Let w be the set of all vectors ⎡⎣⎢xyx+y⎤⎦⎥ with x and y real. find a basis of w⊥.

Mathematics

2 answers:

andrew11 [14]3 years ago

5 0

[-1,-1,1]
or
[-2,-2,2]
.....
Dot Product = 0

Daniel [21]3 years ago

3 0

The basis of ${W^ \bot }$ is $\boxed{{\text{span}}\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\1 \end{array}} \right]} \right\}}$

Further explanation:

Given:

The vector is,

$\left[ {\begin{array}{*{20}{c}}x\\y\\{x + y} \end{array}} \right]$

Explanation:

Consider the set of all vectors can be expressed as follows,

$\left[{\begin{array}{*{20}{c}}x\\y\\ {x + y} \end{array}} \right] = x\left[ {\begin{array}{*{20}{c}}1\\0\\1\end{array}} \right] + y\left[ {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right]$

The spanned vectors of $W$ are $\left[ {\begin{array}{*{20}{c}}1 \\ 0\\1 \end{array}} \right]{\text{ and }}\left[ {\begin{array}{*{20}{c}}0 \\1 \\1 \end{array}} \right]$

Consider a vector $\left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right]$ as ${W^ \bot }.$

The dot product of $W$ and ${W^ \bot }$ must be zero.

$\begin{aligned}\left[{\begin{array}{*{20}{c}}1\\0\\1 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}p\\q \\r\end{array}} \right]&= 0 \hfill \\\left[ {\begin{array}{*{20}{c}}1&0&1 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}p \\q\\r\end{array}} \right] &= 0 \hfill\\\end{aligned}$

Further solve the above equation,

$\begin{aligned}p + r &= 0\\p&= - r\\\end{aligned}$

$\begin{aligned}\left[{\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right]&= 0\\\left[{\begin{array}{*{20}{c}}0&1&1\end{array}} \right] \cdot \left[{\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right]&= 0\\\end{aligned}$