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

  • $\begingroup$ 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

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 $G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.