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 $V_1$.


$$\dim(V_1) = n - 1$$ 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 $= n$.


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

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

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

4 Answers 4


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.

  • $\begingroup$ Hey thanks for the breakdown. What does the $nt$ in "then $nt = 0$ so $t = 0$" mean? $\endgroup$
    – eclmist
    Oct 17, 2019 at 16:46
  • $\begingroup$ 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 $t =0$. $\endgroup$ 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 $\dim V_2=1$).


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