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

[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!

 

[fresh_42]

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.

 

[JD_PM]

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

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 $U$.

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.

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

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

 

[FactChecker]

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?

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

Sorry. I got confused. I have deleted that post.

 

[fresh_42]

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.