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Exercise 10

Avg difficulty 3.0

The equation ending Example 2.5

(110−1)−1(3201)(110−1)=(3001)\begin{pmatrix}1&1\\ 0&-1\end{pmatrix}^{-1}\begin{pmatrix}3&2\\ 0&1\end{pmatrix}\begin{pmatrix}1&1\\ 0&-1\end{pmatrix}=\begin{pmatrix}3&0\\ 0&1\end{pmatrix}

is a bit jarring because for PP we must take the first matrix, which is shown as an inverse, and for P−1P^{-1} we take the inverse of the first matrix, so that the two −1-1 powers cancel and this matrix is shown without a superscript −1-1.

  • Check that this nicer-appearing equation holds.

    (3001)=(110−1)(3201)(110−1)−1\begin{pmatrix}3&0\\ 0&1\end{pmatrix}=\begin{pmatrix}1&1\\ 0&-1\end{pmatrix}\begin{pmatrix}3&2\\ 0&1\end{pmatrix}\begin{pmatrix}1&1\\ 0&-1\end{pmatrix}^{-1}
  • Is the previous item a coincidence? Or can we always switch the PP and the P−1P^{-1}?

Hefferon, Linear Algebra, Similarity, Diagonalizability, Exercise 10

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Adapted from Linear Algebra © 2016 Jim Hefferon. Licensed under CC BY-SA 3.0 US. Source: Linear Algebra. Imported at revision df2262e.