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Prove that $$ \mathbf{X}' \mathbf{X} $$ is positive definite if $$ \mathbf{X} $$ is of full column rank.
Solution
By the definition of positive definite matrix, we need to show that $$ \mathbf{c}' \mathbf{X}' \mathbf{X} \mathbf{c} > 0 $$ for $$ \mathbf{c} \neq \mathbf{0} $$. Define $$ \mathbf{z} \equiv \mathbf{X} \mathbf{c} $$. Then $$ \mathbf{c}' \mathbf{X}' \mathbf{X} \mathbf{c} = \mathbf{z}' \mathbf{z} = \sum_{k=1}^{K} z_i^2 $$. If $$ \mathbf{X} $$ is of full column rank, then the column vectors of $$ \mathbf{X} $$ are linearly independent. This means $$ \mathbf{X} \mathbf{c} = \mathbf{0} $$ if and only if $$ \mathbf{c} = \mathbf{0} $$. Then $$ \mathbf{z} \neq \mathbf{0} $$ for any $$ \mathbf{c} \neq \mathbf{0} $$.