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2 years ago

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]2 years ago

5 0

<span>[-1,-1,1]
or
</span><span>[-2,-2,2]
</span>.....
Dot Product = 0

Daniel [21]2 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,

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

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

Further solve the above equation.

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

Therefore, $p = q.$

The matrix ${W^ \bot }$ will be $\left[ {\begin{array}{*{20}{c}}{ - c}\\ { - c}\\c\end{array}} \right]$

The basis of can be obtained as follows,

$\left[{\begin{array}{*{20}{c}}{ - c}\\{ - c}\\c\end{array}} \right] = c\left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\1\end{array}} \right]$

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

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: College

Subject: Mathematics

Chapter: Vectors and matrices

Keywords: W set, all vectors, x, y, x + y, real numbers, perpendicular, matrices, vectors, basis.

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