Let $A$ be a square matrix $n\times n$ of rank $1$. I want to find a necessary and sufficient condition for the diagonalizable of this particular form of $A$.
I know equivalent conditions for diagonalizability of $n\times n$ matrix $A$.
- Its minimal polynomial has no repeated roots.
- It has $n$ linearly independent eigenvectors.
I found interesting posts about the full rank of $A$ in Diagonalizable vs full rank vs nonsingular (square matrix) ….,
Best Answer
First, we have the following lemma
Lemma: $A$ has rank $1$ iff there exist non-zero column vectors $v, w\in \mathbb R^n$ such that $A = vw^T$ (see e.g. MSE).
Theorem: For non-zero vectors $v, w$ in $\mathbb R^n$, $A=vw^T$ is diagonalizable if and only if $v, w$ are non-othorgonal.
Note. $w^Tv\neq 0$ is equivalent to the trace of $A$ is non-zero.
Proof of theorem:
Part 1 ($\Longleftarrow$). Assume that $w^Tv\neq 0$. We have $\lambda=w^Tv\neq 0$ and $v_0=v\neq 0$ is a pair of eigenvalue and eigenvector, since $Av_0 = vw^Tv= \lambda v$. Let $v_1,...,v_{n-1}$ be a basis of $w^{\bot}=\{ u: w^Tu=0\}$, here $\dim w^{\bot}=n-1$ since $w\neq 0$. We know that $\{v_0, v_1,..., v_{n-1}\}$ is a basis since it is linearly independent ($v_0=v\notin w^{\bot}$ and $v_1,...,v_{n-1}\in w^{\bot}$). Then $V= [v_0, v_1,...,v_{n-1}]$ is invertible. Let $D=\text{ diag}(\lambda, 0,....,0)$ be $n\times n$ diagnonal matrices. We have $A=VDV^{-1}$. Hence $A$ is diagonalizable.
Part 2 ($\Longrightarrow$). We now assume that $w^Tv=0$. We aim to prove that $A$ is not diagonalizable by showing that the set of eigenvectors is linearly dependent. To do that, it is sufficient to show that all $n$ eigenvectors belong to $w^{\bot}$ of dim $n-1$. Assume that $u\neq 0$ is an eigenvector not in $w^{\bot}$, i.e. $w^Tu\neq 0$. Then $Au=(w^Tu) v = \lambda u$, where $\lambda$ is the corresponding eigenvalue of $u$. Since $w^Tu\neq 0$, $v\neq 0$ and $u\neq 0$, then $\lambda\neq 0$. Thus, $u = k v$ where $k=w^T u/\lambda$. Then $w^Tu=k(w^Tv)=0$, which is a contradiction.