Question on a real matrix of rank 1

Question

Let $A$ be a real matrix of order $n$ having rank $1$.Then I need to show the following :

1. There exists a real number $a$ such that $A^2$ $\neq$ $aA$
2. If $A$ has a non-zero eigenvalue, then $A$ is diagonalizable.

I can't understand how to do this.

Is there any relation between number of non-zero eigenvalues and rank? Help, please!

Answer 1

rank$(\boldsymbol{A})=1$ is equivalent to $\boldsymbol{A}$ has a representation of $\boldsymbol{A}=\boldsymbol{u}\boldsymbol{v}^T $ for some vectors $\boldsymbol{u},\boldsymbol{v}\in \mathbb{R}^n\setminus \{0\}$ since all lines and columns are linearly dependent.

Now, distinguish the cases $\boldsymbol{v}^T\boldsymbol{u}=0$ and $\boldsymbol{v}^T\boldsymbol{u}\neq0$.

If $\boldsymbol{v}^T\boldsymbol{u}\neq0$, you can set $a=0$ in problem 1. If $\boldsymbol{v}^T\boldsymbol{u}=0$, you can set $a=1$ (or any other number $\neq 0$).

$\boldsymbol{A}\boldsymbol{w}=\lambda\boldsymbol{w}$ is equivalent to $\boldsymbol{u}\boldsymbol{v}^T\boldsymbol{w}=\lambda \boldsymbol{w}$. So, the only non zero eigenvector $\boldsymbol{A}$ can have is $\frac{\boldsymbol{u}}{\left\|\boldsymbol{u}\right\|}$ to the eigenvalue $\frac{\boldsymbol{v}^T\boldsymbol{u}}{\left\|\boldsymbol{u}\right\|}$. The other eigenvectors are the $n-1$ vectors, which are orthogonal to $\boldsymbol{v}$ with eigenvalue $0$. If you know, that there is a non zero eigenvalue, it must hold $\boldsymbol{v}^T\boldsymbol{u}\neq 0$, so you know that the eigenvectors span the space $\mathbb{R}^n$ which implies that $\boldsymbol{A}$ is diagonalizeable.