Construct bases for coordinate, solution and polynomial spaces.
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Try
A plane needs two independent directions to describe every point in it. A list with three spanning vectors can still be useful, but it contains redundancy. A basis has exactly the two required properties: it spans the space and is independent.
Predict
Predict whether unit length is required for a basis. It is not. The ordered list B=((1,1),(1,−1)) is a basis of R² despite its vectors having length √2. Ordering matters because it determines which coefficient is first. Dimension is the number of vectors in a basis.
Experiment
Express (4,2) in B by solving a+b=4 and a−b=2. The answer is coefficients (3,1), different from the standard coordinates (4,2). Verify the reconstruction. Then remove one basis vector and identify a target that is no longer reachable.
The vector v=(4,2) equals 3b₁+b₂ for b₁=(1,1), b₂=(1,−1). Equal scales preserve the geometry. The dotted path adds three copies of b₁ followed by one b₂.
Understand
Worked basis construction: the plane x+y+z=0 has vectors (s,t,−s−t)=s(1,0,−1)+t(0,1,−1). These two directions span the plane and are independent by their first two coordinates, so its dimension is 2. The surrounding R³ has dimension 3; ambient dimension and subspace dimension differ.
In any finite-dimensional vector space, an independent list can be extended to a basis, and a finite spanning list can be reduced to one. The exchange principle shows all bases have the same length: replacing spanning directions with independent ones cannot require more independent vectors than the spanning list contains.
Invalid conclusion: three vectors in R³ automatically form a basis. They might lie in one plane. One must establish both sufficient reach and lack of redundancy.
Apply
For every proposed basis, prove spanning and independence. You may combine these by proving that every target has exactly one representation, but say why. For polynomial spaces, equality of coefficients replaces coordinate equations for geometric vectors.
Continue
Depth: the zero space has the empty basis and dimension zero. In an n-dimensional space, an independent list of n vectors automatically spans, and a spanning list of n vectors is automatically independent. These shortcuts depend on a known dimension, not on appearance.
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What is the dimension of the real polynomial space P₂ of degree at most two?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For basis B=((1,1),(1,−1)), vector (5,1)=a(1,1)+b(1,−1). Enter a.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
A basis of a vector space is:
Four vectors in R³ are:
Find a basis and dimension for W={(x,y,z):x+y+z=0}.
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Prove 1,1+t,1+t+t² form a basis of P₂.
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
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