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

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A square matrix A is invertible if and only if the matrix product A^T*A is also invertible. The proof involves using properties of determinants, where a matrix is invertible if its determinant is non-zero. If A is invertible, then both det(A) and det(A^T) are non-zero, leading to the conclusion that det(A^T*A) is also non-zero. Conversely, if A^T*A is invertible, it implies that det(A^T*A) is non-zero, which indicates that A must be invertible as well. The discussion also touches on the relevance of inner products, although this point is less central to the main proof.
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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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