- #1

thelema418

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What makes Jordan canonical forms significant? Why would they be useful?

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- Thread starter thelema418
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- #1

thelema418

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What makes Jordan canonical forms significant? Why would they be useful?

- #2

coquelicot

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1) you cannot always diagonalize a matrix. This acts as a replacement for it.

For example, if you need to solve a system of linear differential equations, and cannot diagonalize the matrix of the system, then you put it into Jordan canonical form, and this leads to a solution (see textbooks).

Similarly, if you have a system of linear recurrence equations, and you cannot diagonalize the system, you put it into Jordan canonical form and solve the system. In brief, each time you would use diagonalization and you cannot diagonalize, use Jordan canonical form.

2) Theoretical aspects : this is used in several arguments in abstract mathematics

- #3

Stephen Tashi

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For example, suppose we have a matrix [itex] A [/itex] and we want to compute [itex] A^{10}[/itex]. Suppose you find a matrix [itex] B [/itex] such that [itex] B^{-1} A B = J [/itex] where [itex] J [/itex] is in Jordan cannonical form. To compute [itex] J^{10} [/itex] you only have to raise the Jordan blocks to the 10th power. They individual blocks are usually smaller matrices than the matrix [itex] A [/itex]. From [itex] J^{10} [/itex] you can compute [itex] A^{10} [/itex] from [itex] A^{10} = B J^{10} B^{-1} [/itex]

You can think of Jordan cannonical form as a generalization of the the concept of a "diagonal matrix". It's easy to to multiplications and find inverses (if they exist) for diagonal matrices. If you can find a invertible matrix [itex] B [/itex] that makes matrix [itex] B^{-1} A B = D [/itex] where D is diagonal matrix then you can find [itex] A^{10} [/itex] by the procedure described above. To compute [itex] D^{10} [/itex] you only have to take the 10th power of each diagonal entry.

Why not use diagonal matrices and not worry about Jordan cannonical form? - it's because you can't always find an invertible matrix [itex] B [/itex] such that [itex] B^{-1} A B [/itex] is a diagonal matrix. The Jordan Cannonical form is important because it is the most nearly diagonal format that you can get reliably..

In applications of matrix algebra, expressions of the form [itex] B^{-1} [/itex] (...stuff...) [itex] B [/itex] often have a physical interpretation. The indicate "Change the coordinate system. Do the "stuff" operation in the new coordinate system.. Then change the coordinates of the result back to the original coordinate system. (It's analogous to reading a problem posed in cartesian coordinates, changing to polar coordinates to solve it and then changing back to cartesian coordinates to report the answer.)

If you want to compute the matrix product [itex] A^2 C^3 [/itex] then you can daydream about finding an invertible matrix [itex] B [/itex] such that [itex] B^{-1} A B [/itex] and [itex] B^{-1} C B [/itex] are both diagonal. You use [itex] B [/itex] to change coordinates, you do the computation in the coordinate system where the matrices are diagonal. Then use [itex] B^{-1} [/itex] to change back to the original coordinate system.

You can't always find a matrix [itex] B [/itex] that simultaneously diagonalizes two matrices. You can't always find a matrix [itex] B [/itex] that changes coordinates so two matrices have the same Jordan Block structure. It's important to know special cases when you can. So it's of interest to have theorems that say "If ...so-and-so then the two matrices can be simultaneously diagonalized" or "put in compatible Jordan block form".

An even more general concept of expressing matrices in a convenient format is the "singular value decomposition" - but's that's a different topic.

- #4

thelema418

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I do work with partial and ordinary differential equations that often involves using matrix calculations, but I do not see how this would connect. I understand that if I did not have a calculator, that the process of computing A raised to a power on a similar matrix B could be easier. Yet, calculators and computers can calculate large matrixes raised to a power quickly using the standard left multiplication procedure.

Is this procedure efficient? Or, is it just worthy as a trick to handle situations where you have to calculate by hand?

- #5

Stephen Tashi

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I do work with partial and ordinary differential equations that often involves using matrix calculations, but I do not see how this would connect.

Then perhaps you have encountered the exponential function [itex] e^{tA} [/itex] where [itex] A [/itex] is a matrix.

If you put a matrix in diagonal or Jordan cannonical form, you can often see what the result of long calculation will be without doing it - for example, taking the limit of a sequence of matrix operations. The function [itex] e^{tA} [/itex] is defined as a limit of an infinite series involving powers of the matrix [itex] A [/itex].

There are many numerical algorithms for matrices. If ordinary matrix multiplication met the needs of the world and its computers then many of them would be unnecessary.

- #6

coquelicot

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- #7

mathwonk

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| a 1 0 0 0 |

| 0 a 1 0 0 |

| 0 0 a 1 0 |

| 0 0 0 a 1 |

| 0 0 0 0 a |, a classic example of a Jordan block, i.e. a “stretch plus a shift”.

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