5

find a homogeneous system whose solution space is spanned by the following set of 3 vectors: $$(1,-2,0,3,-1) , (2,-3,2,5,-3), (1,-2,1,2,-2)$$

Please help, I've only seen similar questions where there are 4 unknowns not 5

Hirshy
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Pheobe
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  • Just start, if there are 4 or 5 unknowns doesn't really matter. – Hirshy Aug 12 '15 at 07:27
  • i've tried it two times but i just get stuck after doing the row operations to put it into echelon but then when I try to back substitute I get confused because I'm not getting number values just in terms of the other variables – Pheobe Aug 12 '15 at 07:29

1 Answers1

5

Let $$ v_1 = (1, -2, 0, 3, -1),\\ v_2 = (2, -3, 2, 5, -3),\\ v_3 = (1, -2, 1, 2, -2). $$ and $L = \mathop{\mathrm{span}}(v_1, v_2, v_3)$. Let's find $L^\perp = \{x\mid \forall v\in L\colon (v, x) = 0\}$: $$ L^\perp\colon\left\{\begin{aligned} x_1 -2x_2 + 3x_4 -x_5&=0,\\ 2x_1 -3x_2 + 2x_3 + 5x_4 -3x_5&=0,\\ x_1 -2x_2 + x_3 + 2x_4 -2x_5&=0. \end{aligned}\right. $$ Solve this system: $$ \begin{aligned} x_1 &= t_1,\\ x_2 &= t_1 + 4t_2,\\ x_3 &= -t_1-4t_2,\\ x_4 &= t_2,\\ x_5 &= -t_1-5t_2, \end{aligned} $$ or $$ x = \begin{pmatrix}1\\1\\-1\\0\\-1\end{pmatrix} t_1 + \begin{pmatrix}0\\4\\-4\\1\\-5\end{pmatrix} t_2 = u_1t_1 + u_2 t_2. $$ But $L = (L^\perp)^\perp$. Do the same trick and $$ L\colon\left\{\begin{aligned} x_1 + x_2 -x_3 - x_5 &= 0\\ 4x_2-4x_3+x_4-5x_5 &= 0 \end{aligned}\right. $$ is needed system.

Michael Galuza
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