Given $A= $$\left(\begin{matrix} 2 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \\ \end{matrix} \right) $ find an invertible matrix $P$ and a diagonal matrix $D$ such that $D=P^{-1} \cdot A\cdot P$
First we need the Characteristic Polynomial $P(t)= |t\cdot I-A|$ $=$$\left(\begin{matrix} t-2 & 0 & 0 \\ 1 & t-1 & 0 \\ 0 & 0 & t-2 \\ \end{matrix} \right) $ then calculate the determinant and we get - $P(t)= $$\left|\begin{matrix} t-2 & 0 & 0 \\ 1 & t-1 & 0 \\ 0 & 0 & t-2 \\ \end{matrix} \right| $ $=(t-2)^2 \cdot (t-1)$
second thing is we find our eigenvalues so $P(t)=0$ and we will get $t=2$ or $t=1$
now need to find the eigenspace so for $t=2$ we get $P(2 \cdot I -A)=$ $\left(\begin{matrix} 0 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right) $ let $y= \lambda$ and we get that the general solution is $(x,y,z)=(\lambda,\lambda,0)= \lambda (1,1,0)$so $V_2=Sp\{(1,1,0) \}$
now for $t=1$ we get $P(1 \cdot I -A)=$ $\left(\begin{matrix} -1 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \\ \end{matrix} \right) $ from here we get $P(1 \cdot I -A)=$ $\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right) $ so the solution is $(x,y,z)=(0,0,0)$ I did not know how to continue as I noticed that my way is not right according to the textbook , This is the solution in the textbook: for $2 \cdot I -A$ they got the same matrix $\left(\begin{matrix} 0 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right) $ but the span is different the span they found is $V_2 = Sp {(1,1,0),(0,0,1)}$ my first question is why is that the span also has $(0,0,1)$? and then it also says that the set $\{(1,1,0),(0,0,1)\}$ is a basis for $V_2$
and for $t=1$ they got the same matrix but also different span they got $V_1= Sp\{(0,1,0)\}$ and also they said that the set $\{(0,1,0)\}$ is a basis for $V_1$.
now the set $\{(1,1,0),(0,1,0),(0,0,1)\}$ is $\Bbb R^3$ basis that is made of eigenvectors therefore $P=$ $\left(\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{matrix} \right) $ is a Diagonalizable matrix for A and similar to the matrix $D=$ $\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{matrix} \right) $
Can someone explain the solution to me ? I got lost on the spans mainly and the matrix $D$ and sorry for the English translations if they are not correct as I could not be 100% sure of the translations. I also checked similar questions that I found but I did not find them helpful
Thank you!