# Linear algebra matrix inverse identity

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.

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

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 $A (A^TA)^{-1} A^T = \pmatrix{1&0\\0&0}$.