AlephZero said:Prove it by contradition. If the columns of X are linearly dependent, there is a non-zero vector ##y## such that ##X^TXy = 0##.
Dickfore said:Just one remark, instead of "contradiction", maybe you should have used "contraposition".
In linear algebra, a square matrix is considered "non-singular" if it has an inverse, meaning it is invertible and has a unique solution to the equation Ax = b. In the case of A^T*A, this means that the matrix must have a non-zero determinant.
The non-singularity of A^T*A is important because it guarantees that the matrix is full rank, meaning it has linearly independent columns. This allows for unique solutions to be found for systems of equations involving A^T*A.
Yes, it is possible for A^T*A to be non-singular even if A is singular. This is because the transpose of a singular matrix may have linearly independent columns, resulting in a non-singular A^T*A.
The size of A does not affect the non-singularity of A^T*A. As long as A is a square matrix, the resulting A^T*A will also be a square matrix with the same number of rows and columns, and the same rules for non-singularity apply.
One application is in data analysis, where A represents a matrix of independent variables and A^T*A is used in calculating regression coefficients. The non-singularity of A^T*A ensures that the regression model is well-defined and has a unique solution. It is also used in optimization problems and in solving systems of equations in engineering and physics.