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I was wandering to think that is there any easy way to prove Cayley-Hamilton theorem for $A \in M(2, \Bbb R)$, i.e it satisfies its characteristic polynomial.

I was trying to think in this way that

i) If $A$ has two distinct real eigenvalues (say) $\lambda_1, \lambda_2$ then $A$ is similar to $$ \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{pmatrix} $$

ii) If $A$ has two distinct purely imaginary eigenvalues $a \pm ib$ then $A$ is similar to $$ \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} $$

iii) If $A$ has one real eigenvalue (say)$\lambda$ of multiplicity two then $A$ is similar to $$ \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \\ \end{pmatrix} $$ or $$ \begin{pmatrix} \lambda & 0 \\ 1 & \lambda \\ \end{pmatrix} $$ Now characteristic polynomial of similar matrices are same but this does not help me anything next...

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We can see that: $$ \det (A)=\det \begin{bmatrix} a-\lambda&b\\ c&d-\lambda \end{bmatrix}= \lambda^2- (a+d)\lambda+ad-bc= \lambda^2- tr (A)\lambda+det (A)$$

so the caharacteristic equation is $\lambda^2- tr (A)\lambda+det (A)I=0$

Now we can easely werify that $A^2- tr (A)A+det (A)=0$ : see my answer to: Show that a matrix $A=\pmatrix{a&b\\c&d}$ satisfies $A^2-(a+d)A+(ad-bc)I=O$.

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