# Show that the solutions is a subspace of $\mathbb R^5$

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 $$3$$.

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 $$\mathbb{R}^5$$

• missing $s$ in equation 2 Oct 23, 2019 at 14:34
• 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?
– Carl
Oct 23, 2019 at 14:36
• @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 Oct 23, 2019 at 14:43
• 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.
– Carl
Oct 23, 2019 at 14:50

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

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 $$3$$.