I have $A =$ $ \begin{pmatrix} a & 0 & 0 \\ b & 0 & 0 \\ 1 & 2 & 1 \\ \end{pmatrix} $ I know that this matrix $A$ is diagonalizable when $a \ne 0,1$ and when $a=0 , b = 0$ and when $a=1, b = -0.5$
Which means, I need to find now according to the question 3 matrices that are similar to this given diagonalizable $A$. But the problem is I can compute with $P^{-1}AP = diag(a_{11},a_{22},a_{33})$ but it will be pain in the ass to compute so much when I have $3 \times 3$ matrix.
Thus, Is there a fast way to compute it? It's clearly easy to see that
when $a \ne 0,1$ then $B = \begin{pmatrix} a & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}$
When $a=b=0$ then $C = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}$
and when $a=1,b= -0.5$ then $D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}$
But how can I compute that quickly? Because on the exam, I'll have to compute it quickly and efficiently. Thanks!
This question seems to be worded in a very messy way. Apparently, the OP wants to find diagonalizable matrices to which the given matrix $\;A\;$ is similar in the mentioned different cases, namely:
$$\begin{align*}\bullet&\;\;\text{When}\;\;a\neq 0,1:\;\;\text{in this case}\;,\;\;A=\begin{pmatrix}a&0&0\\b&0&0\\1&2&1\end{pmatrix}\implies\;\text{the char. polynomial is}\\{}\\ &p_A(t)=\begin{vmatrix}t-a&\;\;0&0\\-b&\;\;t&0\\-1&-2&t-1\end{vmatrix}=t(t-1)(t-a)\implies A\sim\begin{pmatrix}0&0&0\\0&1&0\\0&0&a\end{pmatrix}\\{}\\ \bullet&\;\;\text{When}\;\;a=b=0:\;\;A=\begin{pmatrix}0&0&0\\0&0&0\\1&2&1\end{pmatrix}\implies\;\text{the char. polynomial is}\\{}\\ &p_A(t)=\begin{vmatrix}t&\;\;0&0\\0&\;\;t&0\\\!\!-1&-2&t-1\end{vmatrix}=t^2(t-1)\implies A\sim\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}\end{align*}$$
This last case is true iff the geometric multiplicity of the eigenvalue zero is two, and this is true if this is the dimension of the solution space of the system
$$-x-2y-z=0$$
which is obviously so.
Now you try to do the last case, taking into account that with enough practice, the above is likely the fastest, or one of the fastest, ways to do this, since in both first cases (at least) you get a triangular matrix whose characteristic polynomial is very easy to calculate and, if you already know under what conditions the mgiven matrix $\;A\;$ is diagonalizable, thing then flow pretty smoothly.