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Given the $2\ \times\ 2 $ matrix $$\boldsymbol{A}=\begin{pmatrix}4 & -2\\\ 1 & 1\end{pmatrix}$$

I am aware that eigenvalues of matrices can be solved from the usual $p(\lambda)=|\boldsymbol{A}-\lambda I|=0$ However, it is also known that eigenvalues can be found using the trace and determinant of the matrix $\boldsymbol{A}$ where: $$λ_1+λ_2=Tr(\boldsymbol{A}) $$ $$λ_1λ_2=\boldsymbol{|A|}$$

Using the above to solve for the eigenvalues of $\boldsymbol{A}:$

$$λ_1+λ_2=5$$$$λ_1λ_2=6$$                        Now: $$\lambda_1=5-\lambda_2$$ Using $\lambda_1$ to solve for $\lambda_2$ yields: $$\lambda_2\left(5-\lambda_2\right)=6$$$$\lambda_2^2-5\lambda_2^2+6=0$$ Solving for $\lambda_2$ gives two solutions: $$\lambda_{2A}=3$$$$\lambda_{2B}=2$$ Using both $\lambda_{2A}$ and $\lambda_{2B}$ to solve for $\lambda_1$: $$\lambda_{1A}=2$$ $$\lambda_{1B}=3$$ After selecting distinct values for the eigenvalues, I have $\lambda_1=2\ \ \ \lambda_2=3$

My question is: Why is the solution for $\lambda_2:$ $\lambda_{2A}$ and $\lambda_{2A}$, provides the complete distinct eigenvalues for $\boldsymbol{A}$? This can be also said to the solutions of $\lambda_1:$ $\lambda_{1A}$ and $\lambda_{1A}$

Also, is there a theorem for the uniqueness of eigenvalues that is similar to this:

Given an $N\ \times\ N$ matrix $\boldsymbol{A}$, if $\boldsymbol{S}_1$ is a complete set of $n$ distinct eigenvalues, and $\boldsymbol{S}_2$ is also a complete set of distinct eigenvalues for $\boldsymbol{A}$, then $\boldsymbol{S}_1=\boldsymbol{S}_2$

wd violet
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1 Answers1

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If $A$ is an $n \times n$ matrix, the only possible way in which the eigenvalues could fail to be unique is if they depend on the choice of basis. Therefore, let $B := U^{-1} A U$, where $U$ is an invertible matrix. Then see here for the uniqueness of eigenvalues: Similar matrices have the same eigenvalues with the same geometric multiplicity

AmorFati
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