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Snezhnost [94]
3 years ago
7

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.
Mathematics
1 answer:
Artyom0805 [142]3 years ago
3 0

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

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

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