I know the definition of complementary subspace:

If the vector space $V = U \oplus W$, where $U,W\subset V$ and $U\cap W = \phi$. Then $U$ and $W$ are complementary subspaces.

I am really confused about the following fundamental stuff:

Suppose I have a vector space $V\in \mathbb{R}^3$ spanned by three vectors $\{v_1,v_2,v_3\}$. And let $$W = \{\alpha(v_1+v_2+v_3)\mid \alpha \in \mathbb{R}\}.$$ So $W$ in a one-dimensional subspace of $V$. How to say that the complemetary subspace of $W$ is the following:

$$U = \{av_1+bv_2+cv_3 \mid a+b+c = 0\}$$

Should I use the inner product of vectors coming from $U$ and $W$ is $0$ to show that?

Please advise, thanks!

  • 1
    $\begingroup$ You can easily see that $W$ is one-dimensional. You should be able to pick some simple basis for $U$ and show that it is two-dimensional. It is also easy to see that $U \cap W = 0$. Then the result will follow by dimension. $\endgroup$ – Joppy Apr 30 '18 at 0:21

You don't want to use "the" with "complementary subspace". There are usually many complementary subspaces to a given subspace. To show that $U$ is complementary to $W$ you just need to show that the intersection is $0$ and that they span $V$. Clearly the intersection is $0$ since that's the only element of $W$ that satisfies the condition for being in $U$ also. (Three equal coordinates that all add to zero must be zero.) On the other hand, $(1,1,1)\in W, (1,-1,0)\in U, (0,1,-1)\in U$ span $V$, so they are complementary.


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