In the previous chapter, we understood what an abstract vector space is. Linear algebra focuses mainly on finite-dimensional vector spaces. But what does 'finite dimension' mean? How do you define dimension? We know that $(2, 1)$ is two-dimensional, $(2, 1, -2)$ is three-dimensional, and $R^n$ is n-dimensional; these are all finite dimensional. But if a vector is not in the form of a list, how do we define its dimension? What does "dimension" actually mean?

Actually, the dimension of a vector space, is defined by "the number of vectors required to span the space".

As an example, $\mathbf{R}^2$ can be spanned by $(1, 0), (0, 1)$. this means that, any point on the $\mathbf{R}^2$ plane, can be decomposed into $(1, 0)$ multiplied by a scalar, plus $(0, 1)$ multiplied by a scalar.

Using the set notation, we have

On top of that, $(1, 0), (0, 1)$ are not the only vectors that work. Pick any 2 vectors in $\mathbf{R}^2$, and you will find that as long as they are not scalar multiples of each other(in the same direction), then they span $\mathbf{R}^2$. So the choice of vectors don't seem to matter.

However, if we reduce the list to 1 vector, it cannot span $\mathbf{R}^2$ anymore, whichever vector you pick. So it seems like the 2D plan requires at least 2 vectors to span, while the 3D space requires at least 3 vectors to span. This gives a basis for an intuition of the definition of dimension.

A 2-dimensional space can be spanned by 2 vectors.

A 3-dimensional space can be spanned by 3 vectors.

That's pretty much what dimension is! We've explained it in a loose, intuitive manner.

Now, let us explore this further and see how linear algebra formally defines and arrives at the definition of "dimension"!

2.A Span and Linear Independence

When we talk about list of vectors, we often omit the surrounding parentheses. For example:

• When we say $(1, 0), (0, 1)$ spans $\mathbf{R}^2$, we actually mean the list $((1, 0), (0, 1))$ spans $\mathbf{R}^2$.
• $(4, 1, 6), (9, 5, 7)$ is a list of length two of vectors in $\mathbf{R}^3$.

So let's get this definition out of the way.

Back to the main topic. We previously touch upon the notion of "the sum of scalar multiples of vectors". We call this a linear combination. Here we formally define it.

For example: $(17, −4, 2)$ is a linear combination of $(2, 1, −3), (1, −2, 4)$, because

However, $(17, −4, 5)$ is not a linear combination of $(2, 1, −3), (1, −2, 4)$, because no $a_1, a_2 \in \mathbf{F}$ exist such that

Now we formally define "span".

For example,

• $(17, −4, 2) \in \operatorname{span}\left((2, 1, −3), (1, −2, 4)\right)$
• $(17, −4, 5) \notin \operatorname{span}\left((2, 1, −3), (1, −2, 4)\right)$

For instance, for any positive integer $n$, $\mathbf{F}^n$ is a finite-dimensional vector space. because:

• Any vector in $\mathbf{F}^n$ can be written as $(x_1, \ldots, x_n)$
• $\left(x_1, \ldots, x_n\right)=x_1(1,0, \ldots, 0)+x_2(0,1,0 \ldots . .0)+\cdots+x_n(0, \ldots, 0,1)$
• Hence $\left(x_1, \ldots, x_n\right) \in \operatorname{span}((1,0, \ldots, 0),(0,1,0, \ldots, 0), \ldots,(0, \ldots, 0,1))$

Polynomials, which we've seen a lot an high school math, can actually also be seen as vectors! Is it finite dimensional or not though? Before we explore that, let's first introduce some notations.

We introduce 2 concepts here:

• First, we look at polynomials from a new perspective: as a function that maps a number to a number($\mathbf{F} \rightarrow \mathbf{F}$)
• Second, we define the notation $\mathcal{P}(\mathbf{F})$ to denote the set of all polynomials

As two examples：

• $x^3 - 7x + 5 \in \mathcal{P}(\mathbf{R})$
• $(3+i)z^2 - iz + 5 \in \mathcal{P}(\mathbf{C})$

Having defined the set $\mathcal{P}(\mathbf{F})$, you might have guess what I'm going to say next. That's right! We claim that $\mathcal{P}(\mathbf{F})$ is a vector space. You are encouraged to verify all the conditions for a vector space. Refer back to the last chapter if needed.

Remember that we mentioned that $\mathbf{F}^{\mathbf{S}}$ is a vector space in the last chapter? This means $\mathbf{F}^{\mathbf{F}}$, being a special case, a subspace of it.

Now we define $\mathcal{P}_m(\mathbf{F})$, the set of all polynomials with degree no more than $m$

It's easy to see $\mathcal{P}_m(\mathbf{F}) = \operatorname{span}(1, z, \ldots, z^m)$. In other words, a finite list of vectors spans $\mathcal{P}_m(\mathbf{F})$, implying $\mathcal{P}_m(\mathbf{F})$ is a finite-dimensional vector space.

This definition may seem a bit repetitive. But let's think through it to see what it means.

• Finite-dimensional is defined to mean "can be spanned by a finite list of vectors"
• Infinite-dimensional is defined to mean "not finite-dimensional"
• Therefore, a vector space that cannot be spanned by a finite list of vectors is infinite-dimensional

As an example, $\mathcal{P}(\mathbf{R})$ is infinite-dimensional. Why?

• Think of any list of members $\mathcal{P}(\mathbf{R})$
• Let $m$ denote the hight degree of the polynomials in this list
• Then every polynomial in the span of this list has degree at most $m$
• Thus $z^{m+1}$ is not in said span
• Hence no list spans $\mathcal{P}(\mathbf{R})$
• Thus $\mathcal{P}(\mathbf{R})$ is infinite dimensional

Linear Independence

We have understood that the 2D plane can be spanned by 2 vectors. But 2 vectors on the same line cannot span the 2D plane. This distinction is formalized by the concept of "linear independence".

A list of vectors $(v_1, \ldots, v_m)$ is linearly independent if and only if

• $0$ has only 1 way to be written as a linear combination(taking all coefficients to be 0)
• Any vector in $\operatorname{span}(v_1, \ldots, v_m)$ has only one representation as a linear combination of $v_1, \ldots, v_m$

Pause and think about why this is the case. Visualize it with the 2D or 3D cases. What does it look like when your list of vectors is linear independent, and when it is linear dependent?

Now we look at a few examples:

1. $(1,0,0,0), (0,1,0,0), (0,0,1,0)$ is linearly independent in $\mathbf{F}^4$. To see why, assume $a_1, a_2, a_3 \in \mathbf{F}$. Then we have $$a_1(1,0,0,0)+a_2(0,1,0,0)+a_3(0,0,1,0)=(0,0,0,0)$$ Hence $$(a_1,a_2,a_3,0)=(0,0,0,0)$$ Thus $a_1=a_2=a_3=0$. So $(1,0,0,0), (0,1,0,0), (0,0,1,0)$ is linearly independent in $\mathbf{F}^4$.
2. $1, z, \ldots, z^m$ is linearly independent in $\mathcal{P}(\mathbf{R})$
3. A list of one vector is linearly independent if and only if it is not the $0$ vector
4. A list of 2 vectors is linearly independent $\iff$ neither of them is a scalar multiple of the other

If a list of vectors is not linearly independent, they are linearly dependent.

Here are a few examples. Make sure to understand each of them:

• $(2, 3, 1), (1, −1, 2), (7, 3, 8)$ is linearly dependent in $\mathbf{F}^3$
• In $V$, if a list of vectors has a vector that is a linear combination of the other vectors, then this list is linearly dependent.(proof: Writing that one vector in the list as equal to a linear combination of the other vectors, move that vector to the other side of the equation to get a linear combination of $0$ that is not all 0 coefficients)
• Every list of vectors in $V$ containing the $0$ vector is linearly dependent.(a special case of the previous bullet point)

The next lemma is a useful tool that we will use a lot for many proofs we will see later.

Proof: This is a straightforward proof. We want to prove that:

1. There exists a number $k$ such that the $k$-th vector is a linear combination of the vectors preceding it, i.e. $v_k \in \operatorname{span}(v_1, \ldots, v_{k-1})$
2. Removing $v_k$ from the list does not change the span

We first prove the first bullet point.

• $v_1, \ldots, v_m$ are linearly dependent, which means there exist $a_1, \ldots, a_m \in F$, not all zero, such that $a_1 v_1 + \ldots + a_m v_m = 0$.
• Suppose $a_k$ is the last non-zero coefficient among $a_1, \ldots, a_m$, then we can rearrange to get $v_k = -\frac{a_1}{a_k} v_1 - \cdots - \frac{a_{k-1}}{a_k} v_{k-1}$.
• This implies $v_k \in \operatorname{span}\left(v_1, \ldots, v_{k-1}\right)$, as desired.

Now we come to the second point.

• Suppose $u$ is a vector in the span, that is, $u \in \operatorname{span}(v_1, \ldots, v_m)$.
• $u$ can be written as a linear combination of these vectors: $u = c_1 v_1 + \cdots + c_m v_m$.
• From the first point, we know that $v_k$ can be replaced by a linear combination of $v_1, \ldots, v_{k-1}$.
• Since $u$ is any vector, it means that any vector in the span can be formed without needing $v_k$, thus proving the point.

We've completed the proof. But how should we interpret this theorem? I essentially says that in a linearly dependent list of vectors, we can find one vector that is a linear combination of the previous elements. Thus even if the throw out this "redundant" vector, the span of the list does not become smaller.

Conversely, when you adjoin to a list a vector already in the span of of the list, the span does not grow bigger. Only linearly independent vectors contribute to the span.

Let us look at a few examples:

example 2.21

$(1, 2, 3), (6, 5, 4), (15, 16, 17), (8, 9, 7)$ are linearly dependent in $\mathbf{R}^3$. To see why,

• If we take $k=1$, the first vector must be 0. But since $(1, 2, 3)$ is not the zero vector, we cannot take $k=1$.
• If we take $k=2$, the second vector must be a multiple of the first vector, but there does not exist $c \in \mathbf{R}$ such that $(6,5,4)=c(1,2,3)$, so we cannot take $k=2$.
• Taking $k=3$, we can find $(15, 16, 17) = 3(1, 2, 3) + 2(6, 5, 4)$, thus this set of vectors is linearly dependent.

With this lemma proven, we now prepared to prove a key result.

Proof： We first assume:

• $u_1, \ldots, u_m$ are linearly dependent in $V$
• $w_1, \ldots, w_n$ spans $V$

We prove $m \leq n$ by the following process. Note that in each step we add one of the $u$'s and remove one of the $w$'s.

Step 1:

• Let $B$ be the list $w_1, \ldots, w_n$ that spans $V$.
• Insert $u_1$ at the beginning of $B$, which will produce a linearly dependent list $$\overbrace{u_1, w, \ldots, w}^{\text{linearly dependent}}$$
• According to 2.19 (the linear dependence lemma), one of the vectors in the list can be expressed as a combination of the preceding vectors, and thus can be removed. This "redundant vector" will not be $u_1$, because $u_1$ is not in $\{0\}$ (where $\{0\}$ is the span of the empty list).
• In other words, one of the $w$'s can be removed without affecting the span of the list.

Step $k$ ($k=2, \ldots, m$)：

• The list obtained from step $k-1$ spans $V$.
• Therefore, we can insert $u_k$ into the list, forming a linearly dependent list (the subscripts of the $w$'s are omitted here). $$\overbrace{u_1, \ldots, u_k, w, \ldots, w}^{\text{linearly dependent}}$$
• Because this list is linearly dependent, we can remove one of the vectors without affecting its span.
• It is important to note that the first part of this list (composed of the $u$'s) is linearly independent and thus will not be removed. Since we need to find a vector that can be composed of the preceding vectors, the one to be removed must be from the $w$ part. $$\overbrace{\underbrace{u_1, \ldots, u_k}_{\mathclap{\text{linearly independent}}}, w, \ldots, w}^{\text{linearly dependent}}$$

Repeat process above, which adds a $u$ and removes a $w$ in each step, and we will find that all of the $u$'s have been moved to the other list, implying there are more $w$'s than $u$'s, as we expect.

This proof may be confusing at first glance, so here is some additional explanation:

1. Every time we insert a $u$, we get a linearly dependent list
2. By lemma 2.19, there MUST be a vector that we can move without changing the span
3. This vector CANNOT be a $u$, because all of the $u$'s are linearly independent. Thus it must be one of the $w$'s
4. This implies there are more $w$'s than $u$'s

In plain English, this theorem states that a linearly independent list of vectors is always shorter than a spanning list, the explanation for which is that once a list already spans the entire space, any new vector adjoined to it must already be in the span, thus forming a linearly dependent list.

This theorem implies an interesting result: in some cases we know without any calculation whether a list of vector is linearly dependent or not.

example 2.23

• The list $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ of length 3 spans $\mathbf{R}^3$.
• This also means that in $\mathbf{R}^3$, no list of length greater than 3 can be linearly independent!
• For example, the list $(1, 2, 3), (4, 5, 8), (9, 6, 7), (−3, 2, 8)$ has a length of four, and therefore cannot be linearly independent!

example 2.24

• The list $(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)$ is of length 4 and is linearly independent.
• Therefore, any list that spans $\mathbf{R}^4$ must have a length greater than 4, meaning no list of fewer than 4 vectors can span $\mathbf{R}^4$.
• For example, the list $(1, 2, 3, -5), (4, 5, 8, 3), (9, 6, 7, -1)$ has a length of 3, and therefore cannot span $\mathbf{R}^4$.

Hand-picked exercises（from Linear Algebra Done Right, Ch2.A）

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 vectors in the list are not scalar multiples of the other.

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

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

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

2.B Bases

Let us look at some examples:

1. The list $(1,0, \ldots, 0),(0,1,0, \ldots, 0), \ldots,(0, \ldots, 0,1)$ is a basis of $\mathbf{F}^n$, called the standard basis of $\mathbf{F}^n$.
2. The list $(1,2),(3,5)$ is a basis of $\mathbf{F}^2$. Note that this list has length two, which is the same as the length of the standard basis of $\mathbf{F}^2$. In the next section, we will see that this is not a coincidence.
3. The list $(1,2,-4),(7,-5,6)$ is linearly independent in $\mathbf{F}^3$ but is not a basis of $F^3$ because it does not span $F^3$.
4. The list $(1,2),(3,5),(4,13)$ spans $\mathbf{F}^2$ but is not a basis of $\mathbf{F}^2$ because it is not linearly independent.
5. The list $(1,1,0),(0,0,1)$ is a basis of $\left\{(x, x, y) \in \mathbf{F}^3: x, y \in \mathbf{F}\right\}$.
6. The list $(1,-1,0),(1,0,-1)$ is a basis of

7. The list $1, z, \ldots, z^m$ is a basis of $\mathcal{P}_m(\mathbf{F})$, called the standard basis of $\mathcal{P}_m(\mathbf{F})$.

Make sure you work through each example in your brain!

This is a bi-directional proof. We first prove the right arrow direction:

$(\implies)$

• Assume $v_1, \ldots, v_n$ is a basis for $V$
• Let $v$ be a vector in $V$
• Because $v_1, \ldots, v_n$ spans $V$ there exist $a_1, \ldots, a_n \in \mathbf{F}$ such that 2.29 holds
• To show that the representation in 2.29 is unique, suppose $c_1, \ldots, c_n$ are scalars such that we also have $v=c_1 v_1+\cdots+c_n v_n$
• Subtracting the last equation from 2.29, we get $0=\left(a_1-c_1\right) v_1+\cdots+\left(a_n-c_n\right) v_n .$
• This implies that each $a_k-c_k$ equals 0 (because $v_1, \ldots, v_n$ is linearly independent).
• Hence $a_1=c_1, \ldots, a_n=c_n$. We have the desired uniqueness, completing the proof in one direction.

Now, the other direction: $(\impliedby)$

• suppose every $v \in V$ can be written uniquely in the form given by 2.29.
• This implies that the list $v_1, \ldots, v_n$ spans $V$.
• To show that $v_1, \ldots, v_n$ is linearly independent, suppose $a_1, \ldots, a_n \in \mathbf{F}$ are such that $0=a_1 v_1+\cdots+a_n v_n$
• The uniqueness of the representation 2.29 (taking $v=0$ ) now implies that $a_1=\cdots=a_n=0$
• Thus $v_1, \ldots, v_n$ is linearly independent and hence is a basis of $V$.

Proof:

We first assume $v_1, \ldots, v_n$ spans $V$. Applying the following algorithm gives a basis.

Step 1:

• If $v_1=0$, then delete $v_1$ from $B$. If $v_1 \neq 0$, then leave $B$ unchanged.

Step $k$：

• If $v_k$ is in $\operatorname{span}\left(v_1, \ldots, v_{k-1}\right)$, then delete $v_k$ from the list $B$.
• If $v_k$ is not in $\operatorname{span}\left(v_1, \ldots, v_{k-1}\right)$, then leave $B$ unchanged.

The process stops after $k$ steps. We denote the remaining list by $B$. $B$ spans $V$ because we only removed vectors that were linear combinations of previous vectors. This algorithm ensures $B$ does not have any "redundant" vector. Hence by 2.19, $B$ is linearly independent, forming a basis for $V$.

• By definition, a finite-dimensional vector space has a spanning list.
• The previous result tells us that each spanning list can be reduced to a basis

Proof:

• Suppose $u_1, \ldots, u_m$ is linearly independent in a finite-dimensional vector space $V$
• Let $w_1, \ldots, w_n$ be a list of vectors in $V$ that spans $V$
• Thus the list $u_1, \ldots, u_m, w_1, \ldots, w_n$ spans $V$.
• Applying the procedure of the proof of 2.30 to reduce this list to a basis of $V$
• produces a basis consisting of the vectors $u_1, \ldots, u_m$ and some of the $w$ 's (none of the $u$ 's get deleted in this procedure because $u_1, \ldots, u_m$ is linearly independent).

As an example in $\mathbf{F}^3$:

• Suppose we start with the linearly independent list $(2,3,4),(9,6,8)$.
• Take $w_1, w_2, w_3$ to be the standard basis of $\mathbf{F}^3$
• Applying the procedure in the proof above produces the list $(2,3,4),(9,6,8),(0,1,0)$ which is a basis of $\mathbf{F}^3$.

Up until now, we have learned the new concept of a "basis". A basis combines the concepts of "linear independence" and "spanning lists" introduced earlier. A basis of $V$

1. spans $V$
2. is linearly independent

Think of it as "the shortest list of vectors that spans a vector space". It has just the right length—not too many to lose linear independence, and not too few to fail to span the space.

Having understood the basis, in the next section, we can define the concept of "dimension"!

2.B Hand-picked exercises（From Linear Algebra Done Right, Ch2.B）

3. (a) Let $U$ be the subspace of $\mathbf{R}^5$ defined by

   $$U=\left\{\left(x_1, x_2, x_3, x_4, x_5\right) \in \mathbf{R}^5: x_1=3 x_2 \text { and } x_3=7 x_4\right\} .$$

   Find a basis of $U$.

   (b) Extend the basis in (a) to a basis of $\mathbf{R}^5$.

   (c) Find a subspace $W$ of $\mathbf{R}^5$ such that $\mathbf{R}^5=U \oplus W$.

6. Prove or give a counterexample: If $p_0, p_1, p_2, p_3$ is a list in $\mathcal{P}_3(\mathbf{F})$ such that none of the polynomials $p_0, p_1, p_2, p_3$ has degree 2 , then $p_0, p_1, p_2, p_3$ is not a basis of $\mathcal{P}_3(\mathbf{F})$.

7. Suppose $v_1, v_2, v_3, v_4$ is a basis of $V$. Prove that $$v_1+v_2, v_2+v_3, v_3+v_4, v_4$$ is also a basis of $V$.

10. Suppose $U$ and $W$ are subspaces of $V$ such that $V=U \oplus W$. Suppose also that $u_1, \ldots, u_m$ is a basis of $U$ and $w_1, \ldots, w_n$ is a basis of $W$. Prove that $$u_1, \ldots, u_m, w_1, \ldots, w_n$$ is a basis of $V$.

2.C Dimension

Now we finally come to define "dimension", which we mentioned in the beginning. Intuitively, the definition should be such that

• $\mathbf{F}^2$ has dimension 2
• $\mathbf{F}^3$ has dimension 3
• $\mathbf{F}^n$ has dimension $n$。

The concept of a basis may lead you to the intuition of

"Maybe we can define dimension as the length of a basis?"

However, a finite-dimensional vector space in general has many different bases. Our attempted definition only makes sense if all bases for a vector space have the same length. Otherwise, we wouldn't have a consistent dimension for a vector space.

Fortunately, the is not an issue. All bases do have the same length, as we now show.

Recall 2.22:

In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

We now show every basis has the same length using 2.22:

• Suppose $V$ is finite-dimensional. Let $B_1$ and $B_2$ be two bases of $V$.
• Then $B_1$ is linearly independent in $V$ and $B_2$ spans $V$, so the length of 𝐵1 is at most the length of $B_2$ (by 2.22).
• Interchanging the roles of $B_1$ and $B_2$, we also see that the length of $B_2$ is at most the length of 𝐵1.
• Thus the length of $B_1$ equals the length of $B_2$, as desired

Now that we know that any two bases of a finite-dimensional vector space have the same length, we are ready to define the dimension of such spaces.

• $\operatorname{dim} \mathbf{F}^n=n$ because the standard basis of $\mathbf{F}^n$ has length $n$.
• $\operatorname{dim} \mathcal{P}_m(\mathbf{F})=m+1$ because the standard basis $1, z, \ldots, z^m$ of $\mathcal{P}_m(\mathbf{F})$ has length $m+1$.
• If $U=\left\{(x, x, y) \in \mathbf{F}^3: x, y \in \mathbf{F}\right\}$, then $\operatorname{dim} U=2$ because $(1,1,0),(0,0,1)$ is a basis of $U$.
• If $U=\left\{(x, y, z) \in \mathbf{F}^3: x+y+z=0\right\}$, then $\operatorname{dim} U=2$ because the list $(1,-1,0),(1,0,-1)$ is a basis of $U$.

By definition, a basis of a vector space $V$ must

1. be linearly independent
2. span $V$

Now we show that if a list of vector has the right length, then one of the properties being true implies the other holds as well!

For example:

• If there are three linearly independent vectors in $\mathbf{R}^3$, then they will definitely span $V$, so they form a basis.
• If there are three vectors that span $\mathbf{R}^3$, then they are definitely linearly independent in $V$, so they form basis.

Now we prove this intuition is true.

Proof

• Suppose $\operatorname{dim} V=n$ and $v_1, \ldots, v_n$ is linearly independent in $V$.
• The list $v_1, \ldots, v_n$ can be extended to a basis of $V$ (by 2.32).
• However, every basis of $V$ has length $n$, so in this case the extension is the trivial one, meaning that no elements are adjoined to $v_1, \ldots, v_n$.
• Thus $v_1, \ldots, v_n$ is a basis of $V$, as desired.

Now we prove that a spanning list of the right length must be linearly independent, and thus is a basis.

Proof

• Suppose $\operatorname{dim} V=n$ and $v_1, \ldots, v_n$ spans $V$.
• The list $v_1, \ldots, v_n$ can be reduced to a basis of $V$ (by 2.30).
• However, every basis of $V$ has length $n$, so in this case the reduction is the trivial one, meaning that no elements are deleted from $v_1, \ldots, v_n$.
• Thus $v_1, \ldots, v_n$ is a basis of $V$, as desired.

The next result gives a formula for the dimension of the sum of two subspaces of a finite-dimensional vector space. The resembles the formula for the number of elements in the union of two sets:

Writing it with formal notations:

We now prove this formula.

Proof:

• Let $v_1, \ldots, v_m$ be a basis of $V_1 \cap V_2$; thus $\operatorname{dim}\left(V_1 \cap V_2\right)=m$.
• Because $v_1, \ldots, v_m$ is a basis of $V_1 \cap V_2$, it is linearly independent in $V_1$.
• Hence this list can be extended to a basis $v_1, \ldots, v_m, u_1, \ldots, u_j$ of $V_1$ (by 2.32). Thus $\operatorname{dim} V_1=m+j$.
• Also extend $v_1, \ldots, v_m$ to a basis $v_1, \ldots, v_m, w_1, \ldots, w_k$ of $V_2$; thus $\operatorname{dim} V_2=m+k$.

We will show that

is a basis of $V_1+V_2$. This will complete the proof, because then we will have

To get started, we know that

• The list 2.44 is contained in $V_1 \cup V_2$ and thus is contained in $V_1+V_2$.
• The span of this list contains $V_1$ and contains $V_2$ and hence is equal to $V_1+V_2$.
• Thus to show that 2.44 is a basis of $V_1+V_2$ we only need to show that it is linearly independent.

To prove that 2.44 is linearly independent, suppose

where all the $a$ 's, $b$ 's, and $c$ 's are scalars.

We need to prove that all the $a$ 's, $b$ 's, and $c$ 's equal 0 . The equation above can be rewritten as

which shows that $c_1 w_1+\cdots+c_k w_k \in V_1$.

All the $w$ 's are in $V_2$, so this implies that $c_1 w_1+\cdots+c_k w_k \in V_1 \cap V_2$.

Because $v_1, \ldots, v_m$ is a basis of $V_1 \cap V_2$, we have

for some scalars $d_1, \ldots, d_m$. Moving all the terms to the same side, we have

But $v_1, \ldots, v_m, w_1, \ldots, w_k$ is linearly independent, so this implies that all the $c$ 's (and $d$ 's) equal 0 . Thus 2.45 becomes the equation

Because the list $v_1, \ldots, v_m, u_1, \ldots, u_j$ is linearly independent, this equation implies that all the $a$ 's and $b$ 's are 0 , completing the proof.

2.C Hand-picked exercises（From Linear Algebra Done Right, Ch2.C）

1. Show that the subspaces of $\mathbf{R}^2$ are precisely $\{0\}$, all lines in $\mathbf{R}^2$ containing the origin, and $\mathbf{R}^2$.

3. (a) Let $U=\left\{p \in \mathcal{P}_4(\mathbf{F}): p(6)=0\right\}$. Find a basis of $U$.

   (b) Extend the basis in (a) to a basis of $\mathcal{P}_4(\mathbf{F})$.

   (c) Find a subspace $W$ of $\mathcal{P}_4(\mathbf{F})$ such that $\mathcal{P}_4(\mathbf{F})=U \oplus W$.

8. Suppose $v_1, \ldots, v_m$ is linearly independent in $V$ and $w \in V$. Prove that

13. Suppose $U$ and $W$ are both five-dimensional subspaces of $\mathbf{R}^9$. Prove that $U \cap W \neq\{0\}$.

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 $V_1, V_2, V_3$ are subspaces of a finite-dimensional vector space, then $$\begin{aligned} \operatorname{dim}\left(V_1+V_2\right. & \left.+V_3\right) \\ = & \operatorname{dim} V_1+\operatorname{dim} V_2+\operatorname{dim} V_3 \\ & -\operatorname{dim}\left(V_1 \cap V_2\right)-\operatorname{dim}\left(V_1 \cap V_3\right)-\operatorname{dim}\left(V_2 \cap V_3\right) \\ & +\operatorname{dim}\left(V_1 \cap V_2 \cap V_3\right) \end{aligned}$$ Then either prove the formula above or give a counterexample.

Summary

We started this chapter with an intuition for dimension and then formally define and prove our way up to its formal definition.

• First, we defined "linear combinations" and "spans"
• Second, we defined "linear dependence"
• Then, we combined both concepts and defined "bases"
• Finally, we proved that all bases have the same length to show our definition makes sense

References

• Axler, S. (2024). Linear Algebra Done Right (4th ed.). Springer.