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Invertible Matrix Theorem

  • Thread starter fk378
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  • #1
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General question regarding the Inv. Matrix Thm:

One part of the theorem states that for an nxn invertible matrix, then there exists at least one solution for each b in Ax=b. Why wouldn't it be "there exists at MOST one solution for each b" since every column/row has a pivot. How would there exist more than one solution for each b if the columns span R_n?
 

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  • #2
Defennder
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Are you sure that's what the theorem says? My book doesn't say "at least one". It says "exactly one". Here's what Wikipedia says:

[PLAIN]http://en.wikipedia.org/wiki/Invertible_matrix_theorem said:
Let[/PLAIN] [Broken] A be a square n by n matrix over a field K (for example the field R of real numbers). Then the following statements are equivalent:

A is invertible.
A is row-equivalent to the n-by-n identity matrix In.
A is column-equivalent to the n-by-n identity matrix In.
A has n pivot positions.
det A ≠ 0.
rank A = n.
The equation Ax = 0 has only the trivial solution x = 0 (i.e., Null A = {0})
The equation Ax = b has exactly one solution for each b in Rn.
The columns of A are linearly independent.
The columns of A span Rn (i.e. Col A = Rn).
The columns of A form a basis of Rn.
The linear transformation mapping x to Ax is a bijection from Rn to Rn.
There is an n by n matrix B such that AB = In.
The transpose AT is an invertible matrix.
The matrix times its transpose, AT × A is an invertible matrix.
The number 0 is not an eigenvalue of A.
 
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  • #3
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Oh, weird. Yeah my book does say "at least one solution". Thanks for showing me the wiki entry though.
 
  • #4
HallsofIvy
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Well, the statement is still true: if a matrix is invertible, then the equation Ax= b has exactly one solution so it is certainly true that there is at least one solution.

If A is not invertible then Ax= b may have no solutions or an infinite number of solutions.

You book may have some reason for emphasizing "the solution exists" right now rather than "the solution is unique"- both of which are true for A invertible.
 
  • #5
HallsofIvy
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Does anyone know how to prove the following theorems:

1) Ax = b is consistent for every n x 1 matrix b

2) Ax = b has exactly one solution for every n x 1 matrix b

3) Ax = b has exactly one solution for at least one n x 1 matrix b

Pls... I need help. It's urgent!!!!
First, don't "hijack" someone else's thread for your own question- that's rude. Use the "new topic" button to start your own thread.

Second, go back and reread the question. You can't prove any of those, they are all false. For example, if A is the 0 matrix, "Ax= b" has NO soution for b non-zero and has an infinite number of solutions if b is 0.
 

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