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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.

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A matrix of rank $1$ $A$ can be written as $UV^T$ with $U,V\in \mathbb{K}^n$ two column vectors.

So $AA = UV^T\cdot UV^T = U(V^TU)V^T=U(\text{tr}A)V^T = (\text{tr} A) A$.

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  • $\begingroup$ Thank you. I need to prove then that V^TU=tr(A) $\endgroup$
    – Ift h
    Dec 30 '20 at 19:02
  • $\begingroup$ We have that $(UV^T)_{ij} = u_i v_j$, so $\text{tr} (UV^T) = \sum_i u_i v_i = V^T U$. $\endgroup$ Dec 30 '20 at 19:03
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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, $$(A - \text{tr}A)A = 0 \implies\boxed{A^2 = \text{tr}A \cdot A}$$

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    $\begingroup$ Oops, I did not notice that you wanted proof without eigenvalues. Maybe I should not delete this as it may be useful for other readers...? $\endgroup$
    – VIVID
    Dec 30 '20 at 19:28

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