- #1

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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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- Thread starter physicsss
- Start date

- #1

- 319

- 0

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

- #2

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Can you use the determinant?

- #3

- 319

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Yes, but how does that help me?

- #4

matt grime

Science Advisor

Homework Helper

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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.

- #5

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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.

- #6

matt grime

Science Advisor

Homework Helper

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snap. why was the title inner products though?

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