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\(\mathbf{X}^{T}\mathbf{X}\) is nonsingular iff the rank…
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\(\mathbf{X}^{T}\mathbf{X}\) is nonsingular iff the rank of \(\mathbf{X}\) equals \(p\)
Bevis
First suppose that \(\mathbf{X}^T\mathbf{X}\) is singular. There exists a nonzero vector u such that \(\mathbf{X}^T\mathbf{X}\mathbf{u}=0\). Multiplying the left-hand side of this equation by \(\mathbf{u}^T\), we have\[0=\mathbf{u}^{T}\mathbf{X}^{T}\mathbf{X}\mathbf{u}=(\mathbf{X}\mathbf{u})^{T}(\mathbf{X}\mathbf{u})\] so \(\mathbf{X}\mathbf{u}=\mathbf{0}\), the columns of \(\mathbf{X}\) are linearly dependent, and the rank of \(\mathbf{X}\) is less than \(p\).
Next, suppose that the rank of \(\mathbf{X}\) is less than \(p\) so that there exists a nonzero vector \(\mathbf{u}\) such that \(\mathbf{Xu}=0\). Then \(\mathbf{X}^T\mathbf{Xu}=0\), and hence \(\mathbf{X}^T\mathbf{X}\) is singular.
(singular matrix = A square matrix that does not have a matrix inverse.
A matrix is singular iff its determinant is 0)
(The rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns.
This corresponds to the maximal number of linearly independent columns of A.
This, in turn, is identical to the dimension of the space spanned by its rows.
is a measure of the nondegenerateness of the system of equations encoded by A)
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A singular matrix has determinant 0 which can be only possible if at least two rows are linearly dependent.
So, the rank of the singular matrix is less than the number of rows or number of columns, whichever is minimum.