Equivalent Statements in Algebra - Essay Example

2) The equation Ax=0 has only the trivial solution because A is invertible. Will have only a trivial solution. Exists, so multiply on the left by A1 to get:
Since A is invertible, the equation has a solution for every vector b (namely ) so the column vectors span Rn.
4) A is row equivalent to the n x n identity matrix because two matrices are row equivalent if we can get from one to the other by applying a finite set of elementary row operations.
After writing the augmented matrix for this system, use elementary row operations to reduce this to a reduced row-echelon form. If the entries in the last column do not affect the values in the entries in the first n columns and if we take the same set of elementary row operations and apply them to A we will get In and so A is row equivalent to In since we can get to In by applying a finite set of row operations to A.
5) In linear algebra, an n-by-n (square) matrix A is called invertible, non-singular, or regular.

If an n-by-n matrix exists the matrix must be square.
6) A is expressible as a product of elementary matrices because every elementary matrix is invertible and its inverse is also an elementary matrix. A square matrix is only invertible if it can be written as a product of an elementary matrix.
7) A family of vectors is linearly independent if none of them can be written as a linear combination of infinitely many other vectors in the collection. An alternative method uses the fact that n vectors in Rn are linearly dependent if and only if the determinant of the matrix formed by the vectors is zero.
8) To prove that columns span Rn, finding the basis of the column space is essential. Then you must reduce the matrix and see if the rank = n; there has to be n linearly independent. columns, which is very easy to see if the matrix is reduced. The basis is the columns of the original matrix, not the columns of the reduced form. An invertible matrix A, determinant (A) != 0 (not equal to zero).
9) The columns of A form a basis of Rn. basis B of a vector space V is a linearly independent subset of V that spans (or generates) V.
In more detail, suppose that B = { v1, …, vn } is a finite subset of a vector space V over a field F. Then, B is a basis, if it satisfies the following conditions:
• the linear independence property,
for all a1, …, an ∈ F, if a1v1 + … + anvn = 0, then necessarily a1 = … = an = 0; and
• the spanning property,
for every x in V it is possible to choose a1, …, an ∈ F such that x = a1v1 + … + anvn.

10) Rank A = n because, in linear algebra, the column rank (row rank respectively) of a matrix A with entries in some field is defined to be the maximal number of columns (rows respectively) of A which are linearly independent.
The column rank and the row rank are equal; this common number is simply called the rank of A. It is commonly denoted by either rk (A) or rank A.