Show a set is not a subspace of R^4

**Oiler** · August 3rd 2011, 05:06 PM

Hi all,

Having a problem trying to show that the set $<br /> \begin{bmatrix} 3a-b\\ ab \\ a+b\\ b+2a \end{bmatrix}<br />$
where a and b are real numbers, is not a subspace in R^4..

**hatsoff** · August 3rd 2011, 06:37 PM

Let $a=1,b=2$ and consider the vector

$2\left[\begin{array}{c}3a-b\\ab\\a+b\\2a+b\end{array}\right]=2\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]=\left[\begin{array}{c}2\\4\\6\\8\end{array}\right]$ .

Then $3a'-b'=2$ , $a'b'=4$ and $a'+b'=6$ . But this system has no solution:

$3a'-b'=2$ implies $b'=3a'-2$ . Then $a'+b'=6$ gives us $a'+(3a'-2)=6$ and therefore $a'=2$ and $b'=4$ . But then $a'b'=8$ , contradicting the condition $a'b'=4$ .

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