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We're given: $$ V = Span\left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \ \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \ \begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix} \right\} \ \ \ \mathrm{and} \ \ \ W = Span \left\{\begin{bmatrix}1\\-3\\2\\1 \end{bmatrix} \right\} $$

and we're asked to find $\mathrm{dim}(V\cap W^{\bot})$

Here's my approach. First, by inspecting the basis of $W$, I managed to construct a basis for $W^{\bot}$, which is the following: $$W^{\bot} = Span \left\{\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right\}$$ Then, for each vector $\textbf{w}$ in the basis of $W^{\bot}$, I tried to see if the system $A\textbf{x}=\textbf{w}$ was compatible or not. In this case, I found out that the system was only compatible with two of the vectors in $W^{\bot}$, thus indicating me that $\mathrm{dim}(V \cap W^{\bot} ) = 2$ (which is correct).

What I do not get, however, is that since the dimension of the intersection is $2$, why aren't two of the vectors in the basis of $V$ orthogonal to the vector which spans $W$ (that was my initial approach, i.e try to see which vectors of the basis of $V$ are orthogonal to the vector that spans $W$).

Also, I was wondering if there was a simpler/quicker way of doing this.

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2 Answers 2

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Imagine an anologous situation in three dimensions, which is easier to visualize: Let $V$ be a plane through the origin and $W$ a nonzero vector that’s not normal to the plane. The orthogonal complement of $W$ is another plane that intersects $V$ in a line that’s orthogonal to $W$. However, by construction there are elements of $V$ that aren’t orthogonal to $W$, and we can find a pair of them that span $V$.

As to a simpler method of solving this, first verify that the generating vectors of $V$ are linearly independent (they are). You’re looking for elements of $V$ that are orthogonal to $W$, which can be expressed by the equation $$[1,-3,2,1](a[2,2,2,1]^T+b[2,1,1,0]^T+c[5,4,1,1]^T)=0.$$ Expanding and simplifying this equation produces $a+b-4c=0$. This has two free variables, so the intersection is two-dimensional.

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  • $\begingroup$ I got it. Thanks $\endgroup$
    – Skyris
    Jan 19, 2018 at 0:46
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At the first find the $\mathrm{dim}(V\cup W^{\bot})$.

And we know that $\mathrm{dim}(V\cup W^{\bot})=\mathrm{dim}(V)+\mathrm{dim}(W^{\bot})-\mathrm{dim}(V\cap W^{\bot})$

Finding the dimension of $(V\cup W^{\bot})$ is also an easy work.

For above one form a matrix $A$ like following one:

$$\begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix}\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}\begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$

Now find the rank of this matrix via using Gaussian Elimination process.

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  • $\begingroup$ I've tried this (i.e using the inclusion-exclusion principle), but I do not know/understand how I could find dim($V\cap W^{\bot}$). $\endgroup$
    – Skyris
    Jan 17, 2018 at 15:43
  • $\begingroup$ @Skyris I have edited the solution. Does it make sense? $\endgroup$
    – GhD
    Jan 17, 2018 at 21:42
  • $\begingroup$ Yes, thank you very much $\endgroup$
    – Skyris
    Jan 20, 2018 at 15:24
  • $\begingroup$ $V\cup W^\perp$ is usually not a subspace, so what do you mean by its dimension? $\endgroup$ Jun 2, 2021 at 7:35

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