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Ok so i have been doing a few questions on 'Diagonalising' defective matrices, the method I've been using to find generalized Eigenvectors is to make the previous Eigenvector the subject. However i have come across a question where i can't seem to use this method (there seems to be no solution) . I'll use below an example of a successful application and then the question i'm struggling with.

Example 1: $$A=\begin{bmatrix}4 & 1 & 0 \\ 1 & 4 & 1 \\ 4 & -4 & 7 \end{bmatrix}$$ Finding the characteristic polynomial to be $(\lambda-5)^3=0$. We see that the only eigenvalue is 5. We can also see that the Algebraic multiplicity is 3. Now considering: $$(A-5I)\bf{u_0} = \bf{0}$$ we find that the eigenspace for $\lambda$ is $t[1,1,0]$ thus the geometric multiplicity is 1, we then need 2 generalized eigenvectors , to find the first let $t=1$ and then solve the following: $$(A-5I)\bf{u_1} = \begin{bmatrix}1\\1\\0\end{bmatrix}$$ This gives us the following eigenspace $t[1,1,0]+[-1,0,2]$ again letting $t = 1$ we then find the last generalized eigenvector by solving: $$(A-5I)\bf{u_3} = \begin{bmatrix}0\\1\\2\end{bmatrix}$$ This gives us the eigenspace $t[1,1,0]+[0,0,1]$ Letting $t=1%$ again we have [1,1,1]. Thus we now have 3 linearly independent vectors we can use for our $\bf{P}$ matrix.

$$P=\begin{bmatrix}1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 2 & 1 \end{bmatrix}$$ our jordan matrix $\bf{J}$ is then found as follows:

$$\bf{J}=\bf{P^{-1}}\bf{A}\bf{P} =\begin{bmatrix}5 & 1 & 0 \\ 0 & 5 & 1 \\ 0 & 0 & 5 \end{bmatrix} $$

Example 2: $$B=\begin{bmatrix}1 & -3 & 1 \\ 1 & 5 & -1 \\ 2 & 6 & 0 \end{bmatrix}$$ Again finding the characteristic polynomial to be $(\lambda - 2)^3$ our only eigenvalue is 2. On finding the first eigenspace we now have $s[-3,1,0]+t[1,0,1]$ thus we have a geometric multiplicity of 2 and so we must find a generalized eigenvector , this is where i come undone letting s and t be real values to get a eigenvector i can never seem to get a solution. I'm not sure why and any help would be great.

Best regards

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In example 2, since the geometric multiplicity of the eigenspace is 2, also $B$ is a 3x3 matrix, so there are 2 Jordan blocks $J_1$ and $J_2$ with different size. Your method adopted in example 1 can only deal with Jordan blocks with same block size.

Without loss of generality, just take the block size of $J_1$ and $J_2$ to be 2 and 1 respectively.

Then $$J=\begin{pmatrix} 2&1&0 \\ 0&2&0\\ 0&0&2 \end{pmatrix}$$ To solve this problem, our first step is to find out $Nul[(B-2I)^2]$ and the dimension of it.

By computation, $(B-2I)^2$= zero matrix, $dim(Nul[(B-2I)^2])=3$, we'll stop here since the max nullity is 3 as $B$ is a 3x3 matrix, it can't be more for higher power.

$$\\$$ Our second step is to let $\underline{b}_1,\underline{b}_2,\underline{b}_3$ to be the $i-$column of the matrix $P$.

Now we are going to find the vectors in the first block $J_1$, where they are $\underline{b}_1,\underline{b}_2$.

We must start to find $\underline{b}_2$ first, because "2" is the largest number on this block. Therefore we need to take $\underline{b}_2=\underline{e}_1$ as $(B-2I)\underline{e}_1 \neq \underline{0}$; if it equals to the zero vector, then we just take $\underline{b}_2=\underline{e}_2$ or other possible $\underline{e}_k$.

After that, we can apply $(B-2I)$ on $\underline{b}_2$, it makes the "2" to be "1", i.e. $$\underline{b}_1=(B-2I)\underline{b}_2$$.

We are now finish the works on $J_1$, let's move on to $J_2$.

Luckily, there is only one vector in $J_2$.

Note that the smallest number $j$ for $\underline{b}_j$ on each block is the eigenvector. So in this case, "3" is the smallest number on $J_2$.

Therefore we can just take $\underline{b}_3$ to be the eigenvector that you found.

Take $$\underline{b}_3= \begin{pmatrix} 1 \\ 0 \\ 1\end{pmatrix} \notin Span\{\underline{b}_1,\underline{b}_2\}$$ to make the matrix $P$ non-sigular. Finally, $$B=\begin{pmatrix} -1&1&1 \\ 1&0&0\\ 2&0&1 \end{pmatrix}\begin{pmatrix} 2&1&0 \\ 0&2&0\\ 0&0&2 \end{pmatrix} \begin{pmatrix} -1&1&1 \\ 1&0&0\\ 2&0&1 \end{pmatrix}^{-1}$$

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  • $\begingroup$ Thank you very kindly for a in depth explanation, much appreciated :) $\endgroup$ – Dead_Ling0 Mar 30 at 18:13
  • $\begingroup$ Is dim(null) the same as dim(ker) $\endgroup$ – Dead_Ling0 Mar 30 at 18:32
  • $\begingroup$ @Dead_Ling0 yes :) $\endgroup$ – Jade Pang Mar 30 at 18:33

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