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Chapter 3: Problem 11

Let \(S\) be the subspace of \(P_{3}\) consisting of all polynomials of the form \(a x^{2}+b x+2 a+3 b\). Find a basis for \(S\).

Short Answer

Expert verified
A basis for the given subspace S is \(\{x^2 + 2, x + 3\}\).

Step by step solution

01

Identify the given subspace and polynomial form

We are given the subspace S of polynomials of degree 3 or less, in the form of \(a x^{2}+b x+2 a+3 b\).
02

Determine the coefficients of the polynomial

The polynomial has the general form \(a x^{2}+b x+2 a+3 b\). Now identify the coefficients of the polynomial: - Coefficient of \(x^2\): a - Coefficient of \(x\): b - Constant term: 2a + 3b
03

Rewrite the polynomial in terms of the coefficients

Rewrite the polynomial as a linear combination of the coefficients of its terms: \(p(x) = a(x^2) + b(x) + (2a + 3b)(1)\)
04

Express the polynomial as linear combinations of basis polynomials

Express the polynomial in terms of the basis polynomials \(x^2, x, 1\): \(p(x) = a(x^2) + b(x) + 2a(1) + 3b(1)\) \(p(x) = a(x^2 + 2) + b(x + 3)\)
05

Form a basis for the subspace

From step 4, we can see that the subspace S can be generated by the polynomials \(x^2 + 2\) and \(x + 3\). We can form a basis for the subspace S using these two polynomials: \(\{x^2 + 2, x + 3\}\).
06

Verify the basis

Verify that the generated set \(\{x^2 + 2, x + 3\}\) is linearly independent and spans the subspace: - Linear independence: There are no scalar multiples or linear combinations of \(x^2 + 2\) and \(x + 3\) that would result in the zero polynomial, which confirms their linear independence. - Spanning the subspace: Given any polynomial in the subspace S, we can express it as a linear combination of \(x^2 + 2\) and \(x + 3\). Therefore, they span the subspace.
07

Conclusion

The basis for the given subspace S is \(\{x^2 + 2, x + 3\}\).

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Let \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{z}\) be vectors in a vector space \(V .\) Prove that if \\[ \mathbf{x}+\mathbf{y}=\mathbf{x}+\mathbf{z} \\] then \(\mathbf{y}=\mathbf{z}\)

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