How to determine the spectral measurement of a matrix

Question

Let $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ a linear operator whose matrix in the canonical base is

$ \begin{pmatrix} 1 & 2 \\ 2 & -3 \end{pmatrix}.$

Find the expression of the spectral measure $\mu_{v}$ of $T$.

Answer 1

The resolvent of $T$ is $$ (\lambda I-T)^{-1}=\left(\begin{array}{cc}\lambda-1 & -2 \\ -2 & \lambda+3\end{array}\right)^{-1} \\ = \frac{1}{(\lambda-1)(\lambda+3)-4}\left(\begin{array}{cc}\lambda+3 & 2 \\ 2 & \lambda-1\end{array}\right) \\ = \frac{1}{(\lambda+1+2\sqrt{2})(\lambda+1-2\sqrt{2})}\left(\begin{array}{cc}\lambda+3 & 2 \\ 2 & \lambda-1\end{array}\right) $$ This has a partial fraction decomposition equal to $$ \frac{1}{\lambda+1+2\sqrt{2}}A+\frac{1}{\lambda+1-2\sqrt{2}}B $$ where $$ A=\lim_{\lambda\rightarrow -1-2\sqrt{2}}(\lambda+1+2\sqrt{2})(\lambda I-T)^{-1}=\frac{1}{-4\sqrt{2}}\left(\begin{array}{cc}2-2\sqrt{2} & 2 \\ 2 & -2-2\sqrt{2}\end{array}\right) \\ B=\lim_{\lambda\rightarrow -1+2\sqrt{2}}(\lambda+1-2\sqrt{2})(\lambda I-T)^{-1}=\frac{1}{4\sqrt{2}}\left(\begin{array}{cc}2+2\sqrt{2} & 2 \\ 2 & -2 + 2 \sqrt{2}\end{array}\right) $$ You can check that $A$, $B$ are orthogonal projections, with $A+B=I$ and $AB=BA=0$. The spectral measure $\mu$ is $$ \mu(S) = \chi_{\{-1-2\sqrt{2}\}}(S)A + \chi_{\{-1+2\sqrt{2}\}}(S)B $$