# Find the basis of $A + B$, given the solution space of both $A$ and $B$

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 $$x_3 = 0$$

• How can basis vectors of $A+B\subseteq\Bbb R^6$ be three dimensional? May 29, 2021 at 12:44
• I'm sorry, does it have to have six columns?
– muw
May 29, 2021 at 12:47
• Yes, they should have six rows. May 29, 2021 at 12:47
• Does that mean my augmented matrix is wrong? What I did was I appended the solution space of $B$ as a row.
– muw
May 29, 2021 at 12:48
• 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 May 29, 2021 at 13:11

(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 $$A + B$$.

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 $$A+B$$.

• This means that the null space consists only of the zero vector, and consequently has no basis?
– muw
Jun 1, 2021 at 17:01
• 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 $A+B$. Jun 1, 2021 at 17:07
• 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.
– muw
Jun 1, 2021 at 17:16
• If my computation is correct, that is $A+B = \mathbb{R}^6$, then the vector is clearly inside $A+B$. I understand $A+B = \{a +b : a \in A, b \in B\}$. Jun 1, 2021 at 17:17
• I computed the RREF of the matrix that I constructed to find a basis and found it to be $\begin{bmatrix}I_6 \\ 0_{2 \times 6} \end{bmatrix}$. Jun 1, 2021 at 17:24