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Determine if the given set is a subspace of \(\mathbb{P}_{n}\) for an appropriate value of \(n .\) Justify your answers. All polynomials of the form \(\mathbf{p}(t)=a t^{2},\) where \(a\) is in \(\mathbb{R}\)

Short Answer

Expert verified
The set is a subspace of \( \mathbb{P}_2 \).

Step by step solution

01

Identify the Polynomial Space

The given polynomials are of the form \( p(t) = at^2 \). These are quadratic polynomials, thus they belong to the space of polynomials of degree 2, denoted \( \mathbb{P}_2 \). Each polynomial is defined by a real number \( a \).
02

Determine Closed Under Addition

To check if the set is a subspace, first consider two polynomials \( p_1(t) = a t^2 \) and \( p_2(t) = b t^2 \). Their sum is \( (a + b)t^2 \). Since \( a+b \) is also a real number, the sum is also within the set, which shows closure under addition.
03

Verify Closed Under Scalar Multiplication

Consider a scalar \( c \in \mathbb{R} \) and a polynomial \( p(t) = a t^2 \). The scalar product is \( c(at^2) = (ca)t^2 \). Since \( ca \) is also a real number, the set is closed under scalar multiplication.
04

Check for Zero Polynomial

Evaluate the set for the zero polynomial. For \( a = 0 \), the polynomial becomes \( 0 \cdot t^2 = 0 \), which is the zero polynomial, contained in the set.
05

Conclusion

Since the set of polynomials \( p(t) = at^2 \) is closed under addition, scalar multiplication, and contains the zero polynomial, it forms a subspace of \( \mathbb{P}_2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Space
When we speak about polynomial space, particularly in this context, we are referring to collections of polynomials that share certain properties, such as degree. Here, the polynomials are of the form \( p(t) = at^2 \), meaning they belong to the space of polynomials of degree 2, denoted as \( \mathbb{P}_2 \). This space includes all quadratic polynomials where the highest power of \( t \) is 2. Each polynomial in this space is uniquely defined by a single real coefficient \( a \), which multiplies \( t^2 \).

Polynomial spaces like \( \mathbb{P}_2 \) are important because they help in understanding and manipulating polynomials using vector space concepts. Essentially, they provide a framework where polynomials can be added together or multiplied by scalars, much like ordinary vectors.
Closure Under Addition
One of the essential properties of a subspace is closure under addition. This means that if you take two polynomials from the set and add them, the result is also within the set. Let's illustrate this with the polynomials \( p_1(t) = a t^2 \) and \( p_2(t) = b t^2 \).

When you add these polynomials together, you get:
  • \( p_1(t) + p_2(t) = at^2 + bt^2 = (a + b)t^2 \)
The result, \( (a+b)t^2 \), is still a polynomial of the form \( at^2 \), meaning it also belongs to our set. Since adding any two polynomials results in another polynomial of the same form, the set is closed under addition. This closure property is crucial for the set to qualify as a subspace.
Closure Under Scalar Multiplication
Another critical property for determining if a set is a subspace is closure under scalar multiplication. This involves taking any polynomial from our set and multiplying it by a scalar from the real numbers to see if the result remains in the set.

Suppose you have a polynomial \( p(t) = at^2 \) and a real number \( c \.\) Multiply them to get:
  • \( c(p(t)) = c(at^2) = (ca)t^2 \)
Here, \( ca \) is just another real number, which means \( (ca)t^2 \) is of the same form as the original polynomial \( at^2 \). So, multiplying by any scalar produces another polynomial within the set. This assures us that our set is closed under scalar multiplication, reinforcing its status as a subspace.
Zero Polynomial
A fundamental requirement for any subspace is that it must include the zero polynomial. This is the polynomial that evaluates to zero for any input, and it serves as the "zero vector" in vector space terms.

In our polynomial set, the zero polynomial occurs when the coefficient \( a = 0 \). Substituting \( a = 0 \) into \( p(t) = at^2 \), we get:
  • \( p(t) = 0 \cdot t^2 = 0 \)
This confirms the presence of the zero polynomial in the set. Including the zero polynomial ensures that the set satisfies all necessary conditions to be a subspace of \( \mathbb{P}_2 \), confirming a complete and perfect polynomial subspace.

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Most popular questions from this chapter

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Show that if \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent in \(V,\) then the set of images, \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\},\) is linearly dependent in \(W .\) This fact shows that if a linear transformation maps a set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) onto a linearly independent set \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{1}\right)\right\\},\) then the original set is linearly independent, too (because it cannot be linearly dependent).

[M] The Demographic Research Unit of the California State Department of Finance supplied data for the following migration matrix, which describes the movement of the United States population during \(2012 .\) In \(2012,\) about 12.5\(\%\) of the total population lived in California. What percentage of the total population would eventually live in California if the listed migration probabilities were to remain constant over many years? $$ \left[\begin{array}{cc}{.9871} & {.0027} \\ {.0129} & {.9973}\end{array}\right] $$

In the vector space of all real-valued functions, find a basis for the subspace spanned by \(\\{\sin t, \sin 2 t, \sin t \cos t\\}\)

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

\([\mathbf{M}]\) Let \(H=\operatorname{Span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) and \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\} .\) Show that \(\mathcal{B}\) is a basis for \(H\) and \(\mathbf{x}\) is in \(H,\) and find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x},\) for $$ \mathbf{v}_{1}=\left[\begin{array}{r}{-6} \\ {4} \\ {-9} \\\ {4}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{8} \\ {-3} \\\ {7} \\ {-3}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}{-9} \\\ {5} \\ {-8} \\ {3}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{4} \\\ {7} \\ {-8} \\ {3}\end{array}\right] $$

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