Let $T: \Bbb{R}^4 \rightarrow \Bbb{R}^4 $ given by $T(x,y,z,w) = (x+y,y+z+w,z+w,z+w)$.
Determine whether T is diagonalizable. If so, determine a basis $\beta$ $\Bbb{R}^3$ formed by eigenvectors of T and determine the matrix $D =$ $\begin{bmatrix} T \end{bmatrix}_\beta$.
My approach
$\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{matrix}$
Eigenvalues are: $\lambda_1 = 2$ ; $\lambda_2 = 1$ ; $\lambda_3 = 0$. Eigenvectors are: $\lambda_1 = (2,2,1,1)$; $\lambda_2 = (1,0,0,0)$; $\lambda_3 = (0,0,-1,1)$.
In relation to algebraic multiplicity for $\lambda_1 = 1; \lambda_2 = 2; \lambda_3 = 1.$ And in relation to geometric multiplicity for $\lambda_1 = 1; \lambda_2 = 1; \lambda_3 = 1.$ Since am $\not=$ gm T is not diagonalizable.
Even if it were diagonalizable, would it be possible to write a 4x4 matrix in a base $\mathbb{R}^3$? Or did I not understand this second part well.