# Finding a complementary subspace ##U## | Linear Algebra

• JD_PM
I just realized that this is off-topic. Sorry.I think the best is to put the question aside for the moment and to revisit it once the Gram-Schmidt procedure is introduced.

#### JD_PM

Homework Statement
Find a complementary subspace ##U## in ##V## i.e. find ##U## such that ##U \oplus W =V##, where ##W = span \{ 1-3X + 2X^2 , 1 + X + 4X^2\}## and ## p_1 (X) = 1 - 7X ##, ## p_2 (X) = 4 - 2X + 6X^2 \in \Bbb R[X]_{\leq 2}##
Relevant Equations
N/A
We only worry about finite vector spaces here.

I have been taught that a subspace ##W## of a vector space ##V## has a complementary subspace ##U## if ##V = U \oplus W##.

Besides, I understand that, given a finite vectorspace ##(\Bbb R, V, +)##, any subspace ##U## of ##V## has a complementary subspace ##W##. Why? Let ##U## be a subspace of ##V## and its basis ##\{u_1, u_2, \dots, u_k \}##. We know that a basis of a subspace can always be expanded to a basis of the vector space i.e. ##\{u_1, u_2, \dots, u_k, w_{k+1}, \dots,w_n \}##. Hence we get ##W = span\{w_{k+1}, \dots,w_n \}##, the complementary subspace of ##U##.

Time to focus on the given exercise.

We note that the vectors spanning ##W## are linearly independent so they conform a basis of ##W##. Hence we can always expand that basis to one of ##V##.

$$\beta = \{ 1-3X + 2X^2 , 1 + X + 4X^2, 3+5X-6X^2\}$$

So the complementary subspace is given by ##U = span \{ 3+5X-6X^2\}##

And the unique decomposition is given by

$$\underbrace{1 - 7X}_{\in V} = \underbrace{4 - 2X + 6X^2}_{\in W} + \underbrace{(-3-5X-6X^2)}_{\in U}$$

Main issue:

1) I chose the third element of the basis of ##V## so that the unique decomposition holds. Is there a way to pick an arbitrary third element and end up with such unique decomposition?

Is the overall idea OK?

Thanks! JD_PM said:
Homework Statement:: Find a complementary subspace ##U## in ##V## i.e. find ##U## such that ##U \oplus W =V##, where ##W = span \{ 1-3X + 2X^2 , 1 + X + 4X^2\}## and ## p_1 (X) = 1 - 7X ##, ## p_2 (X) = 4 - 2X + 6X^2 \in \Bbb R[X]_{\leq 2}##
Relevant Equations:: N/A

We only worry about finite vector spaces here.

I have been taught that a subspace ##W## of a vector space ##V## has a complementary subspace ##U## if ##V = U \oplus W##.

Besides, I understand that, given a finite vectorspace ##(\Bbb R, V, +)##, any subspace ##U## of ##V## has a complementary subspace ##W##. Why? Let ##U## be a subspace of ##V## and its basis ##\{u_1, u_2, \dots, u_k \}##. We know that a basis of a subspace can always be expanded to a basis of the vector space i.e. ##\{u_1, u_2, \dots, u_k, w_{k+1}, \dots,w_n \}##. Hence we get ##W = span\{w_{k+1}, \dots,w_n \}##, the complementary subspace of ##U##.

Time to focus on the given exercise.

We note that the vectors spanning ##W## are linearly independent so they conform a basis of ##W##. Hence we can always expand that basis to one of ##V##.

$$\beta = \{ 1-3X + 2X^2 , 1 + X + 4X^2, 3+5X-6X^2\}$$

So the complementary subspace is given by ##U = span \{ 3+5X-6X^2\}##

And the unique decomposition is given by

$$\underbrace{1 - 7X}_{\in V} = \underbrace{4 - 2X + 6X^2}_{\in W} + \underbrace{(-3-5X-6X^2)}_{\in U}$$

Main issue:

1) I chose the third element of the basis of ##V## so that the unique decomposition holds. Is there a way to pick an arbitrary third element and end up with such unique decomposition?

Is the overall idea OK?

Thanks! If ##\beta ## are linearly independent vectors, which I think they are but haven't checked, then you can take them and choose the bases for ##W## the way you did. I have no idea about the role the ##p_i(x)## play.

Of course, you cannot pick an arbitrary third polynomial since it must not be in ##U##. A general algorithm for such problems is the Gram-Schmidt procedure. It produces orthogonal vectors, which isn't necessary in the case above, but it will be the next step. You can choose any ##\vec{p}\not\in U## as a basis in your case.

FactChecker said:
If you have a subspace defined as the span of two polynomials, ##p_1(X), p_2(X)## can't you just use those polynomials directly to define the compliment?: ##\{X\in R^3 : p_1(X)=0 \textrm{ or } p_2(X)=0 \}##

Mmm no clue, honestly. All I know about complementary subspaces is what is below

JD_PM said:
I have been taught that a subspace ##W## of a vector space ##V## has a complementary subspace ##U## if ##V = U \oplus W##.

Besides, I understand that, given a finite vectorspace ##(\Bbb R, V, +)##, any subspace ##U## of ##V## has a complementary subspace ##W##. Why? Let ##U## be a subspace of ##V## and its basis ##\{u_1, u_2, \dots, u_k \}##. We know that a basis of a subspace can always be expanded to a basis of the vector space i.e. ##\{u_1, u_2, \dots, u_k, w_{k+1}, \dots,w_n \}##. Hence we get ##W = span\{w_{k+1}, \dots,w_n \}##, the complementary subspace of ##U##.

fresh_42 said:
Of course, you cannot pick an arbitrary third polynomial since it must not be in ##U##. A general algorithm for such problems is the Gram-Schmidt procedure. It produces orthogonal vectors, which isn't necessary in the case above, but it will be the next step. You can choose any ##\vec{p}\not\in U## as a basis in your case.

Thanks for the idea but I am not allowed to use the Gram-Schmidt procedure because it comes in later chapters.

It is just that it felt unnatural to fit a third element of the basis of ##V## such that the following equation holds uniquely.

JD_PM said:
$$\underbrace{1 - 7X}_{\in V} = \underbrace{4 - 2X + 6X^2}_{\in W} + \underbrace{(-3-5X-6X^2)}_{\in U}$$

That's why I ask for feedback; doesn't it look odd to you?

https://www.physicsforums.com/goto/post?id=6516205
If you have a subspace defined as the span of two polynomials,
can't you just use those polynomials directly to define the compliment?
JD_PM said:
Mmm no clue, honestly. All I know about complementary subspaces is what is below
Sorry. I got confused. I have deleted that post.

There is no intuition demanding uniqueness. Imagine a plane, say a square, on your desk. Any vector outside this plane, pointing up (or down) allows you to build a complete basis of three-dimensional space, i.e. a coordinate system. Once you have chosen a specific one, your coordinates become unique. However, all others define different coordinate systems.