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

Question

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.

Answer 1

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!

Answer 2

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 $u \mapsto (u,0)$.

Answer 3

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?