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If $A,B \in \mathbb{R^n}$ are two diagnolizeable matrices with same characteristics polynomial then $A$ and $B$ are similar.

I tried to find a solution whether this is true or not.

I think this is true because if both matrices are diagnolizeable then they both have basis consisting of $n$ eigenvectors and since characteristics polynomial is also same so there eigenvalues must also be same $\rightarrow Tr(A) == Tr(B)$ and $det(A) == det(B)$.

The other way i tried is: we can write both $A$ and $B$ as diagonal matrix,such that both $A$ and $B$ only have eigen values as diagonal elements and all other entries should be zero but here i don't know whether every diagonal matrix is similar to everyother diagonal matrix, I could not find any counter example nor i know any proof?

I don't even know whether the lemma is true or not but the process above lead me to yes. Can someone tell me if this is true and how to prove it?

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2 Answers 2

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Hint: The similarity relation $\sim$ is transitive

$$A\sim D\:\:\:{\rm and}\:\:\:B\sim D\Longrightarrow A\sim B$$

Can you find this shared $D$? You almost got it in your second trial.

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You know that $A$ is similar to a diagonal matrix $D_A$. Let $\alpha_1,\ldots,\alpha_n$ be entries of the main diagonal of $D_A$. And know that $B$ is similar to a diagonal matrix $D_B$. Let $\beta_1,\ldots,\beta_n$ be entries of the main diagonal of $D_B$. Then\begin{align}\text{characteristic polynomial of }D_A&=\text{characteristic polynomial of }A\\&=\text{characteristic polynomial of }B\\&=\text{characteristic polynomial of }D_B.\end{align}But the first and the last of these polynomials are$$(x-\alpha_1)\cdots(x-\alpha_n)\text{ and }(x-\beta_1)\cdots(x-\beta_n)$$respectively. These polynomails beaing equal means that the numbers $\alpha_1,\ldots,\alpha_n$ are the numbers $\beta_1,\ldots,\beta_n$ by another order. But then $D_A\sim D_B$ and therefore $A\sim B$.

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