- #1

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- 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!

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!