# Jordan Normal Form Issues

1. Sep 13, 2011

I just can't figure out how you arrive at having a diagonal matrix consisting of Jordan blocks.

Going by Lang, a vector is (A - λI)-cyclic with period n if (A - λI)ⁿv = 0, for some n ∈ℕ.
It can be proven that v, (A - λI)v, ..., (A - λI)ⁿ⁻¹ are linearly independent, & so
{v, (A - λI)v, ..., (A - λI)ⁿ⁻¹} forms a basis, called the Jordan basis, for what is now known
as a cyclic vector space.

Furthermore, for each (A - λI)ⁿv we have that (A - λI)ⁿv = (A - λI)ⁿ⁺¹v + λ(A - λI)ⁿv.

Now for the life of me I just don't see how the matrix associated to this basis is a matrix
consisting of λ on the diagonal & 1's on the superdiagonal.

But assuming that works, I don't see how taking the direct sum of cyclic subspaces can
be represented as a matrix consisting of matrices on the diagonal.

Basically I'm just asking to see explicitly how you form the matrix w.r.t. the Jordan basis
& to see how you form a matrix representation of a direct sum of subspaces, appreciate
any & all help.

2. Sep 13, 2011

### HallsofIvy

Staff Emeritus
Let's take the simple example where the matrix, A, is 3 by 3 and has the single eigenvalue 3. Then the Characteristic equation is $(x- 3)^3= 0$. Since every matrix satisfies its own characteristice equiation, it must be true that for every vector, v, $(A- 3)^3= 0$. It might be the case that (A- 3I)v= 0 for every vector v. In that there are three independent vectors such that Av= 3v and we can use those three vectors as a basis for the vector space. Written in that basis, A would be diagonal:
$$\begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 1\\ 0 & 0 & 3\end{bmatrix}$$.

Or, it might be that (A- 3I)v= 0 only for multiples of a single vector or linear combinations of two vectors. In the first case, it must still be true that $(A- 3I)^3u= 0$ for all vectors so we must have, for some vector, u, $(A- 3I)v= w\ne 0$ but that $(A- 3I)^3v= (A- 3I)^2w= 0$. Of course that is the same as saying that $(A- 3I)^2w$ is a multiple of v, the eigenvector, and so, letting x be (A- 3I)w, (A- 3I)x= v. Those two vectors, x such that (A- 3I)x= v, and w such that (A- I3)w= x, are called "generalized eigenvectors".

Of course, to find the matrix representation of a linear transformation in a given basis, we apply the matrix to each basis vector in turn, writing the result as a linear combination of the basis vectors, so that the coefficients are the columns of the matrix.

(If you are not aware of that [very important!] fact, suppose A is a linear transformation from a three dimensional vector space to itself. Further, suppose $\{v_1, v_2, v_3\}$ is a basis for that vector space. Then $Av_1$ is a vector in the space and so can be written as a linear combination of the basis vectors, say, $Av_1= av_1+ bv_2+ cv_3$. In that basis, $v_1$ itself is written as the column
$$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$$
since $v_1= (1)v_1+ (0)v_2+ (0)v_3[/tex] Since [itex]Av_1= av_1+ bv_2+ cv_3$, we must have
$$Av_1= A\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ b \\ c\end{bmatrix}$$
so obviously, the first column of a must be
$$\begin{bmatrix}a \\ b \\ c\end{bmatrix}$$.)

Here, our basis vectors are v, x, and w such that (A- 3I)v= 0, (A- 3I)x= v, and (A- 3I)w= u. From (A- 3I)v= 0, which is the same as Av= 3v+ 0x+ 0w, we see that the first column of the matrix is
$$\begin{bmatrix}3 \\ 0 \\ 0\end{bmatrix}$$

From (A- 3I)x= v, which is the same as Ax= v+ 3x+ 0w, we see that the second column of the matrix is
$$\begin{bmatrix}1 \\ 3 \\ 0\end{bmatrix}$$

Finally, from (A- 3I)w= x, which is the same as Aw= 0v+ x+ 3w, we see that the third column of the matrix is
$$\begin{bmatrix}0 \\ 1 \\ 3\end{bmatrix}$$

so that, in this ordered basis, the linear transformation is represented by the matrix
$$\begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{bmatrix}$$

Last edited: Sep 13, 2011
3. Sep 13, 2011

### Simon_Tyler

Thanks HallsofIvy, that was an interesting read. I enjoyed the construction of the matrix from the generalized eigenvectors - I don't remember seeing it that way before.

4. Sep 14, 2011