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Since row operations can affect linear dependence relations of the row vectors, such a basis is instead found indirectly using the fact that the column space of A T is equal to the row space of A. Using the example matrix A above, find A T and reduce it to row echelon form:
The column rank of A is the dimension of the column space of A, while the row rank of A is the dimension of the row space of A. A fundamental result in linear algebra is that the column rank and the row rank are always equal. (Three proofs of this result are given in § Proofs that column rank = row rank, below.)
The fact that two matrices are row equivalent if and only if they have the same row space is an important theorem in linear algebra. The proof is based on the following observations: Elementary row operations do not affect the row space of a matrix. In particular, any two row equivalent matrices have the same row space.
These row operations are labelled in the table as +, +. Once y is also eliminated from the third row, the result is a system of linear equations in triangular form, and so the first part of the algorithm is complete. From a computational point of view, it is faster to solve the variables in reverse order, a process known as back-substitution.
A system of linear equations is said to be in row echelon form if its augmented matrix is in row echelon form. Similarly, a system of linear equations is said to be in reduced row echelon form or in canonical form if its augmented matrix is in reduced row echelon form. The canonical form may be viewed as an explicit solution of the linear system.
Decomposition: = where C is an m-by-r full column rank matrix and F is an r-by-n full row rank matrix Comment: The rank factorization can be used to compute the Moore–Penrose pseudoinverse of A , [ 2 ] which one can apply to obtain all solutions of the linear system A x = b {\displaystyle A\mathbf {x} =\mathbf {b} } .
In practice, we can construct one specific rank factorization as follows: we can compute , the reduced row echelon form of .Then is obtained by removing from all non-pivot columns (which can be determined by looking for columns in which do not contain a pivot), and is obtained by eliminating any all-zero rows of .
By the rank-nullity theorem, dim(ker(A−λI))=n-r, so t=n-r-s, and so the number of vectors in the potential basis is equal to n. To show linear independence, suppose some linear combination of the vectors is 0. Applying A − λI, we get some linear combination of p i, with the q i becoming lead vectors among the p i.
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