Lineare Algebra I [Lecture notes]

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Gaussian Elimination 25m. LU Decomposition 30m. Week Two 30m. Video 13 videos. Vector Spaces 7m.


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Linear Independence 9m. Span, Basis and Dimension 10m. Gram-Schmidt Process 13m. Gram-Schmidt Process Example 9m. Null Space 12m. Application of the Null Space 14m. Column Space 9m. Orthogonal Projections 11m. The Least-Squares Problem 10m. Solution of the Least-Squares Problem 15m. Reading 14 readings. Practice: Zero Vector 10m. Practice: Examples of Vector Spaces 10m.

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Practice: Line of Best Fit 10m. Vector Space Definitions 15m. Gram-Schmidt Process 30m. Fundamental Subspaces 30m. Orthogonal Projections 30m. Week Three 50m.

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Two-by-Two and Three-by-Three Determinants 8m. Laplace Expansion 13m. Leibniz Formula 11m. Properties of a Determinant 15m. The Eigenvalue Problem 12m. Finding Eigenvalues and Eigenvectors 1 10m. Finding Eigenvalues and Eigenvectors 2 7m. Matrix Diagonalization 9m. Matrix Diagonalization Example 15m. Powers of a Matrix 5m. Powers of a Matrix Example 6m.

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Concluding Remarks 3m. Reading 20 readings. Practice: Determinant of the Identity Matrix 10m. Practice: Row Interchange 10m. Practice: Determinant of a Matrix Product 10m. Practice: Complex Eigenvalues 10m. Practice: Linearly Independent Eigenvectors 10m. Practice: Invertibility of the Eigenvector Matrix 10m.

Practice: Diagonalize a Three-by-Three Matrix 10m. Practice: Matrix Exponential 10m. Practice: Powers of a Matrix 10m.

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Please Rate this Course 10m. Acknowledgments 1m. Clearly generates. If the are linearly independent then we are done. Otherwise, there is a non-trivial linear relation between them; after shuffling things around, we see that one of the , say , is a rational linear combination of the.

In such a case, becomes redundant, and we may delete it reducing the rank by one. We repeat this procedure; it can only run for at most steps and so terminates with obeying both of the desired properties. In additive combinatorics, one often wants to use results like this in finitary settings, such as that of a cyclic group where is a large prime.

Lecture Notes on Linear Algebra

Now, technically speaking, is not a vector space over , because one only multiply an element of by a rational number if the denominator of that rational does not divide. On the other hand, saying that one element of is a rational linear combination of another set of elements is not a very interesting statement: any non-zero element of already generates the entire space! However, if one again restricts attention to rational linear combinations of bounded height , then things become interesting again.

For instance, the vector can generate elements such as or using rational linear combinations of bounded height, but will not be able to generate such elements of as without using rational numbers of unbounded height. But again, if one restricts attention to rational numbers of bounded height, then independence begins to emerge: for instance, and are independent in this sense.

Of course to make this rigorous, one needs to quantify the two heights here, the one giving the independence, and the one giving the generation. In order to be useful for applications, it turns out that one often needs the former height to be much larger than the latter; exponentially larger, for instance, is not an uncommon request. Fortunately, one can accomplish this, at the cost of making the height somewhat large:. Theorem 2 Finite generation implies finite basis, finitary version Let be an integer, and let be a function.

Let be an abelian group which admits a well-defined division operation by any natural number of size at most for some constant depending only on ; for instance one can take for a prime larger than.


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Then there exists a collection of vectors in , with , as well an integer , such that. Start with initialised to so initially we have , and initialise.

initgegete.tk Clearly generates at this height. If the are linearly independent up to rationals of height then we are done.


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Otherwise, there is a non-trivial linear relation between them; after shuffling things around, we see that one of the , say , is a rational linear combination of the , whose height is bounded by some function depending on and. In such a case, becomes redundant, and we may delete it reducing the rank by one , but note that in order for the remaining to generate we need to raise the height upper bound for the rationals involved from to some quantity depending on.

We then replace by and continue the process. We repeat this procedure; it can only run for at most steps and so terminates with and obeying all of the desired properties.