Proof: A is Invertible if and only if A^T*A is Invertible | Matrix Algebra

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Discussion Overview

The discussion centers on proving the relationship between the invertibility of a square matrix A and the invertibility of the product A^T*A. The scope includes mathematical reasoning and proof techniques related to matrix algebra.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes to show that A is invertible if and only if A^T*A is invertible, mentioning the rank condition for matrices.
  • Another participant suggests using the determinant as a potential method for the proof.
  • A subsequent reply explains that an invertible matrix M has a non-zero determinant and provides properties of determinants, including that det(M) = det(M^T) and det(MN) = det(M)det(N).
  • It is noted that if A is invertible, then det(A) is non-zero, leading to the conclusion that A^T*A is also invertible.
  • Participants discuss the converse of the argument, indicating that the reasoning applies in both directions.
  • One participant questions the relevance of the title regarding inner products, suggesting a potential disconnect with the main topic.

Areas of Agreement / Disagreement

Participants generally agree on the properties of determinants and their implications for matrix invertibility, but the discussion remains focused on the proof without reaching a final consensus on the overall argument structure.

Contextual Notes

Some assumptions about the dimensions and ranks of matrices are implied but not explicitly stated. The discussion does not resolve all mathematical steps necessary for a complete proof.

physicsss
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Suppose that A is a square. Show that A is invertible if and only if A^T*A is invertible.

I know that if A is an m X n matrix, m>=n, and rkA=n, then the n X n matrix A^T*A is invertible, and that rk(A^TA)= rkA, but I'm still not sure how to start the proof...

TIA
 
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Can you use the determinant?
 
Yes, but how does that help me?
 
M is invertible if and only if det(M) is not zero.

det(M) is the same as det(M^t)

det(MN)=det(M)det(N)

if x and y are real (or complex) numbers and xy=0 then one of x or y is zero.
 
det(A) = det(A^t) and det(AB) = det(A)det(B) (for all matrices A, B of the proper size).

If A is invertible, then det(A) != 0, so that det(A^t) != 0, and therefore det(AA^t) = det(A)det(A^t) != 0, i.e. AA^t is invertible.

The converse is similar.
 
snap. why was the title inner products though?
 

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