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2 added 9 characters in body

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$\DeclareMathOperator{\Tr}{Tr}$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,$$that is, XY = \lambda I_k. Therefore \Tr(A)=\Tr(XY)=\lambda k. 1 This is true. Here is an elementary proof: Let \DeclareMathOperator{\Tr}{Tr}$$ \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\$.