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Taylor
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It's a property of the $\text{rank}$ operator when its used on real matrices $\mathbf{A}$: $$ \text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}^T) = \text{rank}(\mathbf{A}^T\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}^T). $$$$ \text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}') = \text{rank}(\mathbf{A}'\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}'). $$

In your case, the data matrix $\mathbf{X}$ is$\mathbf{X} \in \mathbb{R}^{n \times p}$ is usually tall and skinny ($n > p$), so the rank of everything is the number of linearly independent columns/predictors/covariates/independent variables. If everything is linearly independent $\text{rank}(\mathbf{X}) = p$, and so you have $\mathbf{X}'\mathbf{X}$ is invertible. If you have collinearity, or columns that can be written as linear combinations of others, then $\text{rank}(\mathbf{X}) < p$, and you cannot find a unique inverse for $\mathbf{X}'\mathbf{X}$ (you can, however, find generalized inverses for it).

It's a property of the $\text{rank}$ operator when its used on real matrices $\mathbf{A}$: $$ \text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}^T) = \text{rank}(\mathbf{A}^T\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}^T). $$

In your case, the data matrix $\mathbf{X}$ is usually tall and skinny, so the rank of everything is the number of linearly independent columns/predictors/covariates/independent variables.

It's a property of the $\text{rank}$ operator when its used on real matrices $\mathbf{A}$: $$ \text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}') = \text{rank}(\mathbf{A}'\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}'). $$

In your case, the data matrix $\mathbf{X} \in \mathbb{R}^{n \times p}$ is usually tall and skinny ($n > p$), so the rank of everything is the number of linearly independent columns/predictors/covariates/independent variables. If everything is linearly independent $\text{rank}(\mathbf{X}) = p$, and so you have $\mathbf{X}'\mathbf{X}$ is invertible. If you have collinearity, or columns that can be written as linear combinations of others, then $\text{rank}(\mathbf{X}) < p$, and you cannot find a unique inverse for $\mathbf{X}'\mathbf{X}$ (you can, however, find generalized inverses for it).

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Taylor
  • 21.5k
  • 2
  • 49
  • 88

It's a property of the $\text{rank}$ operator when its used on real matrices $\mathbf{A}$: $$ \text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}^T) = \text{rank}(\mathbf{A}^T\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}^T). $$

In your case, the data matrix $\mathbf{X}$ is usually tall and skinny, so the rank of everything is the number of linearly independent columns/predictors/covariates/independent variables.