# Sum of two subspaces: representing it with equations

I found the following excercise:

Let $$W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$$. Let $$W_2$$ be the span of $$S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\}$$. Give a base, a dimension and an equation representation of $$W_1 + W_2$$

I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $$\textbf{x} = (x_1, ..., x_6) \in W_1$$ satisfies

$$\begin{equation*} \begin{cases} x_1 = -x_2 - x_3 \\ x_4 = -x_5 - x_6 \end{cases} \end{equation*}$$

so that its general form is

$$\begin{equation*} \textbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{pmatrix} = \begin{pmatrix} -x_2 - x_3 \\ x_2 \\ x_3 \\ -x_5 - x_6 \\ x_5 \\ x_6 \end{pmatrix} \end{equation*}$$

We also know any $$\textbf{y} \in W_1$$ is of the general form

\begin{align*} \textbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \\ y_6 \end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\ -x + w \\ x + 2y - z \\ -x + y - z \\ x\\ -x + z\end{pmatrix} \end{align*}

Then for generals $$\mathbf{x}, \mathbf{y}$$ we have

\begin{align*} \textbf{x} + \textbf{y} &= \begin{pmatrix} x + y +z + 2w + (-x_2 - x_3)\\ -x + w + x_2\\ x + 2y - z + x_3\\ -x + y - z + (-x_5 - x_6)\\ x + x_5\\ -x + z + x_6 \end{pmatrix} \end{align*}

One can then conclude

$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R} \Big\}$$

But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!

Note that $$(1,0,−1,−1,0,1)\in W_1$$. Therefore, if$$W_3=\operatorname{span}\bigl\{(1,−1,1,−1,1,−1),(1,0,2,1,0,0),(2,1,0,0,0,0)\bigr\},$$then $$W_1+W_2=W_1+W_3$$.

Now, let us see what $$W_1\cap W_3$$ is. Asserting that$$\overbrace{\alpha(1,−1,1,−1,1,−1)+\beta(1,0,2,1,0,0)+\gamma(2,1,0,0,0,0)}^{\text{arbitrary element of }W_3}\in W_1$$means that $$\alpha+3\beta+3\gamma=-\alpha+\beta=0$$. This occurs if and only if $$\beta=\alpha$$ and $$\gamma=-\frac{4\alpha}3$$. Therefore,$$W_1\cap W_3=\left\{\left(-\frac{2 \alpha }{3},-\frac{7 \alpha }{3},3 \alpha,0,\alpha ,-\alpha \right)\,\middle|\,\alpha\in\Bbb R\right\};$$in particular, $$\dim(W_1\cap W_3)=1$$. So,\begin{align}\dim(W_1+W_3)&=\dim(W_1)+\dim(W_3)-\dim(W_1\cap W_3)\\&=4+3-1\\&=6.\end{align}Therefore, $$W_1+W_3$$ is the whole space $$\Bbb R^6$$.

• (+) The solution can be slightly simplified by considering $W_4=span[(1,0,2,1,0,0),(2,1,0,0,0,0)].$ Then $W_1+W_2\supset W_1+W_4.$ This time $\dim W_4=2$ and $W_1\cap W_4=\{0\},$ if I am not mistaken. Dec 26, 2022 at 17:17
• I must say, it was very uncomfortable not to accept this answer as the correct one, since it is impecable. In truth, the reason I accepted Anne's answer over yours is due to a limitation of my own: her's was a bit more beginner friendly and easy to follow for an almost complete ignorante on the subject. I did take the time to read Meta posts on how to deal with multiple, equally satisfying answers and the only workaround I found was to grant you a bounty to show my appreciation. However, this question is not yet eligible for bounties. So for now, just a big thanks! Dec 26, 2022 at 20:16
• I fully agree that her answer is more beginner friendly than mine. Dec 26, 2022 at 20:19

In contrast to Jose's and Anne's answers using dimension, let me show you the rote method you could use. From your step here:

$$\begin{equation*} \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{pmatrix} = \begin{pmatrix} -x_2 - x_3 \\ x_2 \\ x_3 \\ -x_5 - x_6 \\ x_5 \\ x_6 \end{pmatrix}, \end{equation*}$$

you could then find a spanning set for $$W_1$$ like so:

\begin{align*} \mathbf{x} &= \pmatrix{-x_2\\x_2\\0\\0\\0\\0} + \pmatrix{-x_3\\0\\x_3\\0\\0\\0} + \pmatrix{0\\0\\0\\-x_5\\x_5\\0} + \pmatrix{0\\0\\0\\-x_6\\0\\x_6} \\ &= x_2\pmatrix{-1\\1\\0\\0\\0\\0} + x_3\pmatrix{-1\\0\\1\\0\\0\\0} + x_5\pmatrix{0\\0\\0\\-1\\1\\0} + x_6\pmatrix{0\\0\\0\\-1\\0\\1}. \end{align*} The vector $$\mathbf{x}$$ is arbitrary in $$W_1$$, so we've shown that $$W_1 \subseteq \operatorname{span}\left\{\pmatrix{-1\\1\\0\\0\\0\\0},\pmatrix{-1\\0\\1\\0\\0\\0},\pmatrix{0\\0\\0\\-1\\1\\0},\pmatrix{0\\0\\0\\-1\\0\\1}\right\}.$$ Equality holds, because every vector in the spanning set also lies in $$W_1$$. It's also easy to see the above is linearly independent, so we have a basis, but this is not necessary to observe!

Next, once you have spanning sets for $$W_1$$ and $$W_2$$, you can form a spanning set for $$W_1 + W_2$$ by unioning the sets; every vector in $$W_1 + W_2$$ is the sum of a vector in $$W_1$$ (a linear combination of the first spanning set) and a vector in $$W_2$$ (a linear combination of the second spanning set), so the union will span the sum. That is, $$W_1 + W_2 = \operatorname{span}\left\{\pmatrix{-1\\1\\0\\0\\0\\0},\pmatrix{-1\\0\\1\\0\\0\\0},\pmatrix{0\\0\\0\\-1\\1\\0},\pmatrix{0\\0\\0\\-1\\0\\1}, \pmatrix{1\\-1\\1\\-1\\1\\-1},\pmatrix{1\\0\\2\\1\\0\\0},\pmatrix{1\\0\\-1\\-1\\0\\1},\pmatrix{2\\1\\0\\0\\0\\0}\right\}.$$ This is already a technically correct description of $$W_1 + W_2$$, but it's usually best to reduce the spanning set down to a basis. There are two standard ways to do this, both involving row-reduction:

• Place them as rows in a matrix, row-reduce down to row-echelon form (reduced, if you prefer) and keep only the non-zero rows. These rows will be linearly independent, but retain the same span as the original set, thus producing a basis, or
• Place them as columns in a matrix, and row-reduce down to row-echelon form. Note the columns where the pivots (leading $$1$$s) appear, and retain only the vectors from the original set that you placed in those columns.

Either way, we get an element in the basis of $$W_1 + W_2$$ for every pivot in the row-reduced matrix. And, no matter which method you use, you will get $$6$$ pivots, which tells you that the dimension of $$W_1 + W_2$$ is $$6$$, and must be all of $$\Bbb{R}^6$$. Indeed, if you apply the first method, reducing until reduced row-echelon form, you should find that you get back the standard basis!

• My main solution does not use dimension (only a second one at the end does). Dec 27, 2022 at 5:29

You already found that a vector $$(a,b,c,d,e,f)$$ belongs to $$W_1+W_2$$ if and only if there exist real numbers $$x_2,x_3,x_5,x_6,x,y,z,w$$ such that $$\begin{cases}a&=x+y+z+2w-x_2-x_3\\b&=-x+w-x_2\\c&=x + 2y - z + x_3\\ d&=-x + y - z-x_5 - x_6\\ e&=x + x_5\\ f&=-x + z + x_6. \end{cases}$$ Using for instance the 2nd, 3rd, 5th, and 6th equations to first eliminate the $$x_i$$'s, this is equivalent to the existence of real numbers $$x,y,z,w$$ such that $$\begin{cases}a&=3x+3y+w+b-c\\ d&=-x + y-e -f \end{cases}$$ For any $$(a,b,c,d,e,f)\in\Bbb R^6,$$ such numbers $$x,y,z,w$$ always exist (e.g. choose $$x,z$$ arbitrarily and take $$y=d+e+f+x,$$ and $$w=a-b+c-3x-3(d+e+f+x)$$) hence $$W_1+W_2=\Bbb R^6.$$ You could see it more rapidly since you already knew that $$\dim(W_1\cap W_2)=2$$: $$\dim(W_1)$$ is $$6-2=4$$ (since its 2 equations are clearly independent), $$\dim(W_2)=4$$ (since its 4 generating vectors are - less clearly - independent) hence by Grassmann's formula, $$\dim(W_1+W_2)=4+4-2=6.$$

• As usual, a very clear answer. Much appreciated, Anne. Dec 26, 2022 at 20:17
• Are you able to prove the "less clear" independence in the last sentence? Dec 26, 2022 at 20:47
• Already have! :) Dec 26, 2022 at 21:23