Showing uniqueness property over a vector space that is the direct sum of two of its subspaces.

Question

Let $V$ be a vector space over a field $F$, and let $S$ and $T$ be subspaces of $V$ such that $V=S⊕T$.

Show that for every $x∈V$ ,there are unique $y_1∈S$ and $z_1∈T$ such that $x=y_1+z_1$. In other words, show that, if $y_2 ∈ S$ and $z_2 ∈ T$ also satisfy $y_1 +z_1 = x = y_2+z_2$, then $y_1 =y_2$ and $z_1 =z_2$

$P(x):=$unique $y∈S$ such that $x=y+z$ for some $z∈T$

Prove that $P$ is a linear map, and also that we have $P^2(= P ◦ P) = P$. Show also that $Range(P) = S$ and $Ker(P) = T$.

Attempt for the first part:

By definition of direct sum $S \cap T$={$0$}. If $V=S⊕T$ then for every $x \in V$, $x=s+t$ for some $s \in S$ and $t\in T$. Let $n \in S$ and $m \in T$.

Now assume $$n+m=s+t$$ Rearranging we have $0=(s-n)+(t-m)$

Then since $S,T$ are subspaces of $V$ and $s,n \in S$ and $t,m \in T$, then $s-n \in S$ and $t-m\in T$ Therefore both $s-n$ and $t-m$ are equal to $0$ which gives us our uniqueness property.

For the second part I'm stuck I know that this can be rearranged to get $P(x):=$ unique $y ̄ ∈ S$ such that $x − y ∈ T$. But I'm unsure how to continue from here. I know I need to prove the below but I'm confused where to begin to do this with this function and how the range and kernel would be S and T respectively.

$P()=P()$

$P(+)=P()+P()$

Answer 1

First issue:

By definition of direct sum $S \cap T$ is the empty set.

No, $S \cap T = \{0\}$. It's not possible for the intersection of subspaces to be empty. This is an important distinction!

Second issue:

Now assume $$n+m=s+t$$ Rearranging we have $0=(s-n)+(t-m)$

Then since $S,T$ are subspaces of $V$ and $s,n \in S$ and $t,m \in T$, then $s-n \in S$ and $t-m\in T$ Therefore both $s-n$ and $t-m$ are equal to $0$ which gives us our uniqueness property.

I have no issue with the statements here, it's more about what isn't here. There is a small (but vitally important) observation skipped here. Specifically, since $t - m \in T$, then so is $$s - n = -(t - m) \in T.$$ Therefore $s - n \in S \cap T = \{0\}$, so $s - n = 0$. You can conclude that $t - m = 0$ too.

This observation is really the heart of the proof, and where you get to use the assumption that $S$ and $T$ sum directly. If I were marking this proof, and I saw the above two errors, I would mark you down. I would assume that if the above steps were something you properly understood (even if you didn't write them down), then you wouldn't say that $S \cap T = \emptyset$.

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For the second part, the question is tricky because the map $P$ is not defined explicitly, but implicitly. That is, there's no formula to take $x$ to $Px$. Instead, $Px$ satisfies an equation with a unique solution.

Let's start by taking arbitrary vectors $x_1, x_2 \in V$. Since $V = S \oplus T$, there exist unique $s_1, s_2, s_3 \in S$ and $t_1, t_2, t_3 \in T$ such that \begin{align*} x_1 &= s_1 + t_1 \\ x_2 &= s_2 + t_2. \end{align*} Note that $P(x_1) = s_1$ and $P(x_2) = s_2$.

In order to prove that $P(x_1 + x_2) = P(x_1) + P(x_2) = s_1 + s_2$, we need to find a decomposition of $x_1 + x_2$ into a component in $S$ (which hopefully will be $s_1 + s_2$ and a component in $T$. We can simply get this by adding the above equations: $$x_1 + x_2 = (s_1 + s_2) + (t_1 + t_2).$$ By the first part, this decomposition is unique, and so $P(x_1 + x_2)$ is well-defined to be $s_1 + s_2$.

Now try with scalar multiplication!

Answer 2

If $x = a+b, y=c+d$ are the unique decompositions for x and y respectively, then

1. $P(x+y) = P((a+c)+(b+d)) = a+c = P(x) + P(y).$
2. $P(\alpha x) = P(\alpha a+ \alpha b) = \alpha a.$
3. $P(P(a+b)) = P(a) = a.$
4. $S \subseteq Range(P)$ since $a = a+0$ in $V$.
5. $P(a+b) = 0 \Leftrightarrow$ $a=0$. Therefore the kernel is T.

Answer 3

Note that with

$V = S \oplus T \tag 1$

we have

$S \cap T = \{0\}, \tag 2$

and not

$S \cap T = \emptyset. \tag 3$

This distinction is of course consistent with the assertion that the decomposition of any $v \in V$ as

$v = s + t, \; s \in S, t \in T, \tag 4$

is unique, since if

$v = s_1 + t_1 = s_2 + t_2; \; s_1, s_2 \in S; t_1, t_2 \in T, \tag5$

then

$S \ni s_1 - s_2 = t_2 - t_1 \in T, \tag 6$

whence

$s_1 - s_2 = 0 = t_2 - t_1, \tag 7$

or

$s_1 = s_2, t_1 = t_2, \tag 8$

that is, the decompsition is unique. This demonstration is, with minor variations, the same as that given by our OP Mathisoshardlmao in the body of the question itself.

Now for $v \in V$ with

$v = s + t \tag 9$

uniquely, we define

$P:V \to S \tag{10}$

via

$P(v) = s; \tag{11}$

we then have, from (9),

$\alpha v = \alpha(s + t) = \alpha s + \alpha t, \tag{12}$

whence, since

$\alpha s \in S, \; \alpha t \in T, \tag{13}$

we see, again using uniqueness,

$P(\alpha v) = \alpha s = \alpha P(v); \tag{14}$

also, if

$v_1 = s_1 + t_1, \tag{15}$

and

$v_2 = s_2 + t_2 \tag{16}$

are the decompositions of $v_1$ and $v_2$ in accord with (1), then

$P(v_1 + v_2) = P((s_1 + t_1) + (s_2 + t_2))$ $= P((s_1 + s_2) + (t_1 + t_2)) = s_1 + s_2 = P(v_1) + P(v_2). \tag{17}$

(9)-(17) show that $P$ is a linear map.

Now, yet again in accord with (9),

$P^2(v) = P^2(s + t) = P(P(s + t)) = P(s) = s = P(v), \tag{18}$

which shows

$P^2 = P. \tag{19}$

Since with

$s \in S, \tag{20}$

$P(s) = s, \tag{21}$

it follows that

$\text{Range}(P) = S; \tag{22}$

with

$t \in T, \tag{23}$

$P(t) = 0, \tag{24}$

whence

$T \subset \ker P; \tag{25}$

and if

$t \in \ker P, \tag{26}$

$P(t) = 0, \tag{27}$

and if

$t = s + t \tag{28}$

is the unique decomposition of $t$, then

$P(t) = s = 0, \tag{29}$

so

$t \in T, \tag{30}$

and so

$\ker P \subset T; \tag{31}$

together (25) and (30) yield

$\ker P = T, \tag{32}$

and we are done.