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This

$\DeclareMathOperator{\Tr}{Tr}$This is true. Here is an elementary proof: Let
$\DeclareMathOperator{\Tr}{Tr}$
$$ \phi\colon A\cdot R^n \to R^k \quad \text{and}\quad \psi\colon R^k\to A\cdot R^n $$ mutually inverse isomorphisms. Let $X$ be the matrix of the map $R^n\ni v\mapsto \phi(Av)\in R^k$, and $Y$ the matrix of $\psi$ as map from $R^k\to R^n$ (with respect to the canonical bases). Since $\psi(\phi(Av))=Av$, it follows $YX=A$. Thus $\Tr(A)= $\Tr(A)= \Tr(YX)=\Tr(XY)$. Tr(YX)=\Tr(XY).$$ But $XY$ is the matrix of the map $R^k\ni w \mapsto \phi(A\psi(w))\in R^k$. Since $\psi(w)\in A\cdot R^n$, there is $v\in R^n$ with $\psi(w)=Av$. Thus $$XYw= \phi(A\psi(w))=\phi(A\cdot Av) = \phi(\lambda Av)= \lambda \phi(\psi(w)) = \lambda w,$$ that is, $XY = \lambda I_k$. Therefore $\Tr(A)=\Tr(XY)=\lambda k$.
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This is true. Here is an elementary proof: Let $\DeclareMathOperator{\Tr}{Tr}$
$$ \phi\colon A\cdot R^n \to R^k \quad \text{and}\quad \psi\colon R^k\to R^n $$ mutually inverse isomorphisms. Let $X$ be the matrix of the map $R^n\ni v\mapsto \phi(Av)\in R^k$, and $Y$ the matrix of $\psi$ as map from $R^k\to R^n$ (with respect to the canonical bases). Since $\psi(\phi(Av))=Av$, it follows $YX=A$. Thus $\Tr(A)= \Tr(YX)=\Tr(XY)$. But $XY$ is the matrix of the map $R^k\ni w \mapsto \phi(A\psi(w))\in R^k$. Since $\psi(w)\in A\cdot R^n$, there is $v\in R^n$ with $\psi(w)=Av$. Thus $$XYw= \phi(A\psi(w))=\phi(A\cdot Av) = \phi(\lambda Av)= \lambda \phi(\psi(w)) = \lambda w,$$ that is $XY = \lambda I_k$. Therefore $\Tr(A)=\Tr(XY)=\lambda k$.