Polinom uzayı Polinom Uzayında Spanleme

See also: Linear Algebra, basis vectors, column space, null space, solution space

1. The Problem Statement

Objective: Let $V$ be the vector space $P_2$ (polynomials of degree at most 2). Given the set $S=\{{\vec{v}}_1,{\vec{v}}_2\}$ where:

• ${\vec{v}}_1=t^2+2t+1$
• ${\vec{v}}_2=t^2+2$

Question: Does the set $S$ span the entire vector space $V$? (i.e., Is $\text{Span}(S)=P_2$?)

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2. Theoretical Approach

To span $P_2$, any arbitrary polynomial $\vec{v}=A{t}^2+Bt+C$ must be expressible as a linear combination of ${\vec{v}}_1$ and ${\vec{v}}_2$.

$\vec{v}=a{\vec{v}}_1+b{\vec{v}}_2$

Where $a,b$ are scalars to be determined, and $A,B,C$ are arbitrary constants representing any possible polynomial in the universe of $P_2$.

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3. The Derivation

Step A: Expanding the Equation

We substitute the vectors into the linear combination formula:

$A{t}^2+Bt+C=a(t^2+2t+1)+b(t^2+2)$

Regrouping the terms on the right side by powers of $t$:

$A{t}^2+Bt+C=t^2(a+b)+t(2a)+(a+2b)$

Step B: System of Linear Equations

By comparing coefficients on both sides, we generate the system:

1. $t^2$ term: $a+b=A$
2. $t$ term: $2a=B$
3. Constant term: $a+2b=C$

Step C: Solving for Scalars

From equation (2), we isolate $a$: $a=\frac{B}{2}$

Substitute $a$ into equation (1) to find $b$: $\frac{B}{2}+b=A⟹b=A-\frac{B}{2}$

Step D: The Consistency Check (The Trap)

Now, we must verify if these scalar values satisfy the 3rd equation (the constant term). $C=a+2b$ Substitute our found values: $C=\frac{B}{2}+2\left(A-\frac{B}{2}\right)$ $C=\frac{B}{2}+2A-B$ $C=2A-\frac{B}{2}=\frac{4A-B}{2}$

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4. The Conclusion & Counter-Example

The Result: The system is consistent ONLY IF the polynomial’s coefficients satisfy the specific relationship $C=\frac{4A-B}{2}$. Since a general polynomial in $P_2$ (like $t^2+t+100$) does not necessarily obey this rule, the system is generally inconsistent.

Formal Answer: $\text{Span}(S)\ne P_2$

Counter-Example Proof: Consider the polynomial $\vec{p}=t^2+2t+3$.

• Here, $A=1,B=2,C=3$.
• According to our constraint, $C$ must be: $\frac{4(1)-2}{2}=1$.
• However, our actual $C$ is $3$.
• Since $3\ne 1$, this polynomial cannot be built using ${\vec{v}}_1$ and ${\vec{v}}_2$.

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5. Key Takeaway for Engineering

The Dimensionality Rule: The vector space $P_2$ is isomorphic to ${\mathbb{R}}^3$ (it has 3 dimensions: $t^2,t,1$). To span a 3-dimensional space, you generally need at least 3 linearly independent vectors.

In this problem, we were given only 2 vectors. Geometrically, this means $\{{\vec{v}}_1,{\vec{v}}_2\}$ creates a 2D plane cutting through the 3D space of polynomials. We can reach any point on that plane, but we cannot reach points (polynomials) that float “above” or “below” it.

Rule of Thumb: If the number of vectors in your set is less than the dimension of the space ($n<\dim (V)$), the set cannot span $V$. It can only form a subspace.