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Set $H = \mathbb{R}^6$

Let $A$ be the solution space of the system $\begin{cases} 3x_1 + 2x_2 - x_3 + 4x_4 + x_5 - x_6 = 0\\ x_1 + 2x_3 + x_4 -x_5 -x_6 = 0\\ 2x_1 + 4x_2 - 10x_3 + 4x_4 + 6x_5 + 2x_6 = 0\\ \end{cases}$

Let $B$ be the solution space of the system $\begin{cases} 4x_1 + 2x_2 - x_3 + 5x_4 - 2x_6 = 0\\ x_1 + x_2 + 2x_4 + x_5 - x_6 = 0\\ x_1 - x_2 + x_3 - x_4 - 3x_5 + x_6 = 0\\ \end{cases}$

i.) Find a basis for $A+B$.

ii.) Is $[1, 2, 1, -2, 1, 0]^T \in A + B$ ?

My approach:

Starting with $A$, we create the augmented matrix for $A$: \begin{align*} \begin{bmatrix} 3 & 2 & -1 & 4 & 1 & -1\\ 1 & 0 & 2 & 1 & -1 & -1\\ 2 & 4 & -10 & 4 & 6 & 2\\ \end{bmatrix} \end{align*} Performing elementary row operations, the RREF of the augmented matrix for $A$ is \begin{bmatrix} 1 & 0 & 2 & 1 & -1 & -1\\ 0 & 1 & -7/2 & 1/2 & 2 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}

which corresponds to the system $\begin{cases} x_1 + 2x_3 + x_4 - x_5 - x_6 = 0\\ x_2 - 7/2x_3 + 1/2x_4 + 2x_5 + x_6 = 0 \end{cases}$

Hence, $x_1 = -2x_3 - x_4 + x_5 + x_6$ and $x_2 = \dfrac{7x_3}{2} - \dfrac{x_4}{2} - 2x_5 - x_6$

Creating the augmented matrix for $B$, \begin{align*} \begin{bmatrix} 4 & 2 & -1 & 5 & 0 & -2\\ 1 & 1 & 0 & 2 & 1 & -1\\ 1 & -1 & 1 & -1 & -3 & 1\\ \end{bmatrix} \end{align*} Performing elementary row operations, the RREF of the augmented matrix for $B$ is \begin{bmatrix} 1 & 0 & 0 & 1/2 & -1 & 0\\ 0 & 1 & 0 & 3/2 & 2 & -1\\ 0 & 0 & 1 & 0 & 0 & 0\\ \end{bmatrix}

which corresponds to the system $\begin{cases} x_1 + 1/2x_4 - x_5 = 0\\ x_2 + 3/2x_4 + 2x_5 - x_6 = 0\\ x_3 = 0 \end{cases}$

Hence, $x_1 = -\dfrac{x_4}{2} + x_5$, $x_2 = -\dfrac{3x_4}{2} - 2x_5 + x_6$, and $x_3 = 0$

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    $\begingroup$ How can basis vectors of $A+B\subseteq\Bbb R^6$ be three dimensional? $\endgroup$ May 29, 2021 at 12:44
  • $\begingroup$ I'm sorry, does it have to have six columns? $\endgroup$
    – muw
    May 29, 2021 at 12:47
  • $\begingroup$ Yes, they should have six rows. $\endgroup$ May 29, 2021 at 12:47
  • $\begingroup$ Does that mean my augmented matrix is wrong? What I did was I appended the solution space of $B$ as a row. $\endgroup$
    – muw
    May 29, 2021 at 12:48
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    $\begingroup$ Solve the two linear systems one by one so you can find both $A$ and $B$. That can be a first naive idea, which works $\endgroup$
    – Naj Kamp
    May 29, 2021 at 13:11

2 Answers 2

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(i) One way to get bases for $A,B$ is to set the arbitrary variables in your solutions to $0$ except replace one of them with a $1$. Doing that for each arbitrary variable, we get a basis for $A$ $$\{[-2,7/2,1,0,0,0]^T,[-1,-1/2,0,1,0,0]^T,[1,-2,0,0,1,0]^T,[1,-1,0,0,0,1]^T\}$$ and a basis for $B$ $$\{[-1/2,-3/2,0,1,0,0]^T,[1,-2,0,0,1,0]^T,[0,1,0,0,0,1]^T\}.$$ To get a basis for $A+ B$, row-reduce the matrix whose rows are the basis vectors for $A$ and $B$: $$\begin{bmatrix}-2&7/2&1&0&0&0\\-1&-1/2&0&1&0&0\\1&-2&0&0&1&0\\1&-1&0&0&0&1\\-1/2&-3/2&0&1&0&0\\1&-2&0&0&1&0\\0&1&0&0&0&1\end{bmatrix}\to{\begin{bmatrix}1&0&0&0&0&2\\0&1&0&0&0&1\\0&0&1&0&0&1/2\\0&0&0&1&0&5/2\\0&0&0&0&1&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\\end{bmatrix}}.$$ The non-zero rows in the second matrix are a basis for $A + B$.

(ii) We want to know whether $[1,2,1,−2,1,0]^T$ can be expressed as a a linear combination of the vectors in the basis for $A + B$. In other words, are there scalars $c_i$ such that $$c_1[1,0,0,0,0,2]^T +c_2[0,1,0,0,0,1]^T+c_3[0,0,1,0,0,1/2]^T+c_4[0,0,0,1,0,5/2]^T+c_5[0,0,0,0,1,0]^T = [1,2,1,−2,1,0]^T?$$ That equation forces $c_1 = 1$, $c_2 = 2$, $c_3 = 1$, $c_4 = -2$, and $c_5 = 1$. With those values, the sixth coordinate on the left of our linear-combination equation works out to be $$1(2) + 2(1) + 1(1/2) - 2(5/2) + 1(0) = -1/2,$$ which is not equal to the sixth coordinate, $0$, on the right side. Thus, $[1,2,1,−2,1,0]^T$ is not in $A + B$.

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Building on your work:

Let's find a basis for $A$: $x_1 = -2x_3 - x_4 + x_5 + x_6$ and $x_2 = \dfrac{7x_3}{2} - \dfrac{x_4}{2} - 2x_5 - x_6$

Let $x_3=s, x_4=t, x_5=u, x_6=v$,

Then $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{bmatrix}= s \begin{bmatrix} -2 \\ \frac72 \\ 1\\ 0 \\ 0 \\ 0\end{bmatrix} + t\begin{bmatrix} -1 \\ -\frac12 \\ 0 \\ 1\\ 0 \\0\end{bmatrix} + u\begin{bmatrix}1 \\ -2 \\ 0 \\ 0 \\ 1\\ 0 \end{bmatrix} + v\begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \\0 \\ 1\end{bmatrix} $$

Here is a basis for $A$: $\{(-2, \frac72, 1, 0, 0, 0)^T, (-1, -\frac12, 0, 1, 0, 0)^T, (1, -2, 0, 0, 1, 0)^T, (1, -1, 0, 0, 0, 1)^T \}$

Similarly, let's find a basis for $B$: $x_1 = -\dfrac{x_4}{2} + x_5$, $x_2 = -\dfrac{3x_4}{2} - 2x_5 + x_6$, and $x_3 = 0$

A basis of $B$ is $\{(-\frac12, -\frac32, 0, 1,0,0)^T, (1, -2, 0, 0, 1, 0)^T, (0,1,0,0,0,1)^T \}$.

$A+B$ is spanned by the union of the basis, we can find the RREF of

$$\begin{bmatrix} -2 & \frac72 & 1 & 0 & 0 & 0 \\ -1 & -\frac12 & 0 & 1 & 0 & 0 \\ 1 & -2 & 0 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0 & 0 & 1 \\ -\frac12 & - \frac32 & 0 & 1 & 0 & 0\\ 1 & -2 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1\end{bmatrix}$$

to be $$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0.5 \\ 0 & 0 & 0 & 1 & 0 & 2.5 \\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}.$$

That is a basis is $\{ (1,0,0,0,0,2)^T, (0,1,0,0,0,1)^T, (0,0,1,0,0,0.5)^T, (0,0,0,1,0,2.5)^T, (0,0,0,0,1,0)^T \}$.

We check that $2(1) + 1(2) + 0.5(1) + 2.5(-2) + 1(0) \ne 0$, hence $(1,2,1,-2,1,0)^T$ is not in $A+B$.

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  • $\begingroup$ This means that the null space consists only of the zero vector, and consequently has no basis? $\endgroup$
    – muw
    Jun 1, 2021 at 17:01
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    $\begingroup$ When we talk about null space, we have to specify which matrix are we referring to. $A$ is the solution space of the first system. $B$ is the solution space of the second system. I take the basis that I found from each of them and concatenate them together to find a basis of $A+B$. $\endgroup$ Jun 1, 2021 at 17:07
  • $\begingroup$ That makes more sense. Just to check, $[1,2,1,-2,1,0]^T \in A+B$ would be false, since $\dfrac{7\cdot 1}{2} - \dfrac{-2}{2} - 2(1)- 0 \neq 2$? I'm sorry, I'm just testing it out if I understood the concept of the bases you presented. $\endgroup$
    – muw
    Jun 1, 2021 at 17:16
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    $\begingroup$ If my computation is correct, that is $A+B = \mathbb{R}^6$, then the vector is clearly inside $A+B$. I understand $A+B = \{a +b : a \in A, b \in B\}$. $\endgroup$ Jun 1, 2021 at 17:17
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    $\begingroup$ I computed the RREF of the matrix that I constructed to find a basis and found it to be $\begin{bmatrix}I_6 \\ 0_{2 \times 6} \end{bmatrix}$. $\endgroup$ Jun 1, 2021 at 17:24

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