# About the Jordan Form

So i have a few questions about the Jordan form.

Say we have a matrix $A$ and has $λ_1$ $λ_2$..$λ_κ$ eigenvalues.Why is it Usefull to know the index of the matrices $A-λιI$ ?

Also i have seen jordan forms with the 1's below the dianonal is it the same.?

Also im kinda confused on finding a jordan Basis what i have understood so far is that to find jordan basis alla i have to do is connect the basis of the generalized eigenspaces of the eigenvalues(generalized) right or not?.Maybe you could help me out to clarify it.

• This. – Git Gud Aug 21 '15 at 10:58
• There missing pages.But is answered my question on the usefullness of the index.it is the biggest jordan block of the eigenvallue. – Manolis Lyviakis Aug 21 '15 at 11:07
• What do you mean by the "index" of a matrix? – Omnomnomnom Aug 21 '15 at 13:07
• The min value κ so that $(A-λ1I)^{κ})=0$ – Manolis Lyviakis Aug 21 '15 at 13:09
• @ManolisLyviakis The index of a matrix $M$ is the smallest natural number $k$ such that the range of $M^k$ equals the range of $M^{k+1}$.And the index of an eigenvalue $\lambda$ of $M$ is the index of $M-\lambda I$. – Git Gud Aug 21 '15 at 14:51

## 1 Answer

Your questions regarding the usefulness of the index and finding a Jordan basis I will answer together: as with diagonalization one usually finds the Jordan form first and then determine the basis of vectors that transforms the given matrix to Jordan form. One "useful" way to determine the Jordan form of a matrix $A$ is to find the Segre characteristics, which is

1. the eigenvalues
2. the algebraic multiplicity of each eigenvalue
3. what you call the index of each eigenvalue
4. the algebraic multiplicities of each of the remaining similarity invariants of $A$ associated with each eigenvalue.

1 and 2 follows directly from the characteristic polynomial of $A$ as with standard diagonalization. 3 follows directly from the minimum polynomial. 4 can usually be deduced for smaller matrices if 1,2 and 3 is already known, but for larger matrices one can use a theorem stating that

"the number of simple Jordan blocks $J_m(\lambda_i)$ in $J$, with $m \geq k$, is $$\text{rank}(A-\lambda_iI)^{k-1}-\text{rank}(A-\lambda_iI)^{k}, k=1,2,\ldots"$$ (CG Cullen Theorem 5.14 - basically just a way to determine the size of the generalized eigenspaces)

Then how all of this comes together: As with diagonalization the algebraic multiplicity of an eigenvalue equals the amount of times this eigenvalue will appear on the diagonal of the Jordan form. The "index" determines the size of the largest simple Jordan block containing that eigenvalue and the sizes of the remaining simple Jordan blocks associated with the eigenvalue are determined by the remaining similarity invariants for that eigenvalue. Here is a simple example: A matrix has characteristic polynomial $x^3$, and minimum polynomial $x^2$, then it has only the eigenvalue 0, with algebraic multiplicity 3, the size of its largest simple Jordan block is 2, and then we can deduce that it will only have a single remaining Jordan block of size 1. So the matrix will have Jordan form Dg$[J_2(0),0]$.

Once you have the Jordan form and want to determine a basis - this can be hard work by hand - but the method that works best for me: We have $P^{-1}AP=J$ which we can rewrite as $AP=PJ$. Let the columns of $P$ be $X_1, X_2, \ldots$. Then $AP=[AX_1,AX_2,\ldots]$ and lets say $J=$ Dg$[J_{m_1}(\lambda_1), J_{m_2}(\lambda_1), \ldots, J_{n_2}(\lambda_2), \ldots]$. Then we have a series of $m_1$ equations (I am assuming here that the Jordan form has ones below the diagonal): $$AX_1=\lambda_1X_1+X_2,\; AX_2=\lambda_1X_2+X_3,\; \ldots,\; AX_m=\lambda_1X_m.$$ and similarly for each of the remaining simple Jordan blocks. Using these equations one can then solve for each of the $X_i$ which constitutes the Jordan basis.

Finally, yes the Jordan form with ones below the diagonal is "the same" as with ones above the diagonal, as you can use a simple permutation matrix to see that they are similar. I prefer ones below the diagonal as the Jordan form for blocks with zero eigenvalue then agree with the hypercompanion matrices associated with the characteristic polynomials of these matrices (used in the rational canonical form), but this is just a personal preference.

I hope this helps...

• One thing .At step 4 maybe you mean geometric multiplicity?Perfect thanks.!!! – Manolis Lyviakis Aug 22 '15 at 15:12
• Also what does the m mean in Jm blocks – Manolis Lyviakis Aug 22 '15 at 15:56
• for an invariant factor geometric and algebraic multiplicity is the same. $m$ is the size of the Jordan block...so $J_2$ is a $2 \times 2$ Jordan block – Christiaan Hattingh Aug 22 '15 at 16:07
• Say i got an eigenvallue λ1 multiplicity 4 and i know dimN(A-λI)=2 and dimN(A-λI)^2=3 all i can say is that i have 1 block of size >=2 it might be two 2x2 or 1x1,3x3 or one 4x4 cant find out more? – Manolis Lyviakis Aug 22 '15 at 16:32
• you do this repeatedly until you get that the difference is zero...so do the same calculation for $(A-\lambda I)^2$ and $(A-\lambda I)^3$ if you get that there is a block of size $\geq 3$ then you continue...process of elimination. Of course if you have the minimum polynomial then you will already know the size of the largest block. – Christiaan Hattingh Aug 22 '15 at 18:37