If A is row vector 1×n then is $A^tA$ always diagonizable?

Question

If $A$ is row vector 1×n then is $$A^tA$$ always diagonizable?
Multiplication of $A$ transpose times $A$ gives a matrix of $n×n$ and i was able to prove that this matrix is diagonizable over $\mathbb R$ but i cant prove it over finite fields or over $\mathbb C$.I also can't find counter-examples either. Is the statement true?

Answer 1

No. The classical counterexamples are given by $A=(1,i)$ over $\mathbb C$ or $A=(1,1)$ over $GF(2)$. We have $A^tA\ne0$ (because $A\ne0$) but $(A^tA)^2=A^t(AA^t)A=0$. Therefore $A^tA$ is a nonzero nilpotent matrix, which is non-diagonalisable.

In general, for any two nonzero column vectors $u$ and $v$ of the same size, the matrix $uv^T$ is diagonalisable if and only if $v^Tu\ne0$. We have essentially explained the necessity of $v^Tu\ne0$ above: if $v^Tu=0$, then $uv^T$ will be a nonzero nilpotent matrix and hence it is not diagonalisable. For sufficiency, if $\lambda=v^Tu\ne0$, then $u$ is an eigenvector of $M=uv^T$ corresponding to the nonzero eigenvalue $\lambda$. Since $M$ is rank-one, its nullity is $n-1$. Therefore $u$ together with a basis of $\ker(M)$ form an eigenbasis of $M$.