Finding a Basis for the Complementary Subspace

Oct 13, 2025 by pynkocean.com 7 views

Introduction

In linear algebra, the concept of complementary subspaces plays a crucial role in understanding the structure of vector spaces. Given a subspace $M$ of a vector space $V$ , a complementary subspace $N$ is a subspace such that $V = M \oplus N$ , where $\oplus$ denotes the direct sum. In this article, we will focus on finding a basis for the complementary subspace of $M=(p\in P_4 : p(2x)=p(x+1))$ , where $P_4$ is the space of polynomials of degree at most 4.

Background

To begin with, let's recall some basic concepts from linear algebra and polynomials. A polynomial $p(x)$ is a function of the form $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ , where $a_n \neq 0$ . The space of polynomials of degree at most $n$ is denoted by $P_n$ . In this case, we are dealing with $P_4$ , the space of polynomials of degree at most 4.

Definition of $M$

The subspace $M$ is defined as the set of polynomials $p(x) \in P_4$ such that $p(2x) = p(x+1)$ . This means that for any polynomial $p(x) \in M$ , we have $p(2x) = p(x+1)$ . To understand this condition better, let's consider some examples.

Examples

Let's consider some polynomials in $P_4$ and check if they satisfy the condition $p(2x) = p(x+1)$ .

$p(x) = 1$ : We have $p(2x) = 1$ and $p(x+1) = 1$ . Therefore, $p(x) = 1$ satisfies the condition.
$p(x) = x$ : We have $p(2x) = 2x$ and $p(x+1) = x+1$ . Therefore, $p(x) = x$ does not satisfy the condition.
$p(x) = x^2$ : We have $p(2x) = 4x^2$ and $p(x+1) = (x+1)^2 = x^2 + 2x + 1$ . Therefore, $p(x) = x^2$ does not satisfy the condition.

From these examples, we can see that not all polynomials in $P_4$ satisfy the condition $p(2x) = p(x+1)$ . Therefore, $M$ is a proper subspace of $P_4$ .

Finding a Basis for $M$

To find a basis for $M$ , we need to find a set of linearly independent polynomials that span $M$ . Let's consider the following polynomials:

$p_1(x) = 1$
$p_2(x) = x$
$p_3(x) = x^2$
$p_4(x) = x^3$
$p_5(x) = x^4$

We can check if these polynomials satisfy the condition $p(2x) = p(x+1)$ .

$p_1(x) = 1$ : We have $p_1(2x) = 1$ and $p_1(x+1) = 1$ . Therefore, $p_1(x) = 1$ satisfies the condition.
$p_2(x) = x$ : We have $p_2(2x) = 2x$ and $p_2(x+1) = x+1$ . Therefore, $p_2(x) = x$ does not satisfy the condition.
$p_3(x) = x^2$ : We have $p_3(2x) = 4x^2$ and $p_3(x+1) = (x+1)^2 = x^2 + 2x + 1$ . Therefore, $p_3(x) = x^2$ does not satisfy the condition.
$p_4(x) = x^3$ : We have $p_4(2x) = 8x^3$ and $p_4(x+1) = (x+1)^3 = x^3 + 3x^2 + 3x + 1$ . Therefore, $p_4(x) = x^3$ does not satisfy the condition.
$p_5(x) = x^4$ : We have $p_5(2x) = 16x^4$ and $p_5(x+1) = (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$ . Therefore, $p_5(x) = x^4$ does not satisfy the condition.

From these calculations, we can see that only $p_1(x) = 1$ satisfies the condition. Therefore, a basis for $M$ is $\{p_1(x)\} = \{1\}$ .

Finding a Basis for the Complementary Subspace

To find a basis for the complementary subspace of $M$ , we need to find a set of linearly independent polynomials that span the complementary subspace. Let's consider the following polynomials:

$q_1(x) = x$
$q_2(x) = x^2$
$q_3(x) = x^3$
$q_4(x) = x^4$

We can check if these polynomials are linearly independent.

$\{q_1(x), q_2(x), q_3(x), q_4(x)\}$ is a linearly independent set.

Therefore, a basis for the complementary subspace of $M$ is $\{q_1(x), q_2(x), q_3(x), q_4(x)\} = \{x, x^2, x^3, x^4\}$ .

Conclusion

In this article, we have found a basis for the complementary subspace of $M=(p\in P_4 : p(2x)=p(x+1))$ . We have shown that a basis for $M$ is $\{1\}$ and a basis for the complementary subspace is $\{x, x^2, x^3, x^4\}$ . This result provides a deeper understanding of the structure of the space of polynomials of degree at most 4 and its subspaces.

References

[1] Hoffman, K., & Kunze, R. (1971). Linear Algebra. Prentice-Hall.
[2] Lang, S. (1993). Linear Algebra. Springer-Verlag.
[3] Strang, G. (1988). Linear Algebra and Its Applications. Academic Press.
Complementary Subspace of $M=(p(2x)=p(x+1)) ,p\in P_4$ ===========================================================

Q&A

Q: What is the complementary subspace of $M=(p\in P_4 : p(2x)=p(x+1))$ ? A: The complementary subspace of $M$ is a subspace of $P_4$ that is complementary to $M$ in the sense that $P_4 = M \oplus N$ , where $N$ is the complementary subspace.

Q: How do we find a basis for the complementary subspace of $M$ ? A: To find a basis for the complementary subspace of $M$ , we need to find a set of linearly independent polynomials that span the complementary subspace. We can do this by considering the polynomials that are not in $M$ and checking if they are linearly independent.

Q: What is a basis for the complementary subspace of $M$ ? A: A basis for the complementary subspace of $M$ is $\{x, x^2, x^3, x^4\}$ .

Q: Why is $\{x, x^2, x^3, x^4\}$ a basis for the complementary subspace of $M$ ? A: $\{x, x^2, x^3, x^4\}$ is a basis for the complementary subspace of $M$ because it is a set of linearly independent polynomials that span the complementary subspace. We can check this by showing that the polynomials in this set are linearly independent and that they span the complementary subspace.

Q: What is the significance of the complementary subspace of $M$ ? A: The complementary subspace of $M$ is significant because it provides a deeper understanding of the structure of the space of polynomials of degree at most 4 and its subspaces. It also provides a way to find a basis for the complementary subspace of any subspace of $P_4$ .

Q: How can we use the complementary subspace of $M$ in applications? A: The complementary subspace of $M$ can be used in applications such as finding a basis for a subspace of $P_4$ , solving systems of linear equations, and finding the dimension of a subspace of $P_4$ .

Q: What are some common mistakes to avoid when working with the complementary subspace of $M$ ? A: Some common mistakes to avoid when working with the complementary subspace of $M$ include:

Assuming that the complementary subspace of $M$ is the same as the subspace of $P_4$ that is orthogonal to $M$ .
Assuming that the complementary subspace of $M$ is the same as the subspace of $P_4$ that is spanned by the polynomials that are not in $M$ .
Not checking if the polynomials in the set are linearly independent.

Q: How can we verify that the complementary subspace of $M$ is correct? A: We can verify that the complementary subspace of $M$ is correct by checking if the polynomials in the set are linearly independent and if they span the complementary subspace. We can also check if the dimension of the complementary subspace of $M$ is correct.

Q: What are some tips for working with the complementary subspace of $M$ ? A: Some tips for working with the complementary subspace of $M$ include:

Always checking if the polynomials in the set are linearly independent.
Always checking if the polynomials in the set span the complementary subspace.
Always checking if the dimension of the complementary subspace of $M$ is correct.

Conclusion

In this article, we have provided a comprehensive overview of the complementary subspace of $M=(p\in P_4 : p(2x)=p(x+1))$ . We have discussed how to find a basis for the complementary subspace of $M$ , what a basis for the complementary subspace of $M$ is, and why $\{x, x^2, x^3, x^4\}$ is a basis for the complementary subspace of $M$ . We have also provided some common mistakes to avoid when working with the complementary subspace of $M$ and some tips for working with the complementary subspace of $M$ .