The exercise is quite simple, I did not have too much trouble solving, yet I'm having an issue validating my results, mainly due to a gap in how I was taught the subject.
The exercise is such:
For the following linear operator, determine weather it is diagonalizable or not, if it is, find the diagonalizing base.
$$ T:\mathbb{R}^3\rightarrow\mathbb{R^3} $$ $$ T\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}4a-2b+4c\\b+3c\\b-c\end{pmatrix} $$
I found the eigenvalues and the corresponding eigenvectores.
Afterwards I've placed the eigenvectors I found (3 of them, all have algebraic multiplicity equal to their geometric multiplicity) in a group $B$.
After constructing $[T]_B$ using the eigenvectors I've placed in $B$, I've got that: $[T]_B$ is a 3x3 matrix, yet not a diagonal one.
Only after performing elementary row operations on $[T]_B$ I've got it to a diagonal form.
My question here is such:
Does this mean the group B I've constructed is a diagonalizing base?
If not, does that mean that a group $S$ is a diagonalizing base only when $[T]_S$ is in diagonal form straight off the bat?
Any information/answers are appriciated, thanks in advance!
The matrix of the linear operator $T$ is:
$A = [T] = \left[ \begin{array}{ccc} 4 & -2 & 4 \\ 0 & 1 & 3 \\ 0 & 1 & -1 \end{array} \right] $.
It is easy to verify that $A$ has 3 distinct eigenvalues
$ 4, 2, -2$
and the corresponding independent eigenvectors are:
$ \left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array} \right], \ \left[ \begin{array}{c} 1 \\ 3 \\ 1 \\ \end{array} \right], \ \left[ \begin{array}{c} -1 \\ -1 \\ 1 \\ \end{array} \right]. $
These eigenvectors are the basis vectors you are looking for.
Good luck!