Diagonalization of Linear Transformations Given the linear transformation $\mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $T(x,y,z) = (3x-5z, \frac{1}{5}x - y, x+y-2z)$, find a  basis $B$ for $\mathbb{R}^3$ such that $[T]_B$ is diagonal.


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*So my first instinct is to put the transformation into a matrix, so:


$$\begin{bmatrix}3 &0 & -5 \\ \frac{1}{5} &-1 & 0 \\ 1 &1 &-2 \end{bmatrix}$$


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*From there, I can just find the eigenvalues and eigenspaces of each of those values...

*After that, it's simply defining another matrix made up of those 3 eigenvectors. 

*Is that correct? What's throwing me off is that it's a transformation. How can I verify if the eigenvector matrix is correct? 

 A: Well yes, supposed you found 3 eigenvectors $v_1,v_2,v_3$, your matrix will be just the diagonal of the eigenvalues, and the basis $B$ will be just $v_1,v_2,v_3$.
A first test to verification of eigenvalues: their product equals to the determinant of the matrix.
For this particular case, we can easily spot more conditions on the eigenvalues: first add the 2nd row to the 3rd, then subtract the 5/2th of this new 3rd row from the first one, to get a lower triangle matrix. 
Even without making all the calculations, we can see that the original diagonal elements $-1$ and $-2$ will be among the eigenvalues.
To verify the eigenvectors (the new basis), you can check it by definition: $Av_k=\lambda_kv_k$ ($k=1,2,3$).
A: Once you have represented the transformation in a matrix, it is easy to both find/check to see if there is a basis such that the transformation matrix is diagonal (thus $T$ is "diagonalizable") and verify this. Given this is the field $F$, the real numbers, the only thing you really need to pay heed to is that all the roots are closed in this field, that is they are all real. (why do you need to make sure they are real?)
Once you have the characteristic polynomial, ensure that all eigenvalues of multiplicity $m$ yield $m$ eigenvectors. You'll need at least $3$ linearly independent eigenvectors to create a basis $\beta$ to show that the operator is diagonalizable. (why?)
To check directly, you could simply do a change of basis and calculate using your diagonal matrix $D$ and the original matrix you had $T$ and show they are indeed the same coordinates. That is:
$$[A]_{\beta}^{\alpha}[D]_{\beta}^{\beta}[A]_{\alpha}^{\beta} = [T]_{\alpha}^{\alpha}$$
Note: If the root is not real then what is the eigenvector?
Note: $dim\mathbb{R}^3 = 3$
