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Consider an $m\times n$ matrix $A$ which is full rank. Is $A(A^\top A)^{-1}A^\top= I$ where $I$ is the identity matrix? If so how can this be shown?

Note: it may be assumed that the matrix $A$ has full column rank and therefore $(A^\top A)^{-1}$ exists.

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    $\begingroup$ But in general $A$ and $B$ are not square and so do not have inverses? $\endgroup$ – Dipole Feb 12 '15 at 16:16
  • $\begingroup$ $(A^tA)^{-1}$ may not exist even when $A$ has full rank. $\endgroup$ – Jim Feb 12 '15 at 16:19
  • $\begingroup$ If $A=(1,0)$ then $A^TA$ is not invertible. $\endgroup$ – Casteels Feb 12 '15 at 16:19
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    $\begingroup$ $A^TA$ is necessarily invertible if we assume $m \geq n$ $\endgroup$ – Omnomnomnom Feb 12 '15 at 16:20
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    $\begingroup$ @Jim this question is likely about real matrices. $\endgroup$ – Omnomnomnom Feb 12 '15 at 16:36
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This cannot happen. Assume $m>n$. Let $rank(A)=n$. Then $A^TA \in M_n$ is invertible.

By the properties of the rank $$ rank(A (A^TA)^{-1} A^T) \le \min(rank(A), rank((A^TA)^{-1}), rank(A^T))=n $$ and hence $$ A (A^TA)^{-1} A^T \ne I_m $$ follows.

To get a feeling for this: consider $A=\pmatrix{1\\0}$. Then $A^TA=\pmatrix{1}$, and $A (A^TA)^{-1} A^T = \pmatrix{1&0\\0&0}$.

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