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Let $y \in \mathbb{R}^m$, where $m \ge 2$ so that $||y||=1.$ Here $||.||$ is the Euclidean norm. Define the subspace $$G = \{x \in \mathbb{R}^m: x^Ty =0 \}.$$ For $1 \le i \neq j \le m$, define $w^{(i,j)} \in \mathbb{R}^m$ as follows: $$ w_k^{(i,j)} = \begin{cases} -y_j , \quad k =i\\ y_i, \quad k=j \\ 0, \quad \text{otherwise} \\ \end{cases}$$

Question: Show that $G= H$ where $H$= $ span \{w^{(i,j)}, 1 \le i \neq j \le m\}$. Note that $x_j$ denotes the jth component of a vector $x$

I started with the case in which $m=2$. Observe that if $m =2$, then $$H = span \{w^{(i,j)}, 1 \le i \neq j \le 2 \} = span \{w^{(1,2)}, w^{(2,1)}\}$$ where $$w^{(1,2)} = \begin{pmatrix} w_1^{(1,2)}\\ w_2^{(1,2)}\end{pmatrix} = \begin{pmatrix} -y_2\\ y_1 \end{pmatrix} $$ and $$w^{(2,1)} = \begin{pmatrix} w_1^{(2,1)}\\ w_2^{(2,1)}\end{pmatrix} = \begin{pmatrix} y_2\\ -y_1 \end{pmatrix}. $$

If we take an element $v \in H$, we have that $k$ can be written as $k= aw^{(1,2)}+ bw^{(2,1)}$ such that $a,b \in \mathbb{R}$.

In particular, let $v=w^{(1,2)}$, here $a=1$ and $b=0.$ We see that $$v^Tv^* =0,$$ where $$v^* = \begin{pmatrix} y_1\\ y_2 \end{pmatrix}$$ so $v \in G$. Similarly, we have that $w^{(2,1)} \in G$.

Now in general, let $k= aw^{(1,2)}+ bw^{(2,1)}$ such that $a,b \in \mathbb{R}$. Thus $$k = \begin{pmatrix} -ay_2 +by_2\\ ay_1 -by_1\end{pmatrix}$$ so $k^Tk* =0$ where $$k^* = \begin{pmatrix} (a -b)y_1\\ (a-b)y_2 \end{pmatrix}.$$ Thus $k \in G.$ Therefore, for $m=2$, $H$ is contained in $G$.

Conversely, let $x \in G$. We can write $$x = \begin{pmatrix} x_1\\ x_2\end{pmatrix}.$$ Since $x \in G$, we have that there is a $y= \begin{pmatrix} y_1\\ y_2\end{pmatrix} \in \mathbb{R}^2$ such that $$x^Ty= 0.$$ We have that $$x_1y_1 +x_2y_2 =0.$$ Since our goal is to show that $ x = \begin{pmatrix} x_1\\ x_2\end{pmatrix}$ belongs to the spanning set, it suffices to show that $x$ can be written as the linear combination of elements in $H$. Let $$\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = x = a w^{(1,2)} + b w^{(2,1)} = a\begin{pmatrix} -y_2\\ y_1 \end{pmatrix} + b \begin{pmatrix} y_2\\ -y_1 \end{pmatrix} .$$ After some computations, I found out that $x_1 = (b-a)y_1$ and $x_2 = (a-b)y_1$ but I got stuck here.

I would really appreciate it anyone can please show me the proof to the general case that I asked above. Thanks

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  • $\begingroup$ In the case $n=2$, $w^{(1,2)}$ gives a basis for $G$ and $w^{(2,1)} = -w^{(1,2)}$, so there's nothing further to do. That is, $w^{(1,2)}$ already spans $G$. Indeed, you should be taking the span of $w^{(i,j)}$ for $i<j$, in general, to remove this redundancy. $\endgroup$ – Ted Shifrin Sep 28 '17 at 18:22
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In general, $G$ is an $(m-1)$-dimensional subspace, and so all we need to do is find $m-1$ linearly independent vectors among your $w^{(i,j)}$ and they will be guaranteed to span $G$.

Since $y$ is a unit vector, some entry, say $y_i$, must be nonzero. Then the $m-1$ vectors $w^{(i,j)}$ for $j\ne i$ are linearly independent. With the vectors \begin{align*} w^{(i,1)} &= (y_i,0,\dots, \overbrace{-y_1}^{i\text{th slot}},0,\dots,0) \\ w^{(i,2)} &= (0, y_i, \dots, -y_2,0,\dots,0) \\ &\vdots \\ w^{(i,m)} &= (0,0,\dots, -y_m,0,\dots,y_i) , \end{align*} there is a pivot in every column except the $i$th. Thus, it is immediate that these vectors are linearly independent and therefore span $G$.

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