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I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.

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Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-\lambda)^3$, which only has one root $\lambda=1$ of multiplicity $3$.

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The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.

Consider the matrix $A = \begin{bmatrix}0&1\\0&0\end{bmatrix}$. We have $\text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.

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I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra

Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.

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