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Consider the following matrix $ A = \begin{bmatrix} 0 & 4 & 4 \\ 1 & 1 & 1 \\ 4 & 0 & 4 \\ 4 & 4 & 0 \\ 1 & 1 & 1 \end{bmatrix}$ over $\mathbb{F}_5^{5 \times 3}$.

This matrix has full column rank (rank = 3). When I compute the left inverse, the matrix $A^TA$ is computed to be

$A^TA = \begin{bmatrix} 4 & 3 & 3 \\ 3 & 4 & 3 \\ 3 & 3 & 4 \end{bmatrix}$

This matrix has determinant 0 (and rank 2) and naturally the inverse doesn't exist. So computation of $(A^TA)^{-1}A^T$ is not possible.

Can someone explain to me why even after A having a full column rank failed to have a left inverse?

Added question: The same matrix $A$ when considered over $\mathbb{R}$ (or $\mathbb{Q}$) does have a left inverse. So is the condition for the existence of left (or right) inverse different for matrices over finite fields?

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  • $\begingroup$ In general $\text{rank}(A^TA)\leq \text{rank}(A)$. In fields of characteristic zero this becomes equality, but the proof relies on looking positive definiteness of $\big \Vert Ax \big\Vert_2\geq 0$ which doesn't hold for finite fields. The problem with your computation is $A$ is injective and $A^T$ is surjective so in some sense you have the ordering backwards. if you looked at $AA^T$, you'd see that rank$(AA^T) =$ rank$(A) = 3$ but there isn't a reason for $AA^T$ and $A^TA$ to have the same rank ('comparable' characteristic polynomials isn't enough when you don't have spectral theorem). $\endgroup$ Jan 20, 2020 at 21:27
  • $\begingroup$ to reinforce the point -- (i) in my above comment the transpose naturally should be interpreted as the conjugate transpose if for some reason the scalars were in $\mathbb C$. (ii) A simpler example using a finite field for the OP and @DietrichBurde to consider is $B := \left[\begin{matrix}1 & 0\\0 & 1\\1 & 0\\0 & 1\end{matrix}\right]$ with scalars in $\mathbb F_2$. Then $B$ has full column rank of 2 but $B^T B = \left[\begin{matrix}0 & 0\\0 & 0\end{matrix}\right]$, so $B^T B$ certainly isn't invertible. note: $0 = \text{rank} (B^T B) \lt \text{rank} (BB^T) = \text{rank} (B) = 2$ $\endgroup$ Jan 21, 2020 at 7:22

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It does have a left inverse. $$\begin{pmatrix} 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 4 & 4 & 4 & 0 & 0 \end{pmatrix}$$

The usual proof Why is $A^TA$ invertible if $A$ has independent columns? that $A^TA$ is invertible uses dot products. But dot products don't have good properties over finite fields, because it is possible that $\langle x, x\rangle = 0$ when $x \ne 0$. The columns of $A^TA$ in your example all have this property.

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