Find $U^{\perp}$ based on $U$.

Question

Let $U =\text{Span} \Big\{(3,4,0,0),(3,4,-3,0),(0,0,1,0)\Big\}$. Find $U^{\perp}.$

we know that: $U^{\perp}=\{(x,y,z,t)\in\mathbb{R}^4|\langle(x,y,z,t),u\rangle=0\forall u\in U\}$. so if we take the scalar product for every $u$, we get: $$3x+4y=0\implies y=-\frac 34x$$ $$z=0$$ $$t=t$$ Let $t=\beta$ and $x=\alpha$ so then: $$(x,y,z,t)=(\alpha,-\frac 34\alpha,0,\beta)=\alpha(1,-\frac 34,0,0)+\beta(0,0,0,1)$$

so my $U^{\perp}=Sp\{(1,-\frac 34,0,0),(0,0,0,1)\}$ but shouldn't $\dim_{\mathbb{R}}U+\dim_{\mathbb{R}}U^{\perp}=\dim_{\mathbb{R}}\mathbb{R}^4=4\implies \dim_{\mathbb{R}}U^{\perp} = 1$ but my $\dim_{\mathbb{R}}U^{\perp}=2$ so... what am I doing wrong?

Answer 1

Is it true that $\dim U=3$? I don't think so...

Answer 2

Since $$(3,4,-3,0) = (3,4,0,0)-3(0,0,1,0)$$ we see that $$ U=\text{Span} \Big\{(3,4,0,0),(0,0,1,0)\Big\}$$ Clearly $(3,4,0,0)$ and $(0,0,1,0)$ are linearly independent so $\dim_{\mathbb{R}} U = 2$ and thus $\dim_{\mathbb{R}}U^{\perp} = 2$.