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Two statements q and r are said to be logically equivalent if they have the same truth-value or logical content. Thus, q is true [url=http://www.1855sacramento.com/moncler.php]moncler sito ufficiale[/url] if and only if p is true. An n x n matrix A is invertible if it has an inverse B such that AB = I (identity matrix) where B = A-1. Alternatively, A is invertible iff det (A) ≠ 0 (null matrix) To prove that the ten statements are equivalent you need to show that 1→2→3…..9→ 10→1
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If A has n pivot positions then A has n leading ones in its low echelon form. If Dx =0 where D is the low echelon form of A, D ≠ 0. This implies that x = 0 (trivial solution). Thus if A has n pivotal position then Ax = 0 has only trivial solutions. Alternatively, 3→4. If Ax = 0 has only trivial solution then there exists a matrix b in R^n such that A (x +b) ≠ 0. This implies that Ax + Ab ≠ 0, since Ab ≠ 0 then Ax≠ 0. This indicates that there exists b in R^n such that Ax=b or 4→5.
If Ax=b has at least one solution b in R^n, then the columns of A in low echelon form a basis for R^n. Or, precisely, each of the vectors in R^n can be written as a linear combination of column [url=http://www.ilyav.com/uggpascher.php]boots ugg pas cher[/url] vectors of A. Thus, A spans R^n or 5→6. If [url=http://www.rtnagel.com/airjordan.php]jordan pas cher[/url] the column vectors of A spans R^n, then every vector (matrix) in R^n can be expressed as a linear combination of the column vectors of A and Ax is contained in R^n for all [url=http://www.thehygienerevolution.com/barbour.php]barbour paris[/url] x in [url=http://www.sandvikfw.net/shopuk.php]hollister outlet sale[/url] R^n. Thus, x-->Ax is a one to one and onto function. Therefore, x-->Ax maps R^n onto R^n or 6→7.
If the linear transformation [url=http://www.ilyav.com/uggpascher.php]ugg pas cher[/url] x-->Ax maps R^n onto R^n, then the set of all linear transformations x-->Ax such that Ax is a member of R^n form a Group. All elements in the group have inverses. Therefore, there exists C such that CA=I. In this case, C is the inverse of A. If A and C are n x n matrix such that CA =I, then C is the inverse of A and, thus, CA = I = AC, or AD=I where D is an n x n matrix, thus, 8→9. If D is an n x n matrix such that AD=I the transpose of A is invertible. This shows that the column vectors of A are linearly independent and they span R^n or A forms a basis for R^n, thus, 9→10.
If the columns of A forms a basis of R^n, then they are linearly independent. Additionally, the column vectors of A are non-zero and they span R^n. Thus, the basis of R^n forms an n x n identity matrix. This indicates that there exist an n x n matrix B such that AB= I or A is invertible, thus, 10→1
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