# Proof that the sum of two subspaces of $\mathbb{R}^n$ spans the entire $\mathbb{R}^n$. Is this a valid proof?

I was looking at the question: Linear Algebra Subspaces Proving and I didn't really understand the proposed solutions. I have a solution of my own which I would like verified.

The question is as follows:

Let $$V_1$$ and $$V_2$$ be subspaces of $$\mathbb R^n$$ defined by

$$V_1 = \{(x_1, x_2, ..., x_n) | x_1 + x_2 + \cdots + x_n = 0\}$$ $$V_2 = \{(x_1, x_2, ..., x_n) | x_1 = x_2 = \cdots = x_n\}.$$ Prove that any vector $$v \in\mathbb R^n$$ can be uniquely expressed as $$v = v_1 + v_2$$ such that $$v_1 \in V_1$$ and $$v_2 \in V_2$$.

My proof:

Since $$V_1$$ has $$n$$ elements, but only one linear equation, we can express any vector in $$V_1$$ with $$n - 1$$ parameters, i.e., $$((-x_2 + -x_3 + ... + -x_n), x_2, x_3, ..., x_n)$$. This means that $$\dim(V_1) = n - 1$$, since there are $$n-1$$ parameters.

$$V_2$$ can also be expressed as $$V_2 = \{t * (1, 1, ..., 1) | v \in\mathbb R\}$$

This implies that $$V_2 \not\subseteq V_1$$ since $$t(1) + t(1) + ... + t(1)$$ $$\ne$$ $$0$$ for all $$t \in \mathbb R$$, which means that every vector in $$V_2$$ is linearly independent to every vector in $$V_1$$.

Since

$$\dim(V_1) = n - 1$$ and any vector in $$V_2$$ is linearly independent to every vector in $$V_1$$,

$$\dim(V_1 + V_2) \ge \dim(V_1) + 1 \ge n$$

But since $$\dim(V_1 + V_2)$$ cannot be $$\gt n$$, $$\dim(V_1 + V_2)$$ must be $$= n$$.

Hence,

$$\text{span}(V_1 + V_2) = \mathbb R ^n$$

I'm pretty new to this so this is a pretty messy proof, but is it at all valid?

• "since $t*1+t*1+\cdots+t*1\neq 0, \forall t\in \mathbb{R}$" what about $t=0$? Oct 17, 2019 at 16:22

There are only small errors, but all else is fine.

• Rather than show that $$V_2 \nsubseteq V_1$$, show that $$V_2 \cap V_1 = \{0\}$$, which is easier : If $$(t , t , ... , t) \in V_1 \cap V_2$$ then $$nt = 0$$ so $$t = 0$$. The error you've made there is $$t(1) + t(1) + ... + t(1) = 0$$ can never happen , which is not true because $$t = 0$$ is possible.

• That $$\dim V_1 = n-1$$ needs to be proven more rigorously. Fair enough, it seems that we can express any vector in $$V_1$$ with $$n-1$$ vectors, but why not lesser? The reason is that you can find $$n-1$$ linearly independent vectors in $$V_1$$, for example $$(-1,1,0,...,0), (-1,0,1,0,...,0), (-1,0,0,1,0,...,0),... (-1,0,0,...,0,1)$$ are linearly independent and belong in $$V_1$$, so that shows that the dimension is at least $$n-1$$. What you have shown is that these vectors span $$V_1$$ in your expression so you get the dimension(Also, I am sure $$(-x_2,...,-x_n)$$ is a typo for $$-x_2-...-x_n$$).

I think the rest is fine : note that $$V_1 \cap V_2 = \{0\}$$ and $$\dim V_1 + \dim V_2 = n$$ implies the statement "every $$v \in V$$ can be written uniquely as $$v = v_1 + v_2$$ with $$v_i \in V_i$$, $$i=1,2$$" but you have not shown this in the explanation. I suspect you must already know the proof of that, but if you do not then attempt it.

• Hey thanks for the breakdown. What does the $nt$ in "then $nt = 0$ so $t = 0$" mean? Oct 17, 2019 at 16:46
• the dimension $n$ is a real number, and $t$ is a real number used to represent the entries of $(t,t,...,t)$. Think of it this way : if a vector is in $V_1$ , then all its entries are the same, say $t$. But if the same vector is also in $V_2$, then the sum of all its entries are zero, so the sum of all the $t$s is zero. But the sum of all the $t$s is just $t + t + ... + t$ and there are $n$ of these entries, so $nt$ (the usual product as real numbers) is $0$, so $t =0$. Oct 17, 2019 at 16:48

We know that $$\mathbb{R}^n$$ is a Hilbert space $$\mathfrak H$$, and we have that if $$F$$ is a closed linear subspace of $$\mathfrak H$$ we can write $$\mathfrak H=F\oplus F^{\perp}$$

where $$F^{\perp}=\{h\in\mathfrak H:\langle h,f\rangle=0,\forall f\in F\}$$.

If we take $$F=V_2$$, for any $$v_2=t(1,\cdots,1)\in V_2$$ and $$v_1\in V_1$$

$$\langle v_2,v_1\rangle=t\sum_{j=1}^{n}v_{1,j}=0$$ where $$v_{1,j}$$ denotes the $$j$$-th component of the vector.

This shows that $$V_1$$ and $$V_2$$ and orthogonal, this is $$V_2=F^{\perp}$$. (Notice furthermore that $$V_1\cap V_2=\{0\}$$) Hence we can write that

$$\mathbb{R}^n=V_1\oplus V_2$$

To show existence, consider the decomposition $$(x_1, \ldots, x_n) = \underbrace{\left(x_1 - \sum_{i=1}^n x_i, \ldots, x_n - \sum_{i=1}^n x_i\right)}_{\in V_1} - \underbrace{\left(\sum_{i=1}^n x_i, \ldots, \sum_{i=1}^n x_i\right)}_{\in V_2}$$

To show uniqueness, notice that $$V_1 \cap V_2 = \{0\}$$. Indeed, if $$(x_1, \ldots, x_n) \in V_1 \cap V_2$$ then $$x_1 = \cdots = x_n$$ and $$0 = \sum_{i=1}^n x_i$$ so $$x_1 = \cdots = x_n = 0$$.

Now assume there are $$v_1, v_1' \in V$$ and $$v_2, v_2' \in V_2$$ such that $$x = v_1 + v_2 = v_1' + v_2'$$. Rearranging gives $$\underbrace{v_1 - v_1'}_{\in V_1} = \underbrace{v_2' - v_2}_{\in V_2}$$ so $$v_1 - v_1'$$ and $$v_2' - v_2$$ are both elements of $$V_1 \cap V_2$$ and hence equal to $$0$$. We conclude $$v_1 = v_1'$$ and $$v_2 = v_2'$$.

The intersection $$V_1\cap V_2=\{0\}$$, because a vector in it has to satisfy $$x_1+x_2+\dots+x_n=0,\qquad x_1=x_2=\dots=x_n$$ The subspace $$V_1$$ is the kernel of the nonzero map $$\varphi\colon\mathbb{R}^n\to\mathbb{R}$$, $$\varphi(x_1,\dots,x_n)=x_1+\dots+x_n$$, so it satisfies $$\dim V_1=n-1$$.

Since $$V_2\ne\{0\}$$, we have $$\dim V_2\ge1$$. By Grassmann's formula $$\dim(V_1+V_2)=\dim V_1+\dim V_2+\dim(V_1\cap V_2)\ge n+1-0=n$$ Therefore $$V_1+V_2=\mathbb{R}^n$$ (which also proves that $$\dim V_2=1$$).