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?
On
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
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