2.A

#1

Proof
We're given that $\text{span}(v_1,...,v_4)=V$, so then:

We just need to show that all $v_1,...,v_4$ can be expressed in terms of each of $v_1-v_2,v_2-v_3,v_3-v_4,v_4$. Notice that clearly:

So $v_1\in \text{span}(v_1,...,v_4)$. Similarly:

So then all $v_1,...,v_4\in \text{span}$ of our new list, so then by definition then $\text{span}(v_1,...,v_4)=\text{span}(v_1-v_2,...,v_4)$, so our new list spans $V$.
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#2

We verify each assertion below.

a: A list of one vector $v\in V$ is LI iff $v\ne \overrightarrow{0}$.

Consider the $\to$ direction. Consider some $\{v\}$, where we know the list is LI. Then the only way to write:

is $\alpha =0$. Notice that by contradiction if $v=\overrightarrow{0}$ then a possible $\alpha \ne 0$ to choose is:

which contradicts the list being LI. So then $v\ne \overrightarrow{0}$.

Consider the $\leftarrow$ direction. Say $v\ne \overrightarrow{0}$. Assume for contradiction that there's some nonzero $\alpha \in \mathbb{F}$ such that $\alpha v=\overrightarrow{0}$. Then since $\alpha \ne 0$ we can divide both sides by $\alpha$ to get:

Which contradicts $v\ne \overrightarrow{0}$, so then the only choice we have is that $\alpha =0$, so then by definition then the the only choice of alpha for $\alpha v=\overrightarrow{0}$ is $\alpha =0$, so the list $\{v\}$ is LI.

b: A list of two vectors in $V$ is LI iff neither vector is a scalar multiple of the other

Consider the $\to$ direction. Consider $\{v_1,v_2\}$ where $v_1,v_2\in V$. Say the list is LI, so then the only $a_1,a_2\in \mathbb{F}$ where:

is both $a_1,a_2=0$. Assume for contradiction that one vector is a scalar multiple of the other (which one it is doesn't matter here). So then let's say that $v_1=\alpha v_2$ for some non-zero scalar $\alpha \in \mathbb{F}$.

First, we should show that both $v_1,v_2\ne \overrightarrow{0}$. Notice if one of them was the zero vector, let's say $v_1$, then the linear combination:

wouldn't require that $a_1,a_2=0$ as the choice $a_1=1$ and $a_2=0$ would work. This contradicts the list being LI. As such, then $v_1,v_2\ne \overrightarrow{0}$. Thus, since we are considering $v_1=\alpha v_2$ then $\alpha \ne 0$ as if $\alpha =0$ then $v_1=0v_2=\overrightarrow{0}$ which is another contradiction. Thus, $\alpha \ne 0$.

But notice that:

but we know from above that $a_1=1$ and $-\alpha \ne 0$ so then we've found non-zero scalar that linearly combine $v_1,v_2$ to make the zero-vector, suggesting the list was LD the whole time. This is a contradiction, so then clearly $v_1=\alpha v_2$ is false, so each vector isn't a scalar multiple of the other.

Now, consider the $\leftarrow$ direction, so suppose neither vector of $v_1,v_2$ is a scalar multiple of the other, so no $v_1=\alpha v_2$ nor $v_2=\beta v_1$ for any $\alpha ,\beta \in \mathbb{F}$. Again, we can show that neither $v_1,v_2$ can be the zero vector. If one of them was, say $v_1=\overrightarrow{0}$, then notice that $v_1=0v_2$ which contradicts $v_1$ not being a scalar multiple of $v_2$. WLOG, then $v_2\ne \overrightarrow{0}$ as well.

So we know $v_1,v_2\ne \overrightarrow{0}$. Notice then that their linear combination:

for what values of $a_1,a_2\in \mathbb{F}$? Assume for contradiction that one $a_1,a_2\ne 0$. Then notice then that:

WLOG, let's say that $a_1$ is non-zero. Then we can divide it and see that:

but this contradicts $v_1$ not being a scalar multiple of $v_2$, so then clearly $a_1=0$. WLOG, then $a_2=0$ too. Thus, both $a_1,a_2=0$, showing that the list of $\{v_1,v_2\}$ is LI.

c: $(1,0,0,0),(0,1,0,0),(0,0,1,0)$ is LI in ${\mathbb{F}}^4$.

Let the first vector be $v_1$ the second $v_2$ and the third $v_3$. Consider choices of $a_1,a_2,a_3\in \mathbb{F}$ to make:

This requires, by expanding, that:

Creating 4 equations:

Thus, all $a_1,a_2,a_3=0$, so the list is LI.

d: The list $1,...,z^m$ is LI in $\mathcal{P}(\mathbb{F})$ for each non-negative integer $m\in \mathbb{N}$.

Let $v_1=1,v_2=z,...,v_{m+1}=z^m$. Then consider when:

We can substitute our values in:

equating coefficients yields:

So all $a_i\in \mathbb{F}$ are actually 0, so then the list of vectors above is LI.

#3

We'll find some number $t$ such that:

is NOT LI in ${\mathbb{R}}^3$. We can actually solve for this number by trying to take LC's of the other two vectors to make the new vector:

giving the equations:

Ignore the third equation for now. We can solve the $a_1,a_2$ using matrix algebra:

Notice $det(A)\ne 0$ so we can solve for $A^{-1}$:

so multiply $\overrightarrow{b}$ by our $A^{-1}$ to give:

So then we have $a_1=3,a_2=-2$, so then:

Thus choose $\boxed{{\displaystyle t=2}}$, then notice that:

so by definition then these vectors aren't LI.

#6

Proof
We need to show that the only values $a_i\in \mathbb{F}$ that make:

is that all $a_i=0$. We know that since $v_1,...,v_4$ is LI, then:

only when all $b_i=0$. Notice in our first equation that we can rewrite it to make:

But via our second equation, since each $a_i-a_j\in \mathbb{F}$ then each also is 0, ie all $a_i-a_j=0\Rightarrow a_i=a_j$. Thus, we have:

But notice then that $0=a_1=a_2=a_3=a_4$, so then that shows that our initial list must be LI!
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#7

Proof
Again, we need to show that:

Forces all $a_i=0$. We are given that since $v_1,...,v_m$ is LI then:

only when all $b_j=0$. Taking our first equation, notice that we can rewrite it into:

but by equation (2), then all of the constants in front of each $v_i$ is 0, namely:

Notice then that $a_1=0$ from the first equation we got. Thus, $-4a_1+a_2=0+a_2=0\Rightarrow a_2=0$. The remaining $a_j=0$, so then by definition our original list must be LI.
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#9

This a false proposition. Consider $\{v_1,v_2\}=\{(1,0),(0,1)\}$ and $\{w_1,w_2\}=\{(-1,0),(0,-1)\}$ are LI lists whose contents span ${\mathbb{R}}^2$. Notice then that:

But notice that $\{v_1+w_1,v_2+w_2\}=\{(0,0)\}$ is linearly dependent since we can choose:

is a valid linear combination of vectors in the set that makes the zero-vector.

#10

Proof
Suppose $v_1,...,v_m$ is LI in $V$, and $w\in V$. Suppose $v_1+w,...,v_m+w$ is LD, so there are some $a_i$ not all 0 such that:

Rearranging:

If $a_1+\cdots +a_m=0$ then that would imply that $a_1{v}_1+\cdots +a_m{v}_m=\overrightarrow{0}$ for some non-zero $a_i$ which is a contradiction. Hence, $a_1+\cdots +a_m\ne 0$. Therefore:

So $w\in \text{span}(v_1,...,v_n)$.

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# 11

Proof
Suppose $v_1,...,v_m$ is LI in $V$, and $w\in V$. So then:

only when all $b_i=0$. For this proof, we need to prove both directions.

For $(\to )$ suppose $v_1,...,v_m,w$ is LI. Now, assume for contradiction that $w\in \text{span}(v_1,...,v_m)$, so then there exists at least one non-zero constant $c_i$ such that:

But notice that we can move the $w$ over to create:

But look! At least one $c_i$ isn't 0, and with the $-1$ constant being added to the $w$ vector, it's clear that this implies that $v_1,...,v_m,w$ is not LI, which contradicts our supposition that it is. Therefore, then we must have it that $w\in \text{span}(v_1,...,v_m)$.

Now, consider $(\leftarrow )$, so then suppose that $w\notin \text{span}(v_1,...,v_m)$. Assume for contradiction that instead $v_1,...,v_m,w$ is LD. Then there exist at least one non-zero constant $e_i$ such that:

But look! Consider the cases where $e_{m+1}$ is or isn't 0. If $e_{m+1}=0$ then we have:

But since $v_1,...,v_m$ is LI then all $e_i=0$ which contradicts us having at least one-non zero constant $e_i$ as $v_1,...,v_m,w$ is LD. Now instead consider when $e_{m+1}\ne 0$. Then notice that we can solve for $w$:

But this would contradict $w\notin \text{span}(v_1,...,v_m)$ as this is a valid linear combination of constants to get $w$.

Therefore, no matter what, we have a contradiction, so then we must have it that $v_1,...,v_m,w$ is instead LI.
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#12

We know that there's not a list of 6 polynomials that are linearly independent in ${\mathcal{P}}_4(\mathbb{F})$ since we can make a spanning list that only uses 5 polynomials, so by the lemma that the length of a LI list is $\le$ the length of the spanning list, then at most any LI list of polynomials from this space must have 5 or less vectors.

Now for constructing this spanning list. I chose $\{1,x,x^2,x^3,x^4\}$. Notice that these are LI since if we consider:

then the only solution is all $a_i=0$, so then the list is LI. Furthermore, it is a valid spanning list, since given any arbitrary $p(x)=a+bx+c{x}^2+d{x}^3+e{x}^4\in {\mathcal{P}}_4(\mathbb{F})$ we can choose the constants:

to span this polynomial.

#13

Refer to the LI spanning list for ${\mathcal{P}}_4(\mathbb{F})$. We made a list of 5 items long, so again via the same lemma that the length of a LI list is $\le$ the length of a spanning list, then via our example then any spanning list must be $\ge$ to our LI list which was 5 items long. Thus, it is impossible for a 4 item list to exist that spans our space in question.

#14

Proof
We need to prove both directions.

First the $(\to )$ direction. Suppose that $V$ is infinite-dimensional. Thus, there exists no finite list $v_1,...,v_m$ of vectors from $V$ such that $v_1,...,v_m$ spans the space $V$.

We need to show that given some infinite sequence of vectors $v_1,...$ all in $V$, that cutting off $v_1,...,v_m$ gives a LI list for any positive integer $m\in \mathbb{N}$. Consider the base case of just $v_1$, which is clearly LI by proxy. Let our inductive hypothesis suggest that for all $k\in (1,...,m)$ that $v_1,...,v_k$ is LI. Consider $v_1,...,v_{k+1}$. Since $V$ is infinite-dimensional, then $v_1,...,v_k$ cannot span some new vector $v_{k+1}\in V$, so therefore adding it in gives us a LI list of $v_1,...,v_{k+1}$, completing the inductive step.

For the ($\leftarrow$) direction, suppose that there is an infinite sequence $v_1,v_2,...$ of vectors in $V$ such that $v_1,...,v_m$ is LI for every positive integer $m\in \mathbb{N}$. Notice that via contradiction, if we assumed that $V$ had a finite dimension, then we could make a new sequence $v_1,...,v_n,v_{n+1}$ whereby our supposition, that new list would be LI, and since a spanning list length is $\ge$ a LI list length, then any vectors spanned by $v_1,...,v_n$ clearly missed a vector from $\text{span}(v_1,...,v_n,v_{n+1})$, so then $n$ isn't the dimension of $V$ so $V$ isn't finite dimensional.
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#17

Proof
We'll try to prove this via induction over $m$.

As a base case, consider when $m=0$. The we have $p_0$ such that $p_0(2)=0$. Notice that since we are working in ${\mathcal{P}}_0(\mathbb{R})$, we are only working with constant functions. As such, since $p_0(2)=0=c$ then $p_0(x)=0$, so the list $p_0$ is LD.

Now let's do the inductive step. Let $k\in \mathbb{N}$ be arbitrary. Our inductive hypothesis is that for polynomials $p_0,...,p_k$ in ${\mathcal{P}}_k(\mathbb{F})$ where all $p_j(2)=0$ then $p_0,...,p_k$ is LD. Consider the $k+1$ case, where we have the list $p_0,...,p_k,p_{k+1}$ where the new $p_{k+1}(2)=0$. Consider trying to add the polynomials such as to add to the zero polynomial:

for all $a_j\in \mathbb{F}$. Now, if $p_{k+1}$ is the zero function then we are done as we would have the equation that:

where since $p_0,...,p_k$ is LD via our inductive hypothesis, then we can choose all our constants $a_0,...,a_k$ such that we have at least one non-zero $a_j$, and for $a_{k+1}$ we can set that to any value, even 0. Thus, our new equation would contain non-zero linear combinations of our vectors to get the zero vector, so then we have shown that $p_0,...,p_{k+1}$ is LD.

Now for the other case, suppose that $p_k$ isn't the zero function. We still can just choose $a_{k+1}=0$ so then we'd get:

where again via our inductive hypothesis, then there will be at least one non-zero coefficient $a_j$ to show that our whole $a_0,...,a_{k+1}$ contains at least one non-zero choice. As such, then the whole $p_0,...,p_{k+1}$ must be LD.

This completes the proof via mathematical induction.
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2.B

#2

We'll verify all the assertions from Example 2.28:

a: The list $(1,0,...,0),(0,1,0,...,0),...,(0,...,0,1)$ is a basis of ${\mathbb{F}}^n$, called the standard basis of ${\mathbb{F}}^n$.

Notice that clearly the list spans ${\mathbb{F}}^n$ as any $x\in {\mathbb{F}}^n$, say $x=(x_1,...,x_n)$ can be represented via:

Now, for LI, consider when:

giving the equations:

thus then this list of vectors is LI.

b: The list $(1,2),(3,5)$ is a basis for ${\mathbb{F}}^2$

Again, we can see that LI is made by considering:

solving for $a_1$ gives that $a_1=-3a_2$, so substituting into (2) gives $-6a_2+5a_2=0\Rightarrow a_2=0$ so then $a_1=0$. Thus, the list is LI.

Now for span. Consider an arbitrary vector $x\in {\mathbb{F}}^2$, ie: $x=(x_1,x_2)$. Then:

is an optimal choice, so then clearly it's spanned.

c: The list $(1,2,-4),(7,-5,6)$ is LI in ${\mathbb{F}}^3$ but isn't a basis for it since it doesn't span it

Note we can show LI by noticing that:

only when $a_1+7a_2=0$ and $2a_1-5a_2=0$. Notice that $a_1=-7a_2$ so then substituting into (2) then $-14a_2-5a_2=-19a_2=0\Rightarrow a_2=0$ so then $a_1=0$, showing LI.

Now for lacking span, notice that I can choose the vector $(1,0,0)\in {\mathbb{F}}^3$ but it's not in the span since then:

implying that $a_1+7a_2=1$, $2a_1-5a_2=0$, and $-4a_1+6a_2=0$. From (1) we have $a_1=1-7a_2$ so then putting it into (2) gives:

thus then $a_1=\frac{5}{19}$, but plugging into (3) gives:

thus there's no $a_i$'s that allow for the vector $(1,0,0)$ to be spanned by our list of vectors.

d: The list $(1,2),(3,5),(4,13)$ spans ${\mathbb{F}}^2$ but isn't a basis since it isn't LI

Look at HW 2 - Finite Dimensional Vector Spaces#b A list of two vectors in $V$ is LI iff neither vector is a scalar multiple of the other for a reason why at least the first two vectors span the list. The addition of the third vector doesn't change the span.

For LI, consider:

I claim that $(a_1,a_2,a_3)=(-19,5,1)$ since:

thus the list isn't LI.

e: The list $(1,1,0),(0,0,1)$ is a basis for $\{(x,x,y):x,y\in \mathbb{F}\}$

For span, denote $x\in S$ as arbitrary, so then $x=(i,i,j)$ for $i,j\in \mathbb{F}$. Notice that:

thus since $x$ was arbitrary then these vectors span the set.

For LI, consider:

implying that $a_1=0,a_1=0,a_2=0$, showing LI.

f: The list $(1,-1,0),(1,0,-1)$ is a basis for $\{(x,y,z)\in {\mathbb{F}}^3:x+y+z=0\}$

First let's talk span. Consider $x\in S$, so then $x=(a,b,c)$ such that $a+b+c=0$. Now, notice that:

since $a=-b-c$, so then since $x$ was arbitrary, then the vectors span the space.

Now for LI. Consider when:

we have it that $a_1+a_2=0$, $-a_1=0$ and $-a_2=0$, so then $a_1=a_2=0$ thus the vectors are LI.

g: The list $1,...,z^m$ is a basis for ${\mathcal{P}}_m(\mathbb{F})$

Consider span. Let $x\in {\mathbb{P}}_m(\mathbb{F})$ be arbitrary, so $x=x_1+x_{2}z+...+x_{m+1}{z}^m$. Then notice that:

thus since $x$ was arbitrary the vectors span our space.

For LI, consider when:

equating coefficients yeilds that all $a_i=0$ showing LI.

#3

a

Let $U$ be a subspace of ${\mathbb{R}}^5$ defined by:

a valid basis for $U$ is:

which we'll show. First, for span, let $x\in U$ be arbitrary, so then $x=(3x_2,x_2,7x_4,x_4,x_5)$. Then notice that:

where $v_1$ is the first vector mentioned before, and so on. Since $x$ was arbitrary it follows that the vectors span $U$.

For LI, consider when:

Expanding gives us the equation that $3a_1=0,a_1=0,7a_2=0,a_2=0,a_3=0$, which are only satisfied when $a_1,a_2,a_3=0$ showing LI.

b

A valid basis for ${\mathbb{R}}^5$ using our previous basis would be:

where $v_1$ is the first vector, $v_2$ is the second vector, and so on. First, we show that some $x\in {\mathbb{R}}^5$ is spanned by these vectors. Then $x=(x_1,x_2,x_3,x_4,x_5)$. Notice though that:

is a linear combination of vectors from our basis to compose $x$, so it definitely spans it.

For LI, consider:

Then notice that $3a_1=0,a_1+a_2=0,7a_3=0,a_3+a_4=0,a_5=0$. Notice then that $a_1,a_3,a_5=0$, so then $0+a_2=0\Rightarrow a_2=0$ and similarly $a_4=0$. Thus, the list is LI. This is a valid basis for ${\mathbb{R}}^5$.

c

Let $W=\text{span}(v_2,v_4)$ from the previous problem, so it's the vectors spanned by $v_2=(0,1,0,0,0)$ and $v_4=(0,0,0,1,0)$.

First, notice that $U+W$ are the vectors spanned by $\text{span}(v_1,...,v_5)$ which we saw in the (b) that that spanned all of ${\mathbb{R}}^5$ as it's a basis. So ${\mathbb{R}}^5=U+W$.

Now, to show it's a direct sum, we'll show that $U\cap W=\{\overrightarrow{0}\}$. Let $x\in U\cap W$ be arbitrary. Then both $x\in U$ and $x\in W$. Thus, $x$ is represented as some:

Furthermore, we know that $x$ can be represented as some:

But equating terms says that:

Solving shows that all $x_i=0$, so in reality then $x=\overrightarrow{0}$ all along, completing the proof.

#5

Proof
Consider $p_0=1$, $p_1=x$, $p_2=x^3+x^2$, $p_3=-x^3$. Notice that none of these have a degree of 2.

To show they're LI, notice that if we consider:

clearly $a_0,a_1=0$. We also notice that $a_2-a_3=0\Rightarrow a_2=a_3$, and $a_2=0$ equating the $x^2$ terms, so then $a_3=0$ in suit. Thus, this list is LI.

Now for span. Let $p=a+bx+c{x}^2+d{x}^3$ be arbitrary. Then notice that:

so $p$ is in the span of our new list. Thus the new list is a basis of ${\mathcal{P}}_3(\mathbb{F})$ while not having any polynomials of degree 2.
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#6

Proof
Suppose $v_1,...,v_4$ is a basis of $V$. First, we show that the new list is LI.

Consider what constants make:

true. But notice that we can reorder them to make:

and then further rewritten into:

where since $v_1,...,v_4$ is LI (being a basis), then it follows that all our $a_1=0,a_2+a_1=0\Rightarrow a_2=0,....,a_4=0$, making this new list also LI.

Now for span. Let $x\in V$ be arbitrary. Since $v_1,...,v_4$ is a basis, $x\in \text{span}(v_1,...,v_4)$, so then:

But notice we can rewrite this to see that:

So clearly $x$ is in the span of the new list.
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#7

This is false by the following counterexample. Say $V={\mathbb{R}}^4$, where $v_1=(1,0,0,0),v_2=(0,1,0,0),v_3=(0,0,1,1),v_4=(0,0,0,1)$. Then let $U=\{((a,b,c,0):a,b,c\in \mathbb{R}\}$. Then clearly $v_1,v_2\in U$ and $v_3,v_4\notin U$ but $v_1,v_2$ isn't a basis for $U$.

#8

Proof
We need to show that $u_1,...,u_m,w_1,...,w_n$ is both LI and spans $V$.

First, span. Notice by definition that:

so given any $v\in V$, then there's some representation via:

so clearly $v$ is in the span of the new list.

Now for LI. Consider when we have:

Notice that we can group elements together: