Proving Invertibility of A^TA with Linearly Independent Columns

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SUMMARY

To prove that the matrix A^TA is invertible when A is an m x n matrix with linearly independent columns, one must consider the implications of linear independence on the rank of A. Since the columns of A are linearly independent, the rank of A equals n, which in turn implies that the rank of A^TA is also n. Consequently, A^TA is a square matrix of full rank, confirming its invertibility.

PREREQUISITES
  • Understanding of linear independence in matrices
  • Familiarity with matrix multiplication and properties
  • Knowledge of matrix rank and its implications
  • Basic concepts of linear transformations
NEXT STEPS
  • Study the properties of matrix rank and its relationship to invertibility
  • Learn about the implications of linear independence in higher dimensions
  • Explore the concept of orthogonality in relation to A^TA
  • Investigate the Singular Value Decomposition (SVD) and its applications
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Students of linear algebra, mathematicians, and anyone involved in theoretical mathematics or applied fields requiring matrix analysis.

eunhye732
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Linear Algebra PLS HELP!

I need help on this problem and been trying to figure it out for awhile.
Let A be an m x n matrix with linearly independent columns. Show that A^TA is invertible.
anything will help. Thanks
 
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I assume you meant A^tA? Anyway, consider the size of that matrix to begin with, that's over half the solution right there. Ah and for future reference, maybe you should post questions like this in the other homework help forum since Linear Algebra is typically taught beyond calculus. Maybe there are some people who exclusively look in that thread to answer problems.
 
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What is the rank of A? What is the rank of A^tA?
 
I'm feeling stupid right now. Disregard my earlier suggestions--while they do work eventually, they are more trouble than they are worth. A good way to approach this problem is by considering what would happen if one of the columns of the product was a linear combination of the other columns, writing out what that would mean, and proceeding to a contradiction.
 
Disregard my earlier suggestions--while they do work eventually, they are more trouble than they are worth.
I don't think they're bad. I think you can make the problem much simpler by doing column operations to write A = A'C where C is the matrix representing the column operations, and A' is of a special form. Of course, it means you have to pick a good special form, but hey! Math is an art. :biggrin:
 
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