# Diagonalizable A, computing fast.

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!

• What exactly do you want to compute (do you need $P$ explicitly?), and which step in producing the displayed result cost you too much time? – Marc van Leeuwen Jun 6 '14 at 12:43
• If I compute P, then compute $P^-1$, then compute $P^-1AP$ until I get the first matrix. I thought I could say all these matrices have zeros in the right top corner so I can say their eigenvalues must be in the diagonal, thus I can easily find similar matrices using those eigenvalues with the 3 statements one by one. – Ilan Aizelman WS Jun 6 '14 at 13:34

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.

• I solved the third one! It's $t(t-1)^2$, thank you so much. Your answers are so professional. I love your answers! Thanks again!!! – Ilan Aizelman WS Jun 6 '14 at 14:37
• Also, Can I just say that the top right corner are all zero's so I can say that the eigenvalues are on the diagonal of the matrix, and any matrix that has all zero's and those eigenvalues on the diagonal is similar to A? For e.g. when $a \ne 0,1$ then its diagonal values are a,0,1 then I can write any matrix which its diagonal values have a,0,1 and all other values in it are zero will be similar to A? – Ilan Aizelman WS Jun 6 '14 at 14:39