Invertibility of a product of invertible matrices

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Discussion Overview

The discussion centers on the invertibility of the product of three invertible matrices, A, B, and C, specifically whether the product ABC is also invertible. The scope includes theoretical aspects of linear algebra and properties of matrix multiplication.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • Some participants assert that if A and B are invertible, then their product AB is also invertible, and consequently, the product ABC will be invertible if C is also invertible.
  • One participant highlights that the non-zero determinant of invertible matrices implies that the product of these matrices will also have a non-zero determinant, suggesting that ABC is invertible.
  • Another participant demonstrates the invertibility of the product by showing that (AB)(B^{-1}A^{-1}) equals the identity matrix I, using properties of matrix multiplication and associativity.
  • A later reply draws an analogy between matrix multiplication and the composition of bijective linear maps, stating that the composition of bijective maps remains bijective, thus supporting the claim of invertibility.

Areas of Agreement / Disagreement

Participants generally agree on the principle that the product of invertible matrices is invertible, though the discussion includes various approaches and explanations without explicit consensus on all details.

Contextual Notes

The discussion does not address potential limitations or assumptions regarding the dimensions of the matrices beyond being nxn or the specific properties of the matrices involved.

Who May Find This Useful

Readers interested in linear algebra, matrix theory, and properties of invertible matrices may find this discussion relevant.

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If A,B, and C are each nxn invertible matrices, will the product ABC be invertible?
 
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yea, if A, B are nxn invertible then AB is invertible. Since Matrix Multiplication is associative, (AB)C is invertible provided c is nxn invertible.
 
If your linear algebra is better than your basic set/function theory, remark that A,B,C have non-zero determinant, hence their product also.
 
If A and B are both invertible then (AB)(B^{-1}A^{-1})= A(B^{-1}B)A^{-1} because, as jakncoke says, matrix multiplication is associative. Of course, B^{-1}B= I so that becomes (AB)(B^{-1}A^{-1})= AA^{-1}= I. Simlarly, (B^{-1}A^{-1})(AB)= B^{-1}(A^{-1}A)B^{-1}= B^{1}B= I

You can extend that to any number of factors by induction and repeated use of associativity.
 
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As Landau referenced, if you like to think about functions, matrix multiplication is equivalent to applying a linear map, and invertible matrices define bijective linear maps (and vice versa), and a composition of bijective linear maps is a bijective linear map. so yes.
 

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