031 Review Part 3, Problem 4

Let  $W={\text{Span }}{\Bigg \{}{\begin{bmatrix}2\\0\\-1\\0\end{bmatrix}},{\begin{bmatrix}-3\\1\\0\\0\end{bmatrix}}{\Bigg \}}.$  Is  ${\begin{bmatrix}2\\6\\4\\0\end{bmatrix}}$  in  $W^{\perp }?$  Explain.

Foundations:
Recall that if  $W$  is a subspace of  $\mathbb {R} ^{n},$  then
$W^{\perp }=\{{\vec {v}}\in \mathbb {R} ^{n}~:~{\vec {v}}\cdot {\vec {w}}=0{\text{ for all }}w\in W\}.$

Solution:

Step 1:
To determine whether the vector
${\begin{bmatrix}2\\6\\4\\0\end{bmatrix}}$
is in  $W^{\perp },$  it suffices to see if this vector is orthogonal to
the basis elements of  $W.$
Notice that we have
${\begin{array}{rcl}\displaystyle {{\begin{bmatrix}2\\6\\4\\0\end{bmatrix}}\cdot {\begin{bmatrix}2\\0\\-1\\0\end{bmatrix}}}&=&\displaystyle {2(2)+6(0)+4(-1)+0(0)}\\&&\\&=&\displaystyle {0.}\end{array}}$

Step 2:
Additionally, we have
${\begin{array}{rcl}\displaystyle {{\begin{bmatrix}2\\6\\4\\0\end{bmatrix}}\cdot {\begin{bmatrix}-3\\1\\0\\0\end{bmatrix}}}&=&\displaystyle {2(-3)+6(1)+4(0)+0(0)}\\&&\\&=&\displaystyle {0.}\end{array}}$
Hence, we conclude
${\begin{bmatrix}2\\6\\4\\0\end{bmatrix}}\in W^{\perp }.$

Final Answer:
${\begin{bmatrix}2\\6\\4\\0\end{bmatrix}}\in W^{\perp }$

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