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Let $F= \begin{bmatrix} -1 & -2 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$ is a matrix of a linear transformation $f:\mathbb{R^3}\rightarrow \mathbb{R^3}$ in standard basis. Find a basis in which $F$ will be diagonal matrix.

Matrix $F$ is diagonalizable $\Rightarrow J=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$ and a basis in which $F$ is diagonal are column vectors of $J$.

Is this correct?

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    $\begingroup$ $J$ is correct, but for the basis it is false. Why do you say that? You have to find a basis of eigenvectors. $\endgroup$ – Bernard Jan 16 '16 at 0:19
  • $\begingroup$ the new basis are found as the eigenvectors of $F$ $\endgroup$ – janmarqz Jan 16 '16 at 0:21
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Write $F = Q\Lambda Q^{-1}$, where $\Lambda$ is a diagonal matrix, and Q contains the eigenvectors as its columns. Then Q is a set of basis for which F is a scaling transformation. To see this, $Q\Lambda Q^{-1}$ is applied from right to left, and means:

$Q^{-1}$ maps any vector in the original space into vector represented by basis Q

$\Lambda$ performs a scaling (diagonal) transformation

$Q$ maps the vector represented in basis $Q$ back to original space.

So if you think about this decomposition of transformation mean, you can see why Q is the basis for which the transformation F can be represented by a diagonal matrix

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To calcuate the eigenvalues consider the matrix $T= \begin{bmatrix} -1-t & -2 & 2 \\ 0 & 1-t & 0 \\ 0 & 0 & 1-t \\ \end{bmatrix}$

The characteristic polynomial is $(1-t)^2(-1-t) $ . The eigenvalues are $1,1,-1$. Now consider the matrix $T-tI $ before with $t=1$ and with $t=-1$ and consider the $\ker T-tI $ to find the eigenvectors

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