Hanson Hanson
01/11/2023 · Senior High School
3.- Dada la siguiente matriz, hallar sus autovalores, el subespacio asociado a cada uno de ellos y sus autovectores: \[ \boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{1} & \boldsymbol{a - 1} & \boldsymbol{a}-\mathbf{2} \\ \mathbf{0} & -\mathbf{1} & \boldsymbol{a}-\mathbf{3} \\ \mathbf{0} & \mathbf{0} & \mathbf{- 1}\end{array}\right) \] Decir si A es diagonalizable (justificar), en ese caso hallar la matriz P que la diagonaliza. Justificar.
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El autovalor de la matriz \(\boldsymbol{A}\) es \(\lambda = -1\). El subespacio asociado tiene dimensión 1 y un autovector \(\boldsymbol{v} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\). La matriz \(\boldsymbol{A}\) no es diagonalizable porque no tiene un conjunto completo de autovectores linealmente independientes.
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