0
$\begingroup$

Show that if $S$ is a subspace of $\Bbb R^n$ and $\bf x$ is any vector in $\Bbb R^n$, then $\bf x = \bf u + \bf v$ for some vector $\bf u$ in $S$ and some vector $\bf v$ in $S^\bot$.

I was given a hint saying projection but I am not sure where to go from there.

  • 1
    Do you know how to project onto subspaces? I think that hint already says it all... Are you aware of what the notation $\text{proj}_S(x)$ means? In what space does the element $\text{proj}_S(x)$ reside? What might you use $\text{proj}_S(x)$ for in this context? What is the remaining vector you need to find? What must it be equal to in order for $x=u+v$ to be true? Will it be an element of the space you needed it to be? – 2017-01-26
  • 0
    Take an orthonormal basis for $S$, extend it to an orthonormal basis of $\mathbb{R}^n$ and consider the representation of $\mathbf{x}$ in this basis. The part which has basis vectors from $S$ will be $\mathbf{u}$ and the rest will be your $\mathbf{v}$. – 2017-01-26
  • 0
    You might be missing the setup that defines "orthogonality" in $\mathbb{R}^n$ as an inner product space. – 2017-01-27

2 Answers 2

1

Let $S$ be a subspace, then $S=\text{span}(s_1,s_2, ...,s_k)$ for some $s_j \in \mathbb{R}^n$. For $x \in \mathbb{R}^n$ consider the projection $u = x_{pr} \in S$. Then $\langle x_{pr},x-x_{pr}\rangle=0$. Hence $v = x-x_{pr} \in S^\bot$ and $u+v=x_{pr}+x-x_{pr}=x$.

0

$\DeclareMathOperator{\pr}{proj}\pr_S(x)$ denotes the projection onto the subspace $S$, so it is defined in the following way: let $(e_1,\dots e_r)$ be an orthogonal basis of the subspace $S$. Then $$\pr_S(x)=\frac{\langle x,e_1\rangle}{\langle e_1,e_1\rangle}\,e_1+\dots+\frac{\langle x,e_r\rangle}{\langle e_r,e_r\rangle}\,e_r.$$