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By Kallenrode

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This has to do with finite number fields (try F2 ), and the only available example is rather dull. 5 All bases have equally many vectors We have seen that any linearly independent set of n vectors in an n-dimensional space is a basis. The following statement shows that a basis cannot have fewer than n vectors. The proof is somewhat long and can be skipped unless you would like to gain more facility with coordinate-free manipulations. Theorem: In a finite-dimensional vector space, all bases have equally many vectors.

10) j=1 j=1 where λj , µj are some constants, not all equal to zero. Suppose all λj are zero; then fl+1 would be a linear combination of other fj ; but this cannot happen for a basis {fj }. Therefore not all λj , 1 ≤ j ≤ k are zero; for example, λs = 0. This gives us the index s. Now we can replace es in the set S by fl+1 ; it remains to prove that the resulting set T defined by Eq. 9) is linearly independent. 2 Linear maps in vector spaces where ρj , σj are not all zero. In particular, σl+1 = 0 because otherwise the initial set S would be linearly dependent, l k s−1 σj fj = 0.

However, it does not mean that matrices are always needed. e. g. ” In this book I concentrate on general properties of linear transformations, which are best formulated and studied in the geometric (coordinate-free) language rather than in the matrix language. Below we will see many coordinate-free calculations with linear maps. In Sec. 8 we will also see how to specify arbitrary linear transformations in a coordinate-free manner, although it will then be quite similar to the matrix notation. Exercise 1: If V is a one-dimensional vector space over a field K, prove that any linear operator Aˆ on V must act simply as a multiplication by a number.

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A new approach to linear filtering and prediction problems by Kallenrode


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