# Find m satisfied $\dim(U^\perp) = 2$

Given $U = \operatorname{span} \{(1; -1; 2; 0), (3, 0, 1, m), (1, 2, -3, 0)\}$ If $\dim(U^\perp) = 2\}$

Find $m$.

I don't sure how to solve this problem. Here is my idea:

Because $\dim(U^\perp) + \dim(U) = n$, if $U$ is a subspace of $\mathbb R^n$ And I guess $U$ is subspace of $\mathbb R^4$, so, $\dim(U)$ will be equal to $2$. After that, so after transform U to reduce echelon matrix, will contain exactly $2$ leading one.

I don't sure this way is true or wrong.

Thanks :)

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If you write "dim U" in LaTeX, you get $dim U$, but if you write \dim U, you get $\dim U$. Not only is $\dim$ not italicized but also there's proper spacing before and after it. This is standard usage in LaTeX. I also did a few other TeX edits. – Michael Hardy Nov 6 '12 at 20:09
You know that $\dim(U)=2$, so those three vectors are linearly dependent. The rest is simple. – wj32 Nov 6 '12 at 20:13

Your approach is good. Your set already contains two linearly independent vectors. If $m$ was chosen so that the third vector is also linearly independent then the complement will be of dimension only $1$. You need to choose $m$ so that the vector lies in the span of the other two.