Opening the book…
Find a basis in \(\mathbb{P}_2\) for the subspace \[\text{span}\{ 1+x+x^2 ,\: 1+2x,\: 1+5x−3x^2\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
- Hint
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This is the situation in which you have a spanning set and you want to cut it down to form a linearly independent set which is also a spanning set. Use the same isomorphism above. Since \(T\) is an isomorphism, it preserves all linear relations so if such can be found in \(\mathbb{R}^3\), the same linear relations will be present in \(\mathbb{P}_2\).
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ 1+x−x^2 +x^3 ,\: 1+2x+3x^3 ,\:−1+3x+5x^2 +7x^3 ,\: 1+6x+4x^2 +11x^3\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
- Answer
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Basis: \(\left\{ 1 + x - x^2 + x^3,\; 1 + 2x + 3x^3 \right\}\)
\(-1 + 3x + 5x^2 + 7x^3 = -5(1 + x - x^2 + x^3) + 4(1 + 2x + 3x^3)\)
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ 1+x−x^2 +x^3 ,\: 1+2x+3x^3 ,\:−1+3x+5x^2 +7x^3 ,\: 1+6x+4x^2 +11x^3\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −2x^2 +x+2,\: 3x^3 −x^2 +2x+2,\: 7x^3 +x^2 +4x+2,\: 5x^3 +3x+2\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 +2x^2 +x−2,\: 3x^3 +3x^2 +2x−2,\: 3x^3 +x+2,\: 3x^3 +x+2\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −5x^2 +x+5,\: 3x^3 −4x^2 +2x+5,\: 5x^3 +8x^2 +2x−5,\: 11x^3 +6x+5\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −x^2 +x+1,\: 3x^3 +2x+1,\: 4x^3 +x^2 +2x+1,\: 3x^3 +2x−1\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
- Answer
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Basis: \(\left\{ x^3 - x^2 + x + 1,\; 3x^3 + 2x + 1,\; 4x^3 + x^2 + 2x + 1,\; 3x^3 + 2x - 1 \right\}\). No vector is a linear combination of the others — they form a basis for a 4-dimensional subspace of \(\mathbb{P}_3\), which in fact is \(\mathbb{P}_3\).
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −x^2 +x+1,\: 3x^3 +2x+1,\: 13x^3 +x^2 +8x+4,\: 3x^3 +2x−1\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −3x^2 +x+3,\: 3x^3 −2x^2 +2x+3,\:−5x^3 +5x^2 −4x−6,\: 7x^3 +4x−3\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −2x^2 +x+2,\: 3x^3 −x^2 +2x+2,\: 7x^3 −x^2 +4x+4,\: 5x^3 +3x−2\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −2x^2 +x+2,\: 3x^3 −x^2 +2x+2,\: 3x^3 +4x^2 +x−2,\: 7x^3 −x^2 +4x+4\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −4x^2 +x+4,\: 3x^3 −3x^2 +2x+4,\:−3x^3 +3x^2 −2x−4,\:−2x^3 +4x^2 −2x−4\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 +2x^2 +x−2,\: 3x^3 +3x^2 +2x−2,\: 5x^3 +x^2 +2x+2,\: 10x^3 +10x^2 +6x−6\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 +x^2 +x−1,\: 3x^3 +2x^2 +2x−1,\: x^3 +1,\: 4x^3 +3x^2 +2x−1\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
Find a basis in \(\mathbb{P}_3\) for the subspace \[\text{span}\{ x^3 −x^2 +x+1,\: 3x^3 +2x+1,\: x^3 +2x^2 −1,\: 4x^3 +x^2 +2x+1\}\nonumber\] If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.
- Answer
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Basis: \(\left\{ x^3 - x^2 + x + 1,\; 3x^3 + 2x + 1,\; 4x^3 + x^2 + 2x + 1 \right\}\)
\(x^3 + 2x^2 - 1 = -2(x^3 - x^2 + x + 1) + 1(3x^3 + 2x + 1)\)
Here are some vectors. \[\{ x^3 +x^2 −x−1,\: 3x^3 +2x^2 +2x−1\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Here are some vectors. \[\{ x^3 −2x^2 −x+2,\: 3x^3 −x^2 +2x+2\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Here are some vectors. \[\{ x^3 −3x^2 −x+3,\: 3x^3 −2x^2 +2x+3\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Here are some vectors. \[\{ x^3 −2x^2 −3x+2,\: 3x^3 −x^2 −6x+2,\:−8x^3 +18x+10\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Here are some vectors. \[\{ x^3 −3x^2 −3x+3,\: 3x^3 −2x^2 −6x+3,\:−8x^3 +18x+40\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Here are some vectors. \[\{ x^3 −x^2 +x+1,\: 3x^3 +2x+1,\: 4x^3 +2x+2\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Here are some vectors. \[\{ x^3 +x^2 +2x−1,\: 3x^3 +2x^2 +4x−1,\: 7x^3 +8x+23\}\nonumber\] If these are linearly independent, extend to a basis for all of \(\mathbb{P}_3\).
Let \(\mathbb{P}_3\) be the polynomials of degree no more than \(3\). Determine which of the following are bases for this vector space.
- \(\{ x+1,\: x^3 +x^2 +2x,\: x^2 +x,\: x^3 +x^2 +x\}\)
- \(\{ x^3 +1,\: x^2 +x,\: 2x^3 +x^2 ,\: 2x^3 −x^2 −3x+1\}\)
- Answer
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- The given vectors are linearly independent. Since \(\text{dim}(\mathbb{P}_3)=4\) and we are given 4 linearly independent vectors, they form a basis.
- This set of vectors are also linearly independent. Hence, by the same logic as in part a, they form a basis.
Suppose \(V\) is a finite dimensional vector space. Based on the Exchange Theorem, any two bases have the same number of vectors in them. Give a different proof of this fact using the earlier material in the book.
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Suppose \(\{\vec{x}_1,\cdots ,\vec{x}_n\}\) and \(\{\vec{y}_1,\cdots , \vec{y}_m\}\) are two bases with \(m < n\). Then define \[φ : \mathbb{R}^n \mapsto V,\: ψ :\mathbb{R}^m\mapsto V\nonumber\] by \[φ (\vec{a}) = \sum\limits_{k=1}^n a_k\vec{x}_k ,\: ψ(\vec{b}) =\sum\limits_{j=1}^m b_j\vec{y}_j\nonumber\] Consider the linear transformation, \(ψ^{−1}\circ φ\). Argue it is a one to one and onto mapping from \(\mathbb{R}^n\) to \(\mathbb{R}^m\). Now consider a matrix of this linear transformation and its reduced row-echelon form.
This page titled 7.6.E: Exercises for Section 7.6 was last modified on Sat, 06 Sep 2025 05:04:43 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.
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