# Proof that $\dim(U \times V) = \dim U + \dim V$.

The following theorem in Serge Lang's Linear Algebra is left as an exercise, namely,

Let $U$ and $V$ be finite dimensional vector spaces over a field $K$, where $\dim U = n$ and $\dim V = m$. Then $\dim W = \dim U + \dim V$, where $W = U \times V$, the direct product of the two vector spaces $U$ and $V$. Namely, $W$ contains the set of all ordered pairs $(u,w)$ such that $u \in U$ and $v \in V$. The usual axioms for such a direct product are:

1) Addition is defined component wise, namely if $(u_1,v_1),(u_2,v_2) \in W$, then $(u_1,v_1)+(u_2,v_2) = (u_1 + u_2, v_1 + v_2)$;

2)If $c \in K$, then $c(u_1,w_1) = (cu_1,cw_1)$.

To prove it, let $(u_1, u_2 \ldots u_n)$ be a basis for $U$ and $(v_1,v_2 , \ldots v_m)$ a basis for $V$.

So by definition, every element of $W$ can be written in the form

$(a_1u_1 + \ldots a_nu_n, b_1v_1 + \ldots b_mv_m)$, where the $a_i's$ and $b_j's$ belong to the field $K$.

Using the above axioms this can be rewritten as:

$a_1(u_1,0) + a_2(u_2,0) + \ldots a_n(u_n,0) + b_1(0,v_1) + \ldots b_m(0,v_m)$.

Doubt: If we view all the $(u_i,0)$ ordered pairs as being the "basis" vectors of $U$ and similarly for the $(0,v_j)$ ordered pairs of $V$, then there are $n+m$ number of them and so proving the linear independence of these objects should suffice. But I'm confused because I know that $u_i's$ by themselves are the basis vectors of $U$, but now we are talking about ordered pairs $(u_i,0)$. How can I get out of such a situation?

Perhaps one can define some linear map between say a $u_i$ and the ordered pair $(u_i,0)$.

$\textbf{Edit}:$ First it is easy to see that $(U \times \{0\}) \cap (\{0\} \times V)$ is the ordered pair $(0,0)$. The linear independence of the basis vectors as stated above then follows.

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If it confuses you to "view all the ($u_i$,0) ordered pairs as being the basis vectors of $U$", then don't. The $\{u_i\}$ are a basis for $U$. Use them as you've outlined with a basis for $V$ to get a basis for $U \times V$. It's pretty clear the combined set spans the direct product. Focus on showing linear independence. –  hardmath Jul 9 '11 at 13:49
@hardmath You're saying that the last line of mathematics I wrote spans the direct product, but you see I learned that the dimension of a vector space is the number of linearly independent vectors that span the space, so I need to show that each of the "vectors" $(u_i,0)$ and $(0,v_j)$ are linearly independent (If it means anything at all for these things to be linearly independent). –  user38268 Jul 9 '11 at 14:24

In mathematics, certain unequal objects may be identified. For example, the set of real numbers $\mathbb{R}$ can be identified with the subset $\mathbb{R}\times \{0\}=\{(x,0):x\in\mathbb{R}\}$ of the plane $\mathbb{R}^2$. Similarly, the set of real numbers $\mathbb{R}$ can be identified with the subset $\{0\}\times \mathbb{R}=\{(0,y):y\in\mathbb{R}\}$ of the plane $\mathbb{R}^2$.

Let me explain in more detail what this identification actually means. If you visualize the plane $\mathbb{R}^2$, then you may visualize the set of real numbers $\mathbb{R}$ as being the $x$-axis in the plane. However, this is not logically correct since the $x$-axis in the plane is more properly the set $\{(x,0):x\in\mathbb{R}\}$ and the set of real numbers is the set $\mathbb{R}$. Note that these two sets are not equal but can be identified. The idenfitication is in this case, as it usually is, a bijection between the sets in question. The most natural bijection is the map $f:\mathbb{R}\to \{(x,0):x\in\mathbb{R}\}$ defined by the rule $f(t)=(t,0)$.

Similarly, we can naturally identify the set $\mathbb{R}$ with the set $\{(0,y):y\in\mathbb{R}\}$ by the bijection $g:\mathbb{R}\to \{(0,y):y\in\mathbb{R}\}$ defined by the rule $g(t)=(0,t)$.

In fact, these identifications are relevant to your question. In essence, an element of the plane $\mathbb{R}^2$ (or a vector, if you prefer) is an ordered pair $(x,y)$ where $x,y\in\mathbb{R}$. We can write (using vector addition, if you prefer) $(x,y)=(x,0)+(0,y)$. Note that this sum is not a sum of real numbers since neither $(x,0)$ nor $(0,y)$ is a real number; rather, they are ordered pairs whose coordinates are real numbers. Nonetheless, we can identify them with real numbers using the bijections $f$ and $g$ above. More precisely, $f(x)=(x,0)$ and $g(y)=(0,y)$.

Let us now return to your question. An element of $W=U\times V$ is an ordered pair $(u,v)$ where $u\in U$ and $v\in V$. We can write $(u,v)=(u,0)+(0,v)$. However, this sum is not a sum of a vector in $U$ and a vector in $V$. Nonetheless, for most purposes this is not necessary since we have identifications $f:U\to U\times \{0\}=\{(u,0):u\in U\}$ and $g:V\to \{0\}\times V=\{(0,v):v\in V\}$ defined by the rules $f(a)=(a,0)$ and $f(b)=(0,b)$ for $a\in U$ and $b\in V$. In fact, as you correctly observe, these identification maps are actually linear transformations.

The following exercises might be useful (and are relevant to the result you have noted):

Exercise 1: Let $A,B$ be vector spaces (over a field $\mathbb{F}$) and let $T:A\to B$ be an injective linear transformation (i.e., the kernel (or the null space, if you prefer) of $T$ is ${0}$). If $(v_1,\dots,v_n)$ is a linearly independent tuple of vectors in $A$, prove that $(T(v_1),\dots,T(v_n))$ is a linearly independent tuple of vectors in $B$.

Exercise 2: Let $W=U\times V$ where $U$ and $V$ are vector spaces. Let us recall the identification maps $f:U\to W$ and $g:V\to W$ defined above by the rules $f(u)=(u,0)$ and $g(v)=(0,v)$. Prove that $f$ and $g$ are injective linear transformations.

Exercise 3: If $(u_1,\dots,u_n)$ and $(v_1,\dots,v_m)$ are bases of $U$ and $V$, respectively, prove that $((u_1,0),\dots,(u_n,0))$ and $((0,v_1),\dots,(0,v_m))$ are linearly independent tuples in $W=U\times V$. (Hint: use Exercise 1 and Exercise 2.)

Exercise 4: Prove that if $f(u)+g(v)=0$ for $u\in U$ and $v\in V$, then $f(u)=g(v)=0$. (Of course, $0$ is the zero vector of $W$ in this case.)

Exercise 5: Finally, prove that the tuple $((u_1,0),\dots,(u_n,0),(0,v_1),\dots,(0,v_m))$ is linearly independent in $W=U\times V$ if $(u_1,\dots,u_n)$ and $(v_1,\dots,v_m)$ are bases of $U$ and $V$, respectively. (Hint: use Exercise 3 and Exercise 4.)

Warning: Solutions Below

Solution to Exercise 1: Let $(v_1,\dots,v_n)$ be a linearly independent tuple of vectors in $A$. We wish to prove that the tuple $(Tv_1,\dots,Tv_n)$ is linearly independent in $B$. Let us assume $c_1Tv_1+\cdots+c_nTv_n=0$ for some scalars $c_i\in \mathbb{F}$ ($1\leq i\leq n$); we wish to prove that $c_i=0$ for all $1\leq i\leq n$. We can use the linearity of $T$ and write $T(c_1v_1+\cdots+c_nv_n)=0$. Since $T$ is injective and since $T0=0$, it follows that $c_1v_1+\cdots+c_nv_n=0$. However, the tuple $(v_1,\dots,v_n)$ is linearly independent in $A$ by assumption. Therefore, $c_i=0$ for all $1\leq i\leq n$ and the proof is complete since the scalars $c_i\in\mathbb{F}$ ($1\leq i\leq n$) were arbitrary.

Solution to Exercise 2: If $f(u_1)=f(u_2)$ for $u_1,u_2\in U$, then $(u_1,0)=(u_2,0)$; by the definition of equality in $W=U\times V$, it follows that $u_1=u_2$. Therefore, $f$ is injective; the proof that $g$ is injective is similar.

I hope this helps!

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I was just about to ask if one identifies the map that you have said then we have to prove that the objects you identified above in exercise 3 are linearly independent. –  user38268 Jul 9 '11 at 14:46
I'll type out the proofs of these exercises in an answer. –  user38268 Jul 9 '11 at 14:49
@D Lim: Exercise 1 contains most of the work (but the solution is not difficult). The rest of the exercises mostly require working carefully with the notation but it is worth doing them all since they provide an answer to your question. –  Amitesh Datta Jul 9 '11 at 14:55
I really like your exercise-based approach to answering questions –  ItsNotObvious Jul 12 '11 at 12:12
@3Sphere Thank you very much! It is nice to hear that my exercises are useful. –  Amitesh Datta Jul 12 '11 at 12:29

All you do seems to be ok with me, so I'm not sure to understand your question. But maybe, this can be an answer: the pairs $(u_i, 0)$, $i= 1, \dots , n$, are NOT a basis of $U$, but of the "copy" of $U$ inside $U\times V$. Namely, $U \times \left\{ 0 \right\}$. And the linear map you're talking about -but you needn't it- is the inclusion $U \hookrightarrow U \times \left\{ 0 \right\}$, which is $u \mapsto (u,0)$.

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Roig So these pairs are the "copy" of $U$ inside the direct product, so how does that prove the theorem? –  user38268 Jul 9 '11 at 14:25
You still need to prove linear independence of those $(u_i,0)$ and $(0,v_i)$, as hardmath pointed you out. It's easy and you'll be done. (Easier than begin to think why being I-don't-know-what a copy of $U$ inside $U\times V$ might prove your result.) –  a.r. Jul 9 '11 at 19:02

Proof of exercise 1: Say that the $v_i$ are linearly independent. Then

$c_1v_1 + c_2v_2 + \ldots c_nv_n = 0$ only when all the $c_i = 0$. Applying $T$ to both sides we have

$c_1T(v_1) + c_2T(v_2) + \ldots c_nT(v_n) = 0$ by linearity of $T$ and the fact that it is injective.

But the assumption was that all the $c_i$ were zero in order for the first equation to hold.

It follows that $(T(v_1), T(v_2), \ldots T(v_n))$ are a linearly independent set of vectors.

As for exercise 2, is it not clear from what it means for two ordered pairs to be equal that $f$ and $g$ are injective?

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I think one logical implication in your solution of Exercise 1 is incorrect. More precisely, if $c_1v_1+\cdots+c_nv_n=0$, then we can always conclude that $c_1Tv_1+\cdots+c_nTv_n=0$ for any linear transformation $T$; the injectivity of $T$ is not necessary at this stage. However, once we have an equation $T(c_1v_1+\cdots+c_nv_n)=0$, then we can use the injectivity of $T$ to conclude that $c_1v_1+\cdots+c_nv_n=0$ and hence that $c_i=0$ for all $1\leq i\leq n$ (the tuple $(v_1,\dots,v_n)$ is linearly independent). I have written out a solution to Exercise 1 in my answer above. –  Amitesh Datta Jul 10 '11 at 1:24
Yes, Exercise 2 is clear. –  Amitesh Datta Jul 10 '11 at 1:26
I have added a solution to Exercise 1 at the very bottom of my answer above. (The idea of your solution is correct, of course, but I think it could have been worded slightly better.) –  Amitesh Datta Jul 10 '11 at 1:32
Exercise 3 and Exercise 4 are also easy (but you might wish to do them to make sure you understand them). However, Exercise 5 requires you to combine Exercise 3 and Exercise 4 in its solution. You will meet some of the ideas contained in these exercises when you study direct sums in linear algebra. –  Amitesh Datta Jul 10 '11 at 1:36
I have added, for the sake of completeness, a solution to Exercise 2 at the very bottom of my answer above. –  Amitesh Datta Jul 10 '11 at 1:56