LADR Exercises 2A

Problem 1

Find a list of four distinct vectors in ${\mathbb{F}}^3$ whose span equals

$$\{(x,y,z)\in {\mathbb{F}}^3:x+y+z=0\}$$

We can write $z$ as $-x-y$. Then, we just need linearly independent vectors of the form:

$$(x,y,-x-y)$$

We actually only need two to span the given set, since there are only two variables (this is a plane in 3D space).

Here are four distinct vectors:

$$\begin{array}{r}(1,-1,0)\\ (1,0,-1)\\ (1,1,-2)\\ (-1,1,0)\end{array}$$

Since $x+y+z-0$ is true any of the vectors individually, and any scalar multiplications of the individual vectors, any linear combination of these will always have $x+y+z$ as well.

Problem 2

Prove or give a counterexample: If $v_1,v_2,v_3,v_4$ spans $V$, then the list

$$v_1-v_2,v_2-v_3,v_3-v_4,v_4$$

also spans $V$.

If $v_1,v_2,v_3,v_4$ span $V$, that means any $u\in V$ can be expressed as a linear combination of them. Then, we just need to show that we can produce $v_1,v_2,v_3,v_4$ with the linear combinations four given vectors; if we are able to produce each of these, we will be able to produce everything in the span as well.

$$\begin{array}{rl}{v}_1& =(v_1-v_2)+(v_2-v_3)+(v_3-v_4)+v_4\\ v_2& =(v_2-v_3)+(v_3-v_4)+v_4\\ v_3& =(v_3-v_4)+v_4\\ v_4& =v_4\end{array}$$

Problem 3

Suppose $v_1,\dots ,v_m$ is a list of vectors in $V$. For $k\in \{1,\dots ,m\}$, let

$$w_k=v_1+\cdots +v_k$$

Show that $\text{span}(v_1,\dots ,v_m)=\text{span}(w_1,\dots ,w_m)$.

Any element $w_k$ is a linear combination of the vectors in $v_1,\dots ,v_m$. For example, for some $w_a,w_b$, we would just have

$$w_a+w_b=(v_1+\cdots +v_a)+(v_1+\cdots +v_b)$$

Thus, any $w$, and any linear combination of $w$‘s, can be written as a linear combination of $v$‘s; hence, we can say that $\text{span}(w_1,\dots ,w_m)\subseteq \text{span}(v_1,\dots ,v_m)$.

On the other hand, any $v_k$ can be written as:

$$v_k=w_k-w_{k-1}$$

where we have $w_0=0$. As such, we can also express any $v$ as a linear combination of $w$‘s, giving us $\text{span}(v_1,\dots ,v_m)\subseteq \text{span}(w_1,\dots ,w_m)$. Thus,

$$\text{span}(v_1,\dots ,v_m)=\text{span}(w_1,\dots w_m)$$

Problem 4

• (a) Show that a list of length one in a vector space is linearly independent if and only if the vector in the list is not $0$.
• (b) Show that a list of length two in a vector space is linearly independent if and only if neither of the two vectors in the list is a scalar multiple of the other.

(a) If the only vector in a list of length one is $0$, it is not linearly independent, since any $a\in \mathbb{F}$ makes $a(0)=0$. For any other element, we would need $a=0$, which would be linearly independent.

(b) Call our vectors $v_1$ and $v_2$. If they are scalar multiples of each other, we could write $v_1=\lambda v_2$ for some scalar $\lambda \in \mathbb{F}$. Then, we would have:

$$a_1{v}_1+a_2{v}_2=0$$

For linear independence, the only solution to the above is $a_1=a_2=0$. However, in the case of scalar multiples, we could write:

$$\begin{array}{rl}{a}_1{v}_1+a_2\lambda v_1& =0\\ (a_1+a_2\lambda )v_1& =0\end{array}$$

which could be true if we have $a_1=-a_2\lambda$. Thus, the only solution is not when $a_1=a_2=0$.

Problem 5

Find a number $t$ such that

$$(3,1,4),(2,-3,5),(5,9,t)$$

is not linearly independent in ${\mathbb{R}}^3$.

First we can solve:

$$\begin{array}{r}3x+2y=5\\ x-3y=9\end{array}$$

$(\text{Eq 1})-3\cdot (\text{Eq2})$ gives:

$$\begin{array}{rl}11y=-22& ⟶y=-2\\ x-3(-2)=9& ⟶x=3\end{array}$$

which then gives

$$t=4(3)+5(-2)=2$$

Thus, if $t=2$, we can write $3(3,1,4)-2(2,-3,5)=(5,9,2)$. Since one of the vectors is a linear combination of the other ones, we do not have a linearly independent set.

Problem 6

Show that the list $(2,3,1),(1,-1,2),(7,3,c)$ is linearly dependent in ${\mathbb{F}}^3$ if and only if $c=8$.

Suppose $x,y,z\in \mathbb{F}$ exist such that not all are zero and

$$x(2,3,1)+y(1,-1,2)+z(7,3,c)=0$$

Then we have:

$$\begin{array}{rl}2x+y+7z& =0\\ 3x-y+3z& =0\end{array}$$

Eliminating $y$ gives

$$5x+10z=0⟶x=-2z$$

Eliminating $x$ gives

$$\begin{array}{rl}6x+3y+21z& =0\\ 6x-2y+6z& =0\\ \therefore 5y+15z=0& ⟶y=-3z\end{array}$$

Since $x,y,z$ are not all zero, $z$ is not zero either.

We can write the final equation as:

$$\begin{array}{rl}x+2y+cz& =0\\ -2z+2(-3z)+cz& =0\\ (c-8)z& =0\end{array}$$

Hence, in the case that $x,y,z$ are not $0$, we can still write $x(2,3,1)+y(1,-1,2)+z(7,3,c)=0$ if $c=8$, in which case this would be linearly dependent. The only other way to solve would be $x=y=z=0$, which would be lienarly independent.

Problem 7

• (a) Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then the list $1+i,1-i$ is linearly independent.
• (b) Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{C}$, then the list $1+i,1-i$ is linearly dependent.

(a) If we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then our vectors are in $\mathbb{C}$ and scalars are in $\mathbb{R}$, so scalar multiplication is done with real numbers. Thus, for the list $1+i,1-i$ to be linearly independent, we need

$$a(1+i)+b(1-i)=0$$

where $a,b\in \mathbb{R}$ and $a=b=0$ is the only solution. Expanding:

$$\begin{array}{rl}a+a(i)+b-b(i)& =0\\ (a+b)+(a-b)i& =0\end{array}$$

We need both the real and imaginary parts to be zero:

$$\begin{array}{rl}a+b& =0\\ a-b& =0\\ \therefore 2a& =0⟶a=b=0\end{array}$$

Since the only solution is the trivial one, when we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then the list $1+i,1-i$ is linearly independent.

(b) If we think of $\mathbb{C}$ as a vector space over $\mathbb{C}$, then both our vectors and scalars are in $\mathbb{C}$, which means scalar multiplication can be done with a complex number. Thus, for the list $1+i,1-i$ to be linearly independent, we need

$$a(1+i)+b(1-i)=0$$

with $a=b=0$, except that in this case we have $a,b\in \mathbb{C}$. Since this is a list with only two elements, we actually just need to show that one of the elements can be expressed as a scalar multiple of the other element:

$$\begin{array}{rl}1-i& =c(1+i)\\ c& =\frac{1-i}{1+i}=\frac{1-i}{1+i}\frac{1-i}{1-i}=\frac{(1-i{)}^2}{1^2-i^2}=\frac{-2i}{2}=-i\end{array}$$

Since we have shown that one of the vectors can be written as a scalar multiple of the other (remember vectors and scalars are the same thing in this case because we have $\mathbb{C}$ defined over $\mathbb{C}$).

Alternatively, we could also use the above result for $c$ to derive $a$ and $b$:

$$\begin{array}{rl}(-i)(1+i)& =1-i\\ (-i)(1+i)-(1-i)& =0\\ (-i)(1+i)+(-1)(1-i)& =0\end{array}$$

which can be verified by expanding. This gives us $a=-i,b=-1$. Since $a\ne b\ne 0$, we have linear dependence.

Problem 8

Suppose $v_1,v_2,v_3,v_4$ is linearly independent in $V$. Prove that the list

$$v_1-v_2,v_2-v_3,v_3-v_4,v_4$$

is also linearly independent.

Let us consider a linear combination of these vectors:

$$\begin{array}{rl}{a}_1(v_1-v_2)+a_2(v_2-v_3)+a_3(v_3-v_4)+a_4{v}_4& =0\\ a_1{v}_1-a_1{v}_2+a_2{v}_2-a_2{v}_3+a_3{v}_3-a_3{v}_4+a_4{v}_4& =0\\ a_1{v}_1+(-a_1+a_2)v_2+(-a_2+a_3)v_3+(-a_3+a_4)v_4& =0\end{array}$$

Thus we can write the system

$$\begin{array}{rl}{a}_1& =0\\ -a_1+a_2& =0\\ -a_2+a_3& =0\\ -a_3+a_4& =0\end{array}$$

which has the solution $a_1=a_2=a_3=a_4=0$. Thus, since the only solution is the trivial one, this list is linearly independent.

Problem 9

Prove or give a counterexample: If $v_1,v_2,\dots ,v_m$ is a linearly independent list of vectors in $V$, then

$$5v_1-4v_2,v_2,v_3,\dots ,v_m$$

is linearly independent.

We can follow the same procedure as above:

$$\begin{array}{rl}{a}_1(5v_1-4v_2)+a_2{v}_2+a_3{v}_3+\cdots +a_m{v}_m& =0\\ 5a_1{v}_1-4a_1{v}_2+a_2{v}_2+a_3{v}_3+\cdots +a_m{v}_m& =0\\ 5a_1{v}_1+(-4a_1+a_2)v_2+a_3{v}_3+\cdots +a_m{v}_m& =0\end{array}$$

which yields

$$\begin{array}{rl}5a_1& =0\\ -4a_1+a_2& =0\\ a_3& =0\\ & \dots \\ a_m& =0\end{array}$$

$5a_1=0$ yields $a_1=0$, which in turn gives $a_2=0$ from the second equation. We already know $a_3,\dots ,a_m=0$ because it was given that $a_1,\dots ,a_m$ is linearly independent.

Problem 10

Prove or give a counterexample: If $v_1,v_2,\dots ,v_m$ is a linearly independent list of vectors in $V$ and $\lambda \in \mathbb{F}$, then $\lambda v_1,\lambda v_2,\dots ,\lambda v_m$ is linearly independent.

This is true for $\lambda \ne 0$ but false if $\lambda =0$. If $\lambda =0$, any solution of $a$‘s will fold for

$$a_1(\lambda v_1)+a_2(\lambda v_2)+\cdots +a_m(\lambda v_m)=0$$

Problem 11

Prove or give a counterexample: If $v_1,\dots ,v_m$ and $w_1,\dots ,w_m$ are linearly independent lists of vectors in $V$, then the list $v_1+w_1,\dots ,v_m+w_m$ is linearly independent.

This is false. We can have $v_k=-w_k$, in which case any any $v_k+w_k=0$, so it doesn’t matter what scalar $a_k$ is used, we will always have $a_k(v_k+v_k)=0$. As such, the only solution is not trivial.

Problem 12

Suppose $v_1,\dots ,v_m$ is linearly independent in $V$ and $w\in V$. Prove that if $v_1+w,\dots ,v_m+w$ is linearly dependent, then $w\in \text{span}(v_1,\dots ,v_m)$.

For linear independence, we have

$$a_1(v_1+w)+a_2(v_w+w)+\cdots +a_m(v_m+w)=0$$

Expanding and collecting like terms:

$$\begin{array}{rl}{a}_1{v}_1+a_{1}w+a_2{v}_w+a_{2}2+\cdots +a_m{v}_m+a_{m}w& =0\\ \underset{S}{\underbrace{(a_1{v}_1+a_2{v}_2+\cdots +a_m{v}_m)}}+\underset{A}{\underbrace{(a_1+a_2+\cdots +a_m)}}w& =0\\ S+Aw& =0\end{array}$$

If $A\ne 0$, we have

$$w=-\frac{1}{A}S$$

Since $S$ is a linear combination of $v_1,\dots ,v_m$, $w$ is also a linear combination.

If $A=0$, we have $S=0$. Since $S=a_1{v}_1+\cdots +a_m{v}_m$ and we started off with $v_1,\dots ,v_m$ being linearly independent, by definition we must have $a_1=\cdots =a_m=0$. However, this cannot be true because $v_1+w,\dots ,v_m+w$ is linearly dependent (there must be some solution other than the trivial).

Because of this contradiction, we must have $A=0$, which means $w$ is a linear combination of $v_1,\dots ,v_m$, which by definition means that $w\in \text{span}(v_1,\dots ,v_m)$.

Problem 13

Suppose $v_1,\dots ,v_m$ is linearly independent in $V$ and $w\in V$. Show that

$$v_1,\dots ,v_m,w\text{ is linearly independent}⟺w\notin \text{span}(v_1,\dots ,v_m)$$

First, we can show that $w\notin \text{span}(v_1,\dots ,v_m)⟹v_1,\dots ,v_m,w\text{ is linearly independent}$.

Assume that $w\notin (v_1,\dots ,v_m)$. Suppose there exists some linear combination such that

$$a_1{v}_1+\dots a_m{v}_m+a_{w}w=0$$

Then, if $a_w\ne 0$, we have

$$w=-\frac{1}{a_w}(a_1{v}_1+a_m{v}_m)$$

which would make $w$ a linear combination of $v_1,\dots ,v_m$. Therefore, we must have $a_w=0$, which simplifies the original equation to

$$a_1{v}_1+\cdots +a_m{v}_m=0$$

Since $\{v_1,\dots ,v_m\}$ is linearly independent, all $a_1,\dots ,a_m$ must be zero. Thus, all coefficients $a_1,\dots ,a_m,w$ are zero, so $\{v_1,\dots ,v_m,w\}$ is linearly independent.

Then, we can show that $v_1,\dots ,v_m,w$ being linearly independent implies $w\notin \text{span}(v_1,\dots ,v_m)$.

Assume that $w\in \text{span}(v_1,\dots ,v_m)$, so there exist scalars $a_1,a_2,\dots ,a_m$ such that

$$w=a_1{v}_1+a_2{v}_2+\cdots +a_m{v}_m$$

Consider the linear combination:

$$(-a_1)v_1+\cdots +(-a_m)v_m+w=0$$

Not all coefficients are zero, since the coefficient of $w$ is $1$. This implies that $\{v_1,\dots ,v_m,w\}$ is linearly dependent. This contradicts our assumption that $\{v_1,\dots ,v_m,w\}$ is linearly independent, so we must have $w\notin \text{span}(v_1,\dots ,v_m)$. $■$

Problem 14

Suppose $v_1,\dots ,v_m$ is a list of vectors in $V$. For $k\in \{1,\dots ,m\}$, let

$$w_k=v_1+\cdots +v_k$$

Show that the list $v_1,\dots ,v_m$ is linearly independent if and only if the list $w_1,\dots ,w_m$ is linearly independent.

First, we aim to show that independent $v$‘s means independent $w$‘s. If $w_1,\dots ,w_m$ is linearly independent, we have

$$a_1{w}_1+\cdots +a_m{w}_m=0$$

if and only if all $a_1,\dots ,a_m$ are $0$. This can be expanded to

$$a_1(v_1)+\cdots +a_m(v_1+\cdots +v_m)=0$$

Expanding each term:

$$\begin{array}{rl}& a_1{v}_1\\ +& a_2{v}_1+a_2{v}_2\\ +& a_3{v}_1+a_3{v}_2+a_3{v}_3\\ +& \dots \\ +& a_m{v}_1+a_m{v}_2+a_m{v}_3+\cdots +a_m{v}_m\\ =& 0\end{array}$$

Grouping like terms:

$$(a_1+\cdots +a_m)v_1+(a_2+\cdots +a_m)v_2+\cdots +a_m{v}_m=0$$

where each term has one less $a$ term in its coefficient than before.

If the list $v_1,\dots ,v_m$ is linearly independent, then the only solution is where the coefficients are all zero. Thus we know that $a_m=0$. We can recursively work backward for the other coefficients; for example, for the second last term:

$$\begin{array}{r}(a_{m-1}+a_m)a_{m-1}=0\\ a_m=0⟹a_{m-1}=0\end{array}$$

Following this, we have all $a_1,\dots ,a_m=0$. Since all scalar coefficients are zero, $w_1,\dots ,w_m$ are linearly independent.

Second, we aim to show that independent $w$‘s means independent $v$‘s. Suppose that there exists

$$a_1{v}_1+a_2{v}_2+\cdots +a_m{v}_m=0$$

where we want to show that $a_1=\cdots =a_m=0$.

We can express $v$‘s in terms of $w$‘s. For some term $v_k$, we can say that

$$v_k=w_k-w_{k-1}$$

with $w_0=0$. Then, we can substitute:

$$a_1(w_1-w_0)+a_2(w_2-w_1)+\cdots +a_m(w_m-w_{m-1})=0$$

Expanding each term:

$$\begin{array}{rl}{a}_1{w}_1-a_1{w}_0+a_2{w}_2-a_2{w}_1+\cdots +a_m{w}_m-a_m{w}_{m-1}& =0\\ (a_1{w}_1-a_2{w}_1)+(a_2{w}_2-a_3{w}_2)+\cdots +(a_m{w}_m-a_m{w}_{m-1})& =0\end{array}$$

Grouping like terms:

$$(a_1-a_2)w_1+(a_2-a_3)w_2+\cdots +(a_{m-1}-a_{m-2})w_{m-1}+a_m{w}_m=0$$

Since $w_1,\dots ,w_m$ are linearly independent, each of these coefficient terms must be zero. Like what we did before, we can start from $a_m=0$ to algo get $a_1,\dots ,a_{m-1}=0$. Since all $a_1,\dots ,a_m=0$, we’ve shown that $v_1,\dots ,v_m$ are linearly independent.

Thus, we’ve shown that if $v_1,\dots ,v_m$ are linearly independent, then $w_1,\dots ,w_m$ are linearly independent, as well as the other direction.

Problem 15

Explain why there does not exist a list of six polynomials that is linearly independent in ${\mathcal{P}}_4(\mathbb{F})$.

The space ${\mathcal{P}}_4(\mathbb{F})$ consists of all polynomials with coefficients in $\mathbb{F}$ whose degree is at most 4:

$$p(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4$$

where $a_0,a_1,a_2,a_3,a_4\in \mathbb{F}$.

Suppose we have six polynomials such that each polynomial can be written as

$$p_i(x)=a_{i0}+a_{i1}x+a_{i2}{x}^2+a_{i3}{x}^3+a_{i4}{x}^4$$

Now, for linear independence, we write

$$c_1{p}_1(x)+c_2{p}_2(x)+\cdots +c_6{p}_6(x)=0$$

where $a_1,\dots ,a_6\in \mathbb{F}$.

We can expand these as:

$$\sum _{i=1}^6{c}_i(a_{i0}+a_{i1}x+a_{i2}{x}^2+a_{i3}{x}^3+a_{i4}{x}^4)=0$$

which can be written in terms of each power of $x$:

$$\left(\sum _{i=1}^6{c}_i{a}_{i0}\right)+\left(\sum _{i=1}^6{c}_i{a}_{i1}\right)x+\left(\sum _{i=1}^6{c}_i{a}_{i2}\right)x^2+\left(\sum _{i=1}^6{c}_i{a}_{i3}\right)x^3+{\left(\sum _{i=1}^6{c}_i{a}_{i4}\right)}^4=0$$

Each coefficient must be zero, which gives us a system of five linear equations with 6 unknowns. Since we have more unknowns than equations, there are always non-trivial solutions (not all coefficients are zero). Since non-trivial solutions exist, this list is linearly dependent.

Problem 16

Explain why no list of four polynomials spans ${\mathcal{P}}_4(\mathbb{F})$.

A polynomial in ${\mathcal{P}}_4(\mathbb{F})$ has the form:

$$p(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4$$

Since any polynomial in ${\mathcal{P}}_4(\mathbb{F})$ requires five coefficients to be uniquely determined and we have only four polynomials to combine, it’s impossible to represent every polynomial in ${\mathcal{P}}_4(\mathbb{F})$ with just four polynomials.

Problem 17

Prove that $V$ is infinite-dimensional if and only if there is a sequence $v_1,v_2,\dots$ of vectors in $V$ such that $v_1,\dots ,v_m$ is linearly independent for every positive integer $m$.

($⟹$) First suppose $V$ is infinite-dimensional. We can prove by induction that there exists a sequence $v_1,v_2,\dots$ of vectors in $V$ such that for every $m\in {\mathbb{Z}}^{+}$ (positive integers), the first $m$ vectors are linearly independent.

• Base case: Since $V$ is infinite-dimensional, $V$ contains some nonzero vector $v_1$. The list containing only this vector is clearly linearly independent.
• Inductive step: Suppose the list of vectors $v_1,\dots ,v_k$ is linearly dependent for some $k\in {\mathbb{Z}}^{+}$. Since $V$ is infinite-dimensional, these vectors cannot span $V$, and so there exists some $v_{k+1}\in V$ that is outside $\text{span}(v_1,\dots ,v_k)$. Note that $v_{k+1}\ne 0$. But, then $v_1,\dots ,v_{k+1}$ is linearly independent by the Linear Dependence Lemma; if it were linearly independent, the lemma guarantees that there would exist a vector in the list which could be written as a linear combination of its predecessors, which is impossible by our construction.
• By induction, we have shown there exists a list $v_1,v_2,\dots$ such that $v_1,\dots ,v_m$ is linearly independent for every $m\in {\mathbb{Z}}^{+}$.

($⟸$) Second, assume there is a sequence $v_1,v_2,\dots$ of vectors in $V$ such that $v_1,\dots ,v_m$ is linearly independent for every $m\in {\mathbb{Z}}^{+}$. If $V$ were finite-dimensional, there would exist a list $v_1,\dots ,v_n$ for some $n\in {\mathbb{Z}}^{+}$ such that $V=\text{span}(v_1,\dots ,v_n)$. But, by our assumption, the list $v_1,\dots ,v_{n+1}$ is linearly independent since we specified independence for every positive integer $m$. Linearly independent lists must have length no longer than every spanning list, this is a contradiction. Thus $V$ is infinite-dimensional.

Problem 18

Prove that ${\mathbb{F}}^{\infty }$ is infinite-dimensional.

For every integer $k\in \mathbb{Z}$, define the vector $v_k$ such that it has a $1$ in coordinate $k$ and $0$ everywhere else. Then, the list $v_1,\dots ,v_m$ is linearly independent for any choice of $m\in {\mathbb{Z}}^{+}$. By Problem 17, ${\mathbb{F}}^{\infty }$ must be infinite-dimensional.

Problem 19

Prove that the real vector space of all continuous real-valued functions on the interval $[0,1]$ is infinite-dimensional.

• This means that each function $f$ has domain $x\in [0,1]$. And each $x\in [0,1]$, we have $f(x)\in \mathbb{R}$. The difference between this and ${\mathbb{R}}^{[0,1]}$ is that in this case, we are only concerned with real-valued functions. We will call the set $C[0,1]$.

Consider the set of monomials:

$$S=\{1,x,x^2,x^3,\dots \}$$

Each function $x^n$ in this set is a continuous real-valued function on $[0,1]$, so $S\subseteq C[0,1]$.

To show that $S$ is linearly independent, assume there exists a finite linear combination of elements of $S$ that equals the zero function:

$$c_0(1)+c_{1}x+c_2{x}^2+\cdots +c_n{x}^n=0$$

where $c_0,c_1,\dots ,c_n$ are real coefficients. Expanding this, we have

$$c_0+c_{1}t+c_2{t}^2+\cdots +c_n{t}^n=0\text{ for all }t\in [0,1]$$

For this to be zero, all of its coefficients must be $0$. Therefore, the set $S$ is linearly independent as no non-trivial linear combination of the functions in $S$ is zero.

Since any finite subset of $S$ is linearly independent, and $S$ is infinite, there exists an infinite linearly independent set in $S$. Since $S\subseteq C[0,1]$ there also exists an infinite linearly independent set in $C[0,1]$. Thus, $C[0,1]$ cannot be spanned by any finite set of functions and thus must be infinite-dimensional.

Problem 20

Suppose $p_0,p_1,\dots ,p_m$ are polynomials in ${\mathcal{P}}_m(\mathbb{F})$ such that $p_k(2)=0$ for each $k\in \{0,\dots ,m\}$. Prove that $p_0,p_1,\dots ,p_m$ is not linearly independent in ${\mathcal{P}}_m(\mathbb{F})$.

Suppose that $p_0,p_1,\dots ,p_m$ is linearly independent for contradiction.

• First, we show that this implies that $\text{span}(p_0,p_1,\dots ,p_m)={\mathcal{P}}_m(\mathbb{F})$.
• Then, we show that the above leads to a contradiction, because we can find a polynomial in ${\mathcal{P}}_m(\mathbb{F})$ such that is not in $\text{span}(p_0,p_1,\dots ,p)$.
• Therefore, $p_0,p_1,\dots p_m$ must not be linearly independent.

Note that the list $1,z,\dots ,z^m$ spans ${\mathcal{P}}_m(\mathbb{F})$ and has length $m+1$. Hence, every linearly independent list must be have length $m+1$ or less (see here). If $\text{span}(p_0,p_1,\dots ,p_m)\ne {\mathcal{P}}_m(\mathbb{F})$, there exists some $p$ in ${\mathcal{P}}_m(\mathbb{F})$ such that $p\notin \text{span}(p_0,p_1,\dots ,p_m)$ , and thus the list $p_0,p_1,\dots p_m,p$ is linearly independent and has length $m+2$. This presents a contradiction. Thus, we must have $\text{span}(p_0,p_1,\dots ,p_m)={\mathcal{P}}_m(\mathbb{F})$.

Now, define the polynomial $q\equiv 1$. Then, $q\in \text{span}(p_0,p_1,\dots ,p_m)$, which means there must be $a_0,\dots ,a_m\in \mathbb{F}$ such that

$$q=a_0{p}_0+a_1{p}_1+\cdots +a_m{p}_m$$

which in turn implies

$$q(2)=a_0{p}_0(2)+a_1{p}_1(2)+\cdots +a_m{p}_m(2)$$

This would give us $1=0$. Therefore, $p_0,p_1,\dots p_m$ cannot be linearly independent.