Description
Let X = C([0,1]) denote the space of continuous functions defined in the unit interval. Prove that the map.
- Consider a basis of R3 composed of the vectors
(1,0,−1), (1,1,1) and (2,2,0)
find its dual basis.
- Prove that the determinant, interpreted as a transformation
D : Rn2 →R with D(A) = determinant(A)
is linear in each of the rows. That is, if a row R of the matrix A is given by R = αR1 + βR2 with R1,R2 ∈Rn and α,β ∈R, then
D(A) = αD(A1) + βD(A2)
where Ai is the matrix constructed by taking A and replacing row R with tow Ri. This property is denoted as the determinant is a multilinear transformation row by row.
- Prove that the determinant map D : Rn2 →R defined above is alternating, i.e. if rows Ri and Rj in a matrix
R1 R1
… …
Ri, ! Rj, !
A = … are exchanged to obtain a new matrix A˜ = …
Rj Ri
… …
Rn Rn
then D(A) = −D(A˜).
- Prove that for 2×2 matrices the determinant is the only map D : R4 → R that is both multilinear as a function of the 2 rows and alternating, and that takes the value D(I) = 1 at the identity. The proof can be
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done directly, using multilinearity and the alternating property. Just write any row in the matrix as a sum of vectors in the canonical basis.
Note This result, a characterization of the determinant, holds in any dimensions and can be used as an alternative (and equivalent) definition of the determinant.
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