Question

In Exercises 1–4, let S be the collection of vectors cxyd in R2that satisfy the given property. In each case, either prove thatS forms a subspace of R2 or give a counterexample to show that it does not.

Answer 1

Answer:

S is not the subspace of $R^2$

Step-by-step explanation:

Let us suppose two vectors u and v belong to the S such that the property of xy≥0 is verified than

$u=\left[\begin{array}{c}-1 \\0\end{array}\right]$

$v=\left[\begin{array}{c}0 \\1\end{array}\right]$

Both the vectors satisfy the given condition as follows and belong to the  S

$x_u y_u=-1\times 0=0 \geq 0\\x_v y_v=1\times 0=0 \geq 0$

Now S will be termed as subspace of R2 if

• u+v also satisfy the condition
• ku also satisfy the condition

Taking u+v

$u+v=\left[\begin{array}{c}-1 \\0\end{array}\right]+\left[\begin{array}{c}0 \\1\end{array}\right]\\u+v=\left[\begin{array}{c}-1+0 \\0+1\end{array}\right]\\u+v=\left[\begin{array}{c}-1 \\1\end{array}\right]$

Now the condition is tested as

$\\x_{u+v} y_{u+v}=-1\times 1=-1 <0$

This indicates that the condition is not satisfied so S is not the subspace of $R^2$