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Show that the solutions for the linear system of equations:

$$\begin{aligned} 0 + x_2 +3x_3 - x_4 + 2x_5 &= 0 \\ 2x_1 + 3x_2 + x_3 + 3x_4 &= 0 \\ x_1 + x_2 - x_3 + 2x_4 - x_5 &= 0 \end{aligned}$$

is a subspace of $\mathbb R^5$. What is the dimension of the subspace and determine a basis for the subspace?

I really don't know how to solve this problem. I have achieved this augmented matrix through Gaussian elimination:

$$ \begin{bmatrix} 1& 0& -4& 3& -3& 0 \\ 0& 1& 3& -1& 2& 0 \\ 0& 0& 0& 0& 0& 0 \end{bmatrix} $$

Any hints or some steps I've missed?

Edit

My professor says the dimension is $3$.

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3 Answers 3

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You're almost there. Now your free variables are $x_3=s$,$x_4=t$ and $x_5=u$. Using backward substitution we get $$ x_1=4s-3t+3u \\ x_2=-3s+t-2u\\ x_3=s \\ x_4=t \\ x_5=u $$

Therefore we can write every solution as $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}=s\begin{bmatrix} 4 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+t\begin{bmatrix} -3 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}+u\begin{bmatrix} 3 \\ -2 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ with $s,t,u \in \mathbb{R}$.

Thus the subspace has dimension $3$ and a basis is given by $\begin{bmatrix} 4 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 3 \\ -2 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

To prove is a subspace you need:

  • $0$ belongs to it: this is clear by taking $s=t=u=0$

  • It's closed under sums: if $(s,t,u)$ and $(s',t',u')$ gives us two different solutions, the sum of them is given by $(s+s',t+t',u+u')$

  • It's closed under scalar multiplication: if $(s,t,u)$ gives us a solution and we multiply it by $k \in \mathbb{R}$, then we still have a solution given by $(ks,kt,ku)$.

Therefore it is a subspace of $\mathbb{R}^5$

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  • $\begingroup$ missing $s$ in equation 2 $\endgroup$
    – Prototank
    Oct 23, 2019 at 14:34
  • $\begingroup$ Thank for you your answer, this was also thought on this problem. However, my professor says that the Dim = 3, which I can't really see why, or maybe he has made a mistake? Also, how do I argument that the solutions is a subspace of R5? $\endgroup$
    – Carl
    Oct 23, 2019 at 14:36
  • $\begingroup$ @Carl are you sure about your Gaussian elimination then? Because in that case the dimension is $2$. To get dimension $3$ you need that one of the rows becomes of all zeros $\endgroup$
    – user289143
    Oct 23, 2019 at 14:43
  • $\begingroup$ I'm very, very sorry. I messed up on my augmented matrix. I have updated my question and the augmented matrix, although I think I understand the solution now, it would be very generous of you to edit your answer. $\endgroup$
    – Carl
    Oct 23, 2019 at 14:50
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There's already a good answer on what that subspace is and what dimension it has.

To prove that we are indeed talking about a subspace, you must prove that:

  • If $(x_1, x_2, x_3, x_4, x_5)$ and $(y_1, y_2, y_3, y_4, y_5)$ are solutions, then $(x_1 + y_1, x_2 + y_2, x_3 + y_3, x_4 + y_4, x_5 + y_5)$ is also a solution
  • If $(x_1, x_2, x_3, x_4, x_5)$ is a solution and $\lambda \in \mathbb{R}$, then $(\lambda x_1, \lambda x_2, \lambda x_3, \lambda x_4, \lambda x_5)$ is also a solution
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Here is a correct RREF: \begin{align} &\left[\begin{array}{*{5}{r}} 0&1&3&-1&2 \\ 2&3&1&3&0 \\ 1&1&-1&2&-1 \end{array}\right]\rightsquigarrow \left[\begin{array}{*{5}{r}} 1&1&-1&2&-1 \\ 0&1&3&-1&2 \\ 2&3&1&3&0 \end{array}\right]\rightsquigarrow \left[\begin{array}{*{5}{r}} 1&1&-1&2&-1 \\ 0&1&3&-1&2 \\ 0&1&3&-1&2 \end{array}\right]\rightsquigarrow \\[1ex] &\left[\begin{array}{*{5}{r}} 1&1&-1&2&-1 \\ 0&1&3&-1&2 \\ 0&0&0&0&0 \end{array}\right]\rightsquigarrow \left[\begin{array}{*{5}{r}} 1&0&-4&3&-3 \\ 0&1&3&-1&2 \\ 0&0&0&0&0 \end{array}\right]. \end{align} This matrix has rank $2$. Hence, by the rank-nullity theorem, the kernel, i. e. the subspace of solutions (in a $5$-dimensional space), has dimension $3$.

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