I have $X\in \mathbb{R^a}$, and $L\in\mathbb{R}^{b-a}$ which is a linear subspace of $\mathbb{R}^b$ , where $0<a\leq b$. Now let $Y=\{y_1,...,y_b\}$ be a $a\times b$ matrix, where $y_1,...,y_b$ is a basis for $L^{\perp}$. Define $Z=Y^T X$.
Show that $Z=0 \Rightarrow X\in L$ .
We have $Z=0 \Rightarrow Y^TX=0$. Now since the elements in $Y$ is a basis for $L^{\perp}$, can i then use this conclude that $X\in L$?
Any help is appreciated.
Hint: Note that $$ Y^Tx = \pmatrix{y_1 & \cdots & y_b}^Tx = \pmatrix{y_1^Tx\\ \vdots \\ y_b^Tx} $$ Now, if all of these entries are $0$, how do we conclude that $x \in L$?