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| Linear Relations Among Columns Unchanged |
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Proof. Let the vector space be spanned by , and let be a linearly independent sequence. The task is to show .
Since is a spanning set, one has for each , . One may express this very concisely by writing where is the row of elements of , is the row of elements of , and is the matrix .
If , then the reduced row echelon form of can have at most non-zero rows and, therefore, at most pivot columns. So at least one column of is not a pivot column. This means that column is a linear combination of the pivot columns to its left. Hence, if is the column of coefficients of the ensuing linear relation with , then one has which means that cannot be linearly independent, a contradiction made possible by assuming . Hence .
Proof. Certainly a linearly independent spanning set must be a maximal linearly independent set. Conversely if is a maximal linearly independent set, and is any element of , then the set cannot be linearly independent by the maximality of . Hence, there must be a non-trivial linear relation among the members of a finite subset of . The element must be involved with non-zero coefficient in that linear relation since there can be no such relation among finitely many members of . That linear relation can be used to obtain as a linear combination of finitely many members of . Therefore , which was an arbitary member of , lies in the span of .
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On the other hand, the row space of any matrix is the same linear subspace of as the row space of its reduced row echelon form. It is obvious that a basis of the row space of a matrix in reduced row echelon form is given by the set of its non-zero rows.
Since for a matrix in reduced row echelon form, the number of non-zero rows is equal to the number of columns containing leading 's, the dimension of the row space of a given matrix is equal to the dimension of its column space.
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For any matrix : For any in : If is linear, then Given a linear map choose to be the unique matrix with Then for all in : Therefore,
Assume , , and where is an matrix.
Recall that the image of is spanned by the columns of . Thus, the image of is the same subspace of as the column space of . Hence,
Recall that the kernel of is the subspace of consisting of all in such that . When the system of linear equations represented by the matrix equation is put in reduced row echelon form, one is able to solve for the pivot column variables in terms of the other variables. Thus, the non-pivot column variables may be used as parameters for the space of solutions, i.e., for the kernel of . The “space of parameters” has a “standard basis” consisting of value sets for these parameters. Hence,
Since , the question of linear relations among the columns is equivalent to the question of finding solutions of the linear system . Compute the reduced row echelon form: With the reduced row echelon form the non-pivot columns are transparently expressed in terms of the pivot columns: or
Exchange columns and rows. Then follow the method used for exercise 1. The matrix obtained by exchanging columns and rows in a given matrix is called the transpose of the given matrix.
The transpose:
The RREF of the transpose:
Column relations in the tranpose give row relations in the original:
Linear relations among the rows of correspond to equations satisfied by points in the image of the linear map Form the generic augmented matrix Use row operations to maneuver this augmented matrix so that the coefficient matrix portion is in reduced row echelon form, and then extract the linear relations among the coordinates of arising from zero rows in the coefficient portion.