Dot diagrams and Jordan canonical forms

Click For Summary
SUMMARY

The discussion centers on the Jordan canonical form, which is the matrix representation of an operator with respect to a Jordan canonical basis. This basis consists of cycles of generalized eigenvectors, denoted as ##\gamma=\gamma_1\cup\gamma_2\cup\cdots\cup\gamma_n##. The participants explore the relationship between the number of dots in a dot diagram and the lengths of these cycles, specifically proving that ##p_j=\max\{i:r_i\geq j\}## and ##r_i=\max\{j:p_j\geq i\}##. The conversation highlights the significance of dot diagrams in visualizing Jordan blocks and their properties.

PREREQUISITES
  • Understanding of Jordan canonical forms and their properties.
  • Familiarity with generalized eigenvectors and their representation.
  • Knowledge of matrix representation and linear transformations.
  • Basic skills in mathematical induction and proof techniques.
NEXT STEPS
  • Study the properties of Jordan blocks in linear algebra.
  • Learn about the implications of non-diagonalizable matrices in the context of Jordan forms.
  • Explore mathematical induction techniques for proving properties of sequences and functions.
  • Investigate applications of Jordan canonical forms in differential equations and systems theory.
USEFUL FOR

Mathematicians, graduate students in mathematics, and anyone studying linear algebra or matrix theory, particularly those interested in eigenvalues and eigenvectors.

psie
Messages
315
Reaction score
40
Homework Statement
Why is the number of dots ##r_i## in row ##i## of a dot diagram given by ##r_i=\max\{j:p_j\geq i\}##, where ##p_j## are the number of dots in column ##j##?
Relevant Equations
Some familiarity with Jordan canonical forms is required I think. I'll try to explain the rest below.
We know that a Jordan canonical form is simply the matrix representation of an operator (whose characteristic polynomial splits) with respect to a special basis called a Jordan canonical basis. This basis consists of a disjoint union of cycles/chains of generalized eigenvectors. Take all the ##n## cycles that correspond to a certain eigenvalue ##\lambda## and take their union, which we denote ##\gamma=\gamma_1\cup\gamma_2\cup\cdots\cup\gamma_n##. Note that ##\gamma_i## may have different lengths ##p_1,p_2,\ldots,p_n##. To be concrete, suppose ##n=3## and ##p_1=3, p_2=2## and ##p_3=1##. Then \begin{align*}\gamma_1&=\{(T-\lambda I)^{2}(v_1),(T-\lambda I)(v_1),v_1\};\\ \gamma_2&=\{(T-\lambda I)(v_2),v_2\};\\ \gamma_n&=\{v_3\}.\end{align*} Then the matrix representation of ##T## restricted to ##\operatorname{span}(\gamma)## is a so-called Jordan block. We can visualize a Jordan block with the help of a dot diagram as follows: $$\begin{array}{ccc}\bullet&\bullet&\bullet \\ \bullet&\bullet&\\ \bullet \end{array}$$Here the first dots in the first row are the initial vectors in the cycle; thus the bottom dots in each column are ##v_1,v_2,v_3##, from left to right respectively. Dot diagrams are always ordered in decreasing lengths of cycle going from left to right.

Suppose now a dot diagram has ##k## columns and ##m## rows, with ##p_j## dots in column ##j## and ##r_i## dots in row ##i##. I need to show by induction on ##m=p_1## that ##p_j=\max\{i:r_i\geq j\}## for ##1\leq j\leq k## and ##r_i=\max\{j:p_j\geq i\}## for ##1\leq i\leq m##. The induction step is causing me great trouble (the base case I think I manage by myself). Any help would be very appreciated.
 
Physics news on Phys.org
I get it now I think. We have a dot at ##(i,j)## if and only if ##p_j\geq i## and ##r_i\geq j##.
 
For the sake of math trivia, a matrix with a non-trivial Jordan block is an (counter) example of a matrix that can't be diagonalized; not even over ##\mathbb C ##
 
psie said:
Homework Statement: Why is the number of dots ##r_i## in row ##i## of a dot diagram given by ##r_i=\max\{j:p_j\geq i\}##, where ##p_j## are the number of dots in column ##j##?
Relevant Equations: Some familiarity with Jordan canonical forms is required I think. I'll try to explain the rest below.

We know that a Jordan canonical form is simply the matrix representation of an operator (whose characteristic polynomial splits) with respect to a special basis called a Jordan canonical basis. This basis consists of a disjoint union of cycles/chains of generalized eigenvectors. Take all the ##n## cycles that correspond to a certain eigenvalue ##\lambda## and take their union, which we denote ##\gamma=\gamma_1\cup\gamma_2\cup\cdots\cup\gamma_n##. Note that ##\gamma_i## may have different lengths ##p_1,p_2,\ldots,p_n##. To be concrete, suppose ##n=3## and ##p_1=3, p_2=2## and ##p_3=1##. Then \begin{align*}\gamma_1&=\{(T-\lambda I)^{2}(v_1),(T-\lambda I)(v_1),v_1\};\\ \gamma_2&=\{(T-\lambda I)(v_2),v_2\};\\ \gamma_n&=\{v_3\}.\end{align*} Then the matrix representation of ##T## restricted to ##\operatorname{span}(\gamma)## is a so-called Jordan block. We can visualize a Jordan block with the help of a dot diagram as follows: $$\begin{array}{ccc}\bullet&\bullet&\bullet \\ \bullet&\bullet&\\ \bullet \end{array}$$Here the first dots in the first row are the initial vectors in the cycle; thus the bottom dots in each column are ##v_1,v_2,v_3##, from left to right respectively. Dot diagrams are always ordered in decreasing lengths of cycle going from left to right.

Suppose now a dot diagram has ##k## columns and ##m## rows, with ##p_j## dots in column ##j## and ##r_i## dots in row ##i##. I need to show by induction on ##m=p_1## that ##p_j=\max\{i:r_i\geq j\}## for ##1\leq j\leq k## and ##r_i=\max\{j:p_j\geq i\}## for ##1\leq i\leq m##. The induction step is causing me great trouble (the base case I think I manage by myself). Any help would be very appreciated.

"The highest-numbered column tall enough to reach row i is max{j : p_j ≥ i}"

If column heights are [4, 3, 3, 1]:

Row 3 gets dots from columns with height ≥ 3
That's columns 1, 2, and 3 (but not 4, since p₄ = 1 < 3)
So r₃ = 3 = max{j : p_j ≥ 3}
The botton line is - r_i counts how many columns are tall enough to reach row i, which is exactly max{j : p_j ≥ i}.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K