4=0$. Then $a,b,c$ satisfy

$\begin{alignedat}{4} 2a&+b&&+c&=0\\ -3a&+4b&&+37c&=0 \\ a&-2b&&-17c&=0 \end{alignedat}\implies \begin{alignedat}{4} &&5b&&+35c&=0\\ &&b&&+21c&=0 \\ a&&-2b&&-17c&=0 \end{alignedat}\implies \begin{alignedat}{4} &&&&-14c&=0\\ &&b&&+21c&=0 \\ a&&-2b&&-17c&=0 \end{alignedat}$

which implies $a=b=c=0$. Hence $\{u_1, u_2, u_4\}$ is a basis.

Problem 9

For any given $a = (a_1, a_2, a_3, a_4)\in\mathbb{F}^4$ we are asked to find $(c_1, c_2, c_3, c_4)$ such that $c_1u_1+c_2u_2+c_3u_3+c_4u_4=a$, i.e.

$c_1\begin{pmatrix} 1 \\ 1\\ 1\\ 1 \end{pmatrix}+ c_2\begin{pmatrix} 0 \\ 1\\ 1\\ 1 \end{pmatrix}+ c_3\begin{pmatrix} 0 \\ 0\\ 1\\ 1 \end{pmatrix}+ c_4\begin{pmatrix} 0 \\ 0\\ 0\\ 1 \end{pmatrix}= \begin{pmatrix} a_1\\a_2\\a_3\\a_4 \end{pmatrix}.$

This leads to the following linear equations:

$\begin{alignedat}{5} c_1 &&&&=a_1\\ c_1 &+c_2&&&=a_2\\ c_1 &+c_2&+c_3&&=a_3\\ c_1 &+c_2&+c_3&+c_4&=a_4 \end{alignedat}\implies \begin{alignedat}{2} c_1&=a_1&\\ c_2&=a_2-c_1=a_2-a_1\\ c_3&=a_3-c_1-c_2=a_3-a_2\\ c_4&=a_4-c_1-c_2-c_3=a_4-a_3 \end{alignedat}\implies \begin{alignedat}{2} c_1&=a_1&\\ c_2&=a_2-a_1\\ c_3&=a_3-a_2\\ c_4&=a_4-a_3 \end{alignedat}$

Problem 12

Since $\{u,v,w\}$ is a basis for $V$, we know that $V$ is three dimensional. To show that $\{u+v+w, v+w, w\}$ is a basis we have to show that $\{u+v+w, v+w, w\}$ is independent.

Suppose

$a(u+v+w)+b(v+w)+cw=0.$

Then

$(a+b+c)u+(b+c)v+cw=0.$

Since $\{u,v,w\}$ is independent, it follows that $a+b+c=b+c=c=0$. This implies $a=b=c=0$. Therefore $\{u+v+w, v+w, w\}$ is indeed linearly independent.

Problem 13

The problem states that the set of all $(x_1, x_2, x_3)\in\R^3$ that satisfy

$\begin{alignedat}{4} x_1&-2x_2&+x_3 &= 0 \\ 2x_1&-3x_2&+x_3 &= 0 \end{alignedat}$

is a linear subspace of $\R^3$. Solve the equations for $x_1, x_2$ by row reduction:

$\begin{alignedat}{4} x_1&-2x_2&+x_3 &= 0 \\ 2x_1&-3x_2&+x_3 &= 0 \end{alignedat}\implies \begin{alignedat}{4} &x_1&-2x_2&+x_3 &= 0 \\ & & x_2&-x_3 &= 0 \end{alignedat}\implies \begin{alignedat}{4} &x_1& \qquad &-x_3 &= 0 \\ & & x_2&-x_3 &= 0 \end{alignedat}$

We see that the solution set consists of all $\tmat x_1\\x_2\\x_3\trix\in\R^3$ that satisfy $x_1=x_2=x_3$. Hence the solution set is

$\Bigl\{x_3\tmat1\\1\\1\trix \;\big|\; x_3\in\R\Bigr\}.$

Therefore $\Bigl\{\tmat1\\1\\1\trix\Bigr\}$ is a basis for the solution space.

Problem 14

The vector space $W_1$ consists of all $a\in\mathbb{F}^5$ that satisfy $a_1=a_3+a_4$. $W_1$ is a linear subspace of $\mathbb F^5$, so $\dim W_1\leq 5$. We also have $W_1\neq \mathbb{F}^5$, because $e_1=(1, 0,0,0,0)\in\mathbb F^5$, but $e_1\not\in W_1$. Therefore $\dim W_1\leq 4$. The vectors

$\begin{pmatrix} 0\\1 \\0 \\0 \\0 \end{pmatrix},\quad \begin{pmatrix} 1\\0 \\1 \\0 \\0 \end{pmatrix},\quad \begin{pmatrix} 1\\0 \\0 \\1 \\0 \end{pmatrix},\quad \begin{pmatrix} 0\\0 \\0 \\0 \\1 \end{pmatrix},$

all belong to $W_1$ and they are linearly independent. Therefore this is a basis for $W_1$ and $\dim W_1=4$.

The space $W_2$ consists of all $a\in\mathbb F^5$ with $a_2=a_3=a_4$ and $a_1=-a_5$.
It follows that each $a\in W_2$ is given by

$a = \begin{pmatrix} a_1 \\ a_2 \\ a_2 \\ a_2 \\ -a_1 \end{pmatrix} = a_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -1 \end{pmatrix} + a_2 \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}$

The vectors $\left(\begin{smallmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -1 \end{smallmatrix}\right) , \left(\begin{smallmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{smallmatrix}\right)$ are independent, and therefore they are a basis for $W_2$. It follows that $W_2$ is two dimensional.