User:Penitence/sandbox

Linear Algebra Exam 3 Practice, Problem 5
5.a

$$\text{If }a\text{ is an eigenvalue of }\mathrm{A}\text{ and }\mathrm{A}\text{ is invertible, show }\frac{1}{a}\text{ is an eigenvalue of }\mathrm{A}^{-1}\text{.}$$

$${\mathrm{A}\vec x} = {a\vec x} \;\xrightarrow{\text{multiply both sides by }\mathrm{A}^{-1}}\; {\mathrm{A}\mathrm{A}^{-1}\vec x} = {a\mathrm{A}^{-1}\vec x} \;\xrightarrow{\mathrm{AA}^{-1} = \mathrm{I}}\;  {\mathrm{I}\vec x} = {a\mathrm{A}^{-1}\vec x} \;\xrightarrow{{\mathrm{I}\vec x} = \vec x}\; \vec x = {a\mathrm{A}^{-1}\vec x} \;\xrightarrow{\text{divide both sides by }a}\; {(\frac{1}{a})\vec x} = {\mathrm{A}^{-1}\vec x}\xrightarrow{\text{rewrite}}\; {\mathrm{A}^{-1}\vec x} = {(\frac{1}{a})\vec x}$$

5.b

$$\text{Show that if }\mathrm{A}^2\text{ is the zero matrix, then the only eigenvalue of }\mathrm{A}\text{ is 0}$$

$${\mathrm{A}\vec x} = {\lambda\vec x} \;\xrightarrow{\text{multiply both sides by }\mathrm{A}}\; {\mathrm{A}\mathrm{A}\vec x} = {\lambda\mathrm{A}\vec x} \;\xrightarrow{\mathrm{AA} = \mathrm{A}^2}\; {\mathrm{A}^2\vec x} = {\lambda\mathrm{A}\vec x} \;\xrightarrow{\mathrm{A}^2 = \vec 0}\; {0\vec x} = {\lambda\mathrm{A}\vec x} \;\xrightarrow{\text{divide by }\lambda}\; {0\vec x} = {\mathrm{A}\vec x} \;\xrightarrow{{\mathrm{A}\vec x} = {\lambda\vec x}}\; {\lambda\vec x} = 0 \therefore \lambda = 0$$

Syntax: all parameters except for text are optional, with default size of 30 by 10.