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


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, $$B=\begin{pmatrix} -1&1&1 \\ 1&0&0\\ 2&0&1 \end{pmatrix}\begin{pmatrix} 2&1&0 \\ 0&2&0\\ 0&0&2 \end{pmatrix} \begin{pmatrix} -1&1&1 \\ 1&0&0\\ 2&0&1 \end{pmatrix}^{-1}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.