Finding the basis of a Jordan matrix associated to a transformation.

Question

Let $T$ be a linear operator on a vector space $V$ such that its matrix representation with respect to the basis $(v_1, v_2, v_3, v_4, v_5, v_6)$ is

\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 2 \\ \end{bmatrix}

Find a basis for $V$ such that the matrix associated to $T$ is

\begin{bmatrix} 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}

From the first matrix, I can tell the following information: $$T(v_1)=v_1,$$ $$T(v_2)=v_1+v_2,$$ $$T(v_3)=v_3,$$ $$T(v_4)=2v_4+v_5,$$ $$T(v_5)=2v_5+v_6,$$ $$T(v_6)=2v_6.$$

Does that information help me attain any information on the basis associated to the second matrix? These matrices look like they are in Jordan form (almost), but I don't know what to do with that information. Please help me! Thank you.

Answer 1

Notice your first matrix contains a Jordan block of size 2 corresponding to eigenvalue 1, then a Jordan block of size 1 corresponding to eigenvalue 1, then what almost looks like a Jordan block of size 3 corresponding to eigenvalue 2, but the 1's are beneath the diagonal instead of on the super diagonal. You are correct in your observation, this looks very similar to Jordan form. The second matrix is in fact in Jordan form. We can find the basis the second matrix is written with respect to just by manipulating the order of the basis vectors.

To handle the block of size 3, Consider what happens to the matrix if we reverse the order of the vectors corresponding to this block in the basis. i.e. consider: $v_1, v_2, v_3, v_6, v_5, v_4$. Recalling that: $$T(v_6) = 2v_6$$ $$T(v_5) = 2v_5 + v_6$$ $$T(v_4) = 2v_4 + v_5$$

the transformation written with respect to this basis is:

$$\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 2\end{bmatrix}$$

So by reversing the order of the vectors in the basis, we have formed a Jordan block, as desired. This works in general, by reversing the order of basis vectors corresponding to a block where the 1s are beneath the diagonal, the new block will have the 1s on the super diagonal. (and vice-versa)

Now we have the matrix in Jordan form, which is unique up to rearrangement of the blocks. All that remains is to rearrange the blocks. Start with the basis vectors corresponding to the Jordan block of size 3 with eigenvalue 2, then the vector corresponding to the block of size 1 with eigenvalue 1, then the vectors corresponding to the block of size 2 with eigenvalue 1. The second matrix is written with respect to the basis:

$$(v_6, v_5, v_4, v_3, v_1, v_2)$$