Jordan Normal Form Issues

by sponsoredwalk
Tags: form, jordan, normal
sponsoredwalk is offline
Sep13-11, 07:42 AM
P: 530
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⁻ are linearly independent, & so
{v, (A - λI)v, ..., (A - λI)ⁿ⁻} 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⁺v + 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.
Phys.Org News Partner Science news on
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
HallsofIvy is online now
Sep13-11, 09:03 AM
Sci Advisor
PF Gold
P: 38,900
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 [itex](x- 3)^3= 0[/itex]. Since every matrix satisfies its own characteristice equiation, it must be true that for every vector, v, [itex](A- 3)^3= 0[/itex]. 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:
[tex]\begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 1\\ 0 & 0 & 3\end{bmatrix}[/tex].

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 [itex](A- 3I)^3u= 0[/itex] for all vectors so we must have, for some vector, u, [itex](A- 3I)v= w\ne 0[/itex] but that [itex](A- 3I)^3v= (A- 3I)^2w= 0[/itex]. Of course that is the same as saying that [itex](A- 3I)^2w[/itex] 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 [itex]\{v_1, v_2, v_3\}[/itex] is a basis for that vector space. Then [itex]Av_1[/itex] is a vector in the space and so can be written as a linear combination of the basis vectors, say, [itex]Av_1= av_1+ bv_2+ cv_3[/itex]. In that basis, [itex]v_1[/itex] itself is written as the column
[tex]\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}[/tex]
since [itex]v_1= (1)v_1+ (0)v_2+ (0)v_3[/tex]
Since [itex]Av_1= av_1+ bv_2+ cv_3[/itex], we must have
[tex]Av_1= A\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex]
so obviously, the first column of a must be
[tex]\begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex].)

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
[tex]\begin{bmatrix}3 \\ 0 \\ 0\end{bmatrix}[/tex]

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

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
[tex]\begin{bmatrix}0 \\ 1 \\ 3\end{bmatrix}[/tex]

so that, in this ordered basis, the linear transformation is represented by the matrix
[tex]\begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{bmatrix}[/tex]
Simon_Tyler is offline
Sep13-11, 08:39 PM
P: 313
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.

sponsoredwalk is offline
Sep14-11, 05:40 PM
P: 530

Jordan Normal Form Issues

Great stuff, thanks Halls. Figured out the importance of direct sums and invariant subspaces
today & am alright now.

Register to reply

Related Discussions
jordan normal form Calculus & Beyond Homework 2
Jordan Normal Form / Jordan basis Precalculus Mathematics Homework 0
Jordan Normal Form / Jordan basis Calculus & Beyond Homework 3
jordan basis and jordan normal form Calculus & Beyond Homework 12
Jordan Normal Form Linear & Abstract Algebra 2