If $A$ is a diagonalizable matrix, and its characteristic polynomial is $(\lambda-c)^n$ for some real constant $c$, how do I prove that $A=cI$, where $I$ is the identity matrix? All the information I can glean so far is that there is only one eigenvalue, with algebraic and geometric multiplicity n. How can I get to answer from here?

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    $\begingroup$ What does "$A$ is diagonalisable mean"? What's special about the $\lambda$ in the characteristic polynomial? Can you do something with the two? $\endgroup$ – Bill Wallis May 6 '18 at 16:17
  • $\begingroup$ How does $n$ relate to the size of the matrix? $\endgroup$ – amd May 6 '18 at 20:19

If $A$ is similar to a diagonalizable matrix $D$ and if the entries of the main diagonal of $D$ are $d_1,d_2,\ldots,d_n$, then the characteristic polynomial of $A$ is equal to the characteristic polynomial of $D$; in other words,$$(\lambda-d_1)\ldots(\lambda-d_n)=(\lambda-c)^n.$$But this means that each $d_i$ is equal to $c$ and therefore that $D=c\operatorname{Id}$. The only matrix similar to $c\operatorname{Id}$ is $c\operatorname{Id}$ itself.


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