I have to find a basis for the generalized eigenspace $\ker(A-\operatorname{Id})^3$, where $A\in M_n(C)$ is given by : $$ A=\begin{pmatrix} 1 &1+i& 2&3-i \\ 0 & 1+i & 1 & 2-i\\ 0 &-1-i&-1 &-3+i\\ 0&1&1&2 \end{pmatrix} $$ And $X_A(t)=(t-1)^3(t-i)$.
The solution of this problem states that a basis is : $$\{(1,0,0,0)^t,(0,1,-2,1)^t,(0,0,1,0)^t\}$$ But I found : $$\{(1,0,0,0)^t,(0,1,0,1)^t,(0,0,1,0)^t\}$$ Is it the same basis ? Because $(0,1,-2,1)^t=(0,1,0,1)^t-2(0,0,1,0)^t$ ? I'm not sure about this... Any help would be appreciate,
Here's a result you can rely on to verify whether or not your candidate basis is valid.
Proposition. Suppose that $\{v_1,\dotsc, v_d\}$ is a basis of a vector subspace $V\subset\Bbb R^n$ and let $\{w_1,\dotsc, w_d\}\subset\Bbb R^n$. Let $A$ and $B$ be the matrices whose columns are $\{v_1,\dotsc,v_d\}$ and $\{w_1,\dotsc,w_d\}$ respectively, so \begin{align*} A &= \begin{bmatrix}v_1 & \dotsb & v_d\end{bmatrix} & B &= \begin{bmatrix}w_1 & \dotsb & w_d\end{bmatrix} \end{align*} Then $\{w_1,\dotsc,w_d\}$ is a basis of $V$ if and only if there is an invertible matrix $P$ satisfying $AP=B$.
In your situation, the matrices $A$ and $B$ are given by \begin{align*} A &= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 1 & 0 \end{array}\right] & B &= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \end{align*} Is there an invertible $P$ satisfying $AP=B$?
Hint. There is and the matrix looks like this $$ P = \left[\begin{array}{rrr} 1 & \ast & 0 \\ 0 & \ast & 0 \\ 0 & \ast & 1 \end{array}\right] $$ Can you find the second column of $P$?