# If $A \in M_{n \times 1} (\mathbb K)$, then $AA^t$ is diagonalizable.

Let $A \in M_{n \times 1} (\mathbb K)$. I’m asked to proof that $AA^t$ is diagonalizable.

My attempt: If $A = 0, \, AA^t = 0$ is diagonal. Let $A = \begin{bmatrix} a_1\\\vdots \\ a_n \end{bmatrix} \neq 0$, then $AA^t = \begin{bmatrix} a_1\\\vdots \\ a_n \end{bmatrix} \begin{bmatrix} a_1&… & a_n \end{bmatrix} = \begin{bmatrix} a_1 a_1 & \cdots & a_1 a_n \\ \vdots & & \vdots \\ a_n a_1 & \cdots & a_n a_n \end{bmatrix}$. Each column $v_i = \begin{bmatrix} a_ia_1\\ \vdots \\ a_ia_n \end{bmatrix} = a_i \begin{bmatrix} a_1\\ \vdots \\ a_n \end{bmatrix}$. So, $A$ has rank 1 what implies in $A$ diagonalizable or Nilpotent.

How can I see that $A$ is not Nilpotent?

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#### Solutions Collecting From Web of "If $A \in M_{n \times 1} (\mathbb K)$, then $AA^t$ is diagonalizable."

Also: let $u$ be the column vector
$$u=\begin{pmatrix}a_1\\\vdots \\ a_n\end{pmatrix}$$
and $B=AA^t$. Note that $Bu=\|u\|^2u$ and $Bx=0\cdot x$ when $\langle x,u\rangle=0$. So, the eigenvectors generate the whole space and therefore $B$ is diagonalizable.

You cannot prove that, because $AA^t$ can be nilpotent but nonzero.

In fact, when $A\ne0$, the rank-one matrix $AA^t$ is nilpotent if and only if its nilpotency index is $2$. As $(AA^t)^2=A(A^tA)A^t=(A^tA)\cdot(AA^t)$ (where the dot denotes a scalar multiplication), we see that $AA^t$ is nilpotent (hence non-diagonalisable) precisely when $A^tA=0$. And this can indeed happen, such as when $\mathbb K=\mathbb C$ and $A=\pmatrix{1\\i}$.