# Prove that $A^2=\text{tr}(A)A$ where $A$ is $n\times n$ matrix of rank $1$

Let $$A$$ be a matrix of $$n\times n$$ with rank $$1$$. I need to prove that $$A^2=\text{tr}(A)A$$, without using eigenvalues.

A matrix of rank $$1$$ $$A$$ can be written as $$UV^T$$ with $$U,V\in \mathbb{K}^n$$ two column vectors.
So $$AA = UV^T\cdot UV^T = U(V^TU)V^T=U(\text{tr}A)V^T = (\text{tr} A) A$$.
• We have that $(UV^T)_{ij} = u_i v_j$, so $\text{tr} (UV^T) = \sum_i u_i v_i = V^T U$. Dec 30 '20 at 19:03
Since $$\text{rank}(A) = 1$$, it has eigenvalue $$0$$ with multiplicity of $$n-1$$ and the other eigenvalue $$\text{tr}A$$ (because the sum of eigenvalues is equal to the trace).
Therefore, $$p(x) = (x-\text{tr}A)x^{n-1}$$ where $$p(x)$$ is the characteristic polynomial of $$A$$. Now, observe that the minimal polynomial of $$A$$ divides $$(x-\text{tr}A)x$$, and thus, $$(A - \text{tr}A)A = 0 \implies\boxed{A^2 = \text{tr}A \cdot A}$$