Direct sum of two subspace

Question

Given the definition on textbook: Let $V$ be a subspace of $\mathbb R^n.$ Every vector $u \in \mathbb R^n$ can be written uniquely as $u = n + p.$ I still don't understand what it means because i am ask to do with a question that ask me the prove the uniqueness claim in the definition.

The question is asking:

Let $v\in \mathbb R^n$. Suppose $p_1, p_2 \in V$ and $n_1,n_2 \in V^\bot$ are such that $v=p_1+n_1 = p_2 + n_2.$ Prove that $p_1 = p_2$ and $n_1 = n_2$.

Anyone can help me answer to this question?

Answer 1

Let's look at an easy example. Let $n=2$ and $V= <\begin{pmatrix} 1\\0 \end{pmatrix}>$. Then $V^{\perp}=<\begin{pmatrix} 0\\1 \end{pmatrix}>$. Now pick some $x\in\mathbb{R}^2$, say, $x=\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$. Then $x=n+p$ where $n=x_1\begin{pmatrix} 1\\0 \end{pmatrix}\in V$ and $p=x_2\begin{pmatrix} 0\\1 \end{pmatrix}\in V^{\perp}$. The theorem tells you that these are your only choices of $n$ and $p$ such that $x=n+p$.

Now to the proof: Let $v=n_1+p_1$ and $v=n_2+p_2$. where $n_i\in V$ and $p_i\in V^{\perp}$ for $i=1,2$. Then $$0=(p_1-p_2)+(n_1-n_2)$$. Since $V$ and $V^{\perp}$ are subspaces of $\mathbb{R}^n$, we have that $p_1-p_2\in V$ and $n_1-n_2\in V^{\perp}$. By definition of $V$ and $V^{\perp}$, we get that $p_1-p_2 \perp n_1-n_2$. But if two vectors are orthogonal, they must be linearly independent. Hence we have that 0 is written as a linear combination of two linearly independent vectors, hence these two vectors must be zero, yielding $p_1-p_2=0$ and $n_1-n_2=0$ which proves that they were the same.

Answer 2

Let $V, U$ be subspaces of $\mathbb{R}^n$ such that their direct sum equals $\mathbb{R}^n$ (I use $U$ since I don't know how to write the orthogonality symbol). By definition of the direct sum, $U \cap V = \{0\}$. Suppose that $$p_1 + n_1 = p_2 + n_2$$ then we obtain that $$p_1 - p_2 = n_2 - n_1$$ the left hand side is an element of $V$ by definition of subspace. However, it equals an element in $U$, so $p_1 - p_2 \in U \cap V$. The intersection only contains $0$, so $p_1 - p_2 = 0$. Can you do the same for $n_2 - n_1$?

Hence if you assume that in a direct sum, you can express a vector $v$ as what appear to be two different sums, then those sum are actually the same. Hence each vector has a unique expression as a sum of elements in $V$ and its complementary space.

Answer 3

Since $V\cap V^{\perp}=0$, we can conclude from $(p_1-p_2)=-(n_1-n_2)$ that $p_1-p_2\in V\cap V^{\perp}=0$.