Hi all,
Having a problem trying to show that the set $\displaystyle
\begin{bmatrix} 3a-b\\ ab \\ a+b\\ b+2a \end{bmatrix}
$
where a and b are real numbers, is not a subspace in R^4..
Let $\displaystyle a=1,b=2$ and consider the vector
$\displaystyle 2\left[\begin{array}{c}3a-b\\ab\\a+b\\2a+b\end{array}\right]=2\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]=\left[\begin{array}{c}2\\4\\6\\8\end{array}\right]$.
Then $\displaystyle 3a'-b'=2$, $\displaystyle a'b'=4$ and $\displaystyle a'+b'=6$. But this system has no solution:
$\displaystyle 3a'-b'=2$ implies $\displaystyle b'=3a'-2$. Then $\displaystyle a'+b'=6$ gives us $\displaystyle a'+(3a'-2)=6$ and therefore $\displaystyle a'=2$ and $\displaystyle b'=4$. But then $\displaystyle a'b'=8$, contradicting the condition $\displaystyle a'b'=4$.