Jordan Standard Form

 

The standard form theorem of Jordan is the basic theorem in linear algebra. It seems redundant to write a long article for it: there are too many handouts in this field! What new ideas can be written for an old theorem?

There are two reasons. The first reason is that when I spoke to students about this theorem, I suddenly found that I didn't know how to inspire students. Although I know many proofs of Jordan's standard form theorem, I think it's a bit confusing: How can I clarify the ideas behind the theorem, why do students think that the theorem is justified? So I know that my understanding of this theorem is vague.

The second reason is that the Jordan block has an important algebraic property which is not mentioned in the textbooks. This property is the prototype of an important and common property in substitute mathematics, this is non-decomposition. What corresponds to this is the completeness of the linear transformation that can be divided into diagonal. It is advantageous to allow students to access these phenomena from the very beginning.

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From the middle school, we know that the integer ring and polynomial ring have a unique factorization theorem: Each integer can be uniquely divided into the product of prime numbers, and each (in the domain) polynomials can be uniquely decomposed into the product of non-approximately polynomials. There are many such unique decomposition theorems in mathematics, but now we want to know: Is there a so-called "unique decomposition theorem of linear transformation? We can guess that if such a theorem exists, it can be expressed as follows:

The unique decomposition theorem of linear transformation (rough version ):Set $ V $ to the finite-dimensional vector space of the domain $ F $. $ A $ is a linear transformation on $ V $, then $ A $ can be uniquely divided into several "simple" linear transformations, and these "simple" linear transformations cannot be further decomposed.

This statement is not clear. The decomposition of integers and polynomials is the product of prime factors. So what is the decomposition of linear transformation? What is an unrecoverable linear transformation? The correct concept is straight and:

Set $ T $ to linear transformation on vector space $ V $, if $ V $ can be broken down into some non-trivial sub-spaces and $ v = v_1 \ oplus \ cdots \ oplus V_K $, if $ V_ I $ is a $ T-$ constant sub-space, $ T $ can be decomposed. If $ V $ does not exist, $ T $ is an unrecoverable linear transformation.

In this way, we can accurately express the unique decomposition theorem of linear transformation:

The unique decomposition theorem of linear transformation (corrected version): Set $ V $ to the finite dimension of the domain $ F $.Vector Space, $ T $ is a linear transformation on $ V $, then $ T $ can be uniquely divided into severalUnrecoverable linear variationChange the straight sum.

There is a very serious problem that needs to be explained: Studying the linear transformation of "unrecoverable" in the general domain $ F $ is a tricky problem, to solve this problem, we need to use the rational standard form we will learn later. In the complex field, the problem is much simpler. This is what Jordan standard form does. Therefore, in this article, $ F $ is assumed to be a complex field $ \ mathbb {c} $.

So what kind of linear transformation is unrecoverable linear transformation?

The simplest and most important example is the shift operator: Assume that $ T $ is in a group of base $ \ {v_1, \ cdots, v_n \} $ is a shift to the right:

\ [T: \ quad v_n \ rightarrow V _ {n-1} \ rightarrow \ cdots \ rightarrow v_1 \ rightarrow0. \]

$ T $ is a shift operator. $ T $ matrix under this group

\ [J_0 =\begin {pmatrix} 0 & 1 & \\\ ddots & \ ddots \\\ & 0 & 1 \\& & 0 \ end {pmatrix }. \]

$ J_0 $ is a Jordan block with a feature value of 0. Note that $ T $ is a power zero OPERATOR: $ t ^ n = 0 $, which has only the unique feature value 0.

Of course, it must be noted that the shift operator $ T $ is indeed a linear transformation that cannot be decomposed. If $ v = W \ oplus N $ is the straight sum of two non-trivial $ T-$ unchanged subspaces, $ T $ has a feature vector with the feature value 0 on $ W $ and $ N $, therefore, the solution space of the homogeneous linear equations $ Tx = 0 $ contains at least two linear independent vectors. However, the rank of $ T $ is $ n-1 $, so the space of $ Tx = 0 $ is 1-dimension, which leads to a conflict.

The same method can be used to show that after adding a number multiplication transformation to the shift operator $ T $, the resulting linear transformation is still unrecoverable: set $ \ Lambda \ In \ mathbb {C }$, $ S = T + \ Lambda I $, then $ S $ is also a matrix corresponding to non-decomposed linear transformation.

\ [J _ \ Lambda =\begin {pmatrix} \ Lambda & 1 & \\\& \ ddots & \ ddots &\\& \ Lambda & 1 \\&&& \ Lambda \ end {pmatrix} \]

The value is $ \ Lambda $.

Now we have found a family of unrecoverable linear transformations. Does they constitute all linear transformations? The answer is yes. This is the standard form theorem of Jordan:

Jordan standard form theorem:Set $ T $ to $ \ mathbb {c} $ linear transformation on the finite dimension vector space $ V $, there is a set of bases of $ V $ so that the matrix of $ T $ under this group is the sum of the Jordan blocks: \ [T = J _ {\ lambda_1} \ oplus \ cdots \ oplus J _ {\ lambda_r }. \] This decomposition is unique, this means that if another group of bases of $ V $ exists, the matrix of $ T $ is also the sum of \ [T = J _ {\ mu_1} \ oplus \ cdots \ oplus J _ {\ mu_s }, \] Then $ r = S $ and $ J _ {\ lambda_ I} = J _ {\ mu_ I} $ after appropriate shuffling.

The conclusion of the theorem includes two parts: Existence and Uniqueness. Let's first process the proof of existence.

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Step 1: Convert to power zero

Theorem [generalized feature subspaces decomposition ]:Set the feature polynomial of $ T $ to $ f (x) $, and $ f (x) $ is decomposed into the product of an int in the complex field \ [F (x) = (X-\ lambda_1) ^ {n_1} \ cdots (X-\ lambda_k) ^ {n_k}, \]

Here, $ \ lambda_ I $ is different from each other. Make $ V_ I = \ Ker (t-\ lambda_ I I) ^ {n_ I} $, every $ V_ I $ is a $ T-$ constant sub-space and \ [V = v_1 \ oplus \ cdots \ oplus v_m. \]

In this way, $ V $ is decomposed into the straight sums of $ V_ I $, a constant sub-space. $ T $ is restricted to a single feature value $ \ lambda_ I $ on each $ V_ I $.

Proof:Obviously, $ V_ I $ is a $ T-$ constant sub-space. To prove that $ V $ is their straight sum, let's start with a special conclusion:

There is a polynomial for each $1 \ Leq I \ Leq K $ \ pi_ I (x) $ to make $ \ pi_ I (x) \ equiv1 \ Mod (X-\ lambda_ I) ^ {n_ I} $, but for other $ J \ ne I $, $ \ pi_ I (x) \ equiv0 \ Mod (X-\ lambda_j) ^ {n_j} $. Linear transformation $ \ pi_ I (t) $ is the projection from $ V $ to the sub-space $ V_ I $.

Because the root of all $ (X-\ lambda_ I) ^ {n_ I} $ is different from each other, the two elements are mutually exclusive. Therefore, the $ \ pi_ I (X) $ exists. Obviously, $ \ pi_ I (t) $ is a constant transformation on $ V_ I $, while 0 on the other $ V_j \ ne V_ I $. $ \ Pi (x) = \ pi_1 (x) + \ cdots + \ pi_k (x) $ modulo any $ (X-\ lambda_ I) ^ {n_ I} $ is 1, therefore, $ \ Pi (x)-1 $ can be divisible by $ T $'s feature polynomials $ f (x) $, and thus $ \ Pi (t) -I $ is a zero transformation on $ V $, which proves that $ \ Pi (t) $ is a constant transformation on $ V $. For any $ V \ In V $, \ [V = \ Pi (t) V = \ pi_1 (t) V + \ cdots + \ pi_k (t) v. \]

Let's explain $ \ pi_ I (t) V \ In V_ I $, and then $ v = v_1 + \ cdots + V_K $. This is because $ (X-\ lambda_ I) ^ {n_ I} \ pi_ I (x) $ can be divisible by $ f (x) $, so $ (t-\ lambda_ I) ^ {n_ I} \ pi_ I (t) V = 0 $, which proves $ \ pi_ I (t) V \ In V_ I $.

Let's explain that this is straight and. If $ V_ I \ In V_ I $ meets $ v_1 + \ cdots + V_K = 0 $, use $ \ pi_ I (t) $ to get the result on the left (according to the previous analysis, $ \ pi_ I (t) $ is a constant transformation on $ V_ I $, and 0 on other $ V_j $)

\ [\ Pi_ I (t) v_1 + \ cdots + \ pi_ I (t) V_K = \ pi_ I (t) V_ I = 0, \]

$ V_1 = \ cdots = V_K = 0 $ is obtained from the arbitrariness of $ I $, which proves that it is a straight sum.

It is a common and important technique to construct a special operator (usually a given operator $ T $ polynomial) by using the Chinese Remainder Theorem, the proof here may be a little high-end, but it is the most concise.

Now we only need to consider a single sub-space $ V_ I $. So $ n = T-\ lambda_ I $, then $ N $ is a power-zero linear transformation on $ V_ I $: $ n ^ {n_ I} = 0 $, this problem is caused by the analysis of the structure of power zero linear transformation $ N $.

The simpler reason for power-zero linear transformation is that it can be expressed as the straight sum of the shift operator, while the structure of the shift operator is very simple.

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Step 2: Verify the power zero Transformation

If $ N $ is a zero-power linear transformation on $ V $, it must be proved that a group of bases exist, so that the matrix of $ N $ is the sum of several Jordan blocks. Note that a Jordan block corresponds to a shift track \ [V \ rightarrow NV \ rightarrow \ cdots \ rightarrow n ^ kV \ rightarrow 0. \]

We need to prove that there are several such non-Intersecting tracks. All the non-zero vectors contained in these tracks constitute a group of bases for $ V $.

There are many methods to prove this step, but the difference is not very big. Specifically, you like the method that depends on your own subjective nature. Here we will introduce the simplest and easiest way to be accepted by beginners.

Sum up the dimension $ \ dim v $ of $ V $. When $ \ dim v = 1 $, the conclusion is clearly true.

Assuming that all vector spaces with dimensions less than $ \ dim v $ are valid, we consider $ V $'s image space $ N (v) $. This is a $ n-$ constant sub-space, and because $ N $ is a power zero linear transformation, $ \ dim N (v) <\ dim v $, therefore, we can use the inductive hypothesis for the sub-space $ N (v) $: there is a group of bases for $ N (v) $ as follows, they constitute a $ q $ non-intersecting orbit $ \ mathcal {o} _ 1, \ cdots, \ mathcal {o} _ q $:

\ [\ Begin {array} {L} & V _ {1, 1} \ rightarrow V _ {1, 2} \ rightarrow \ cdots \ rightarrow V _ {1, n_1} \ rightarrow 0. \ & V _ {2, 1} \ rightarrow V _ {2, 2} \ rightarrow \ cdots \ rightarrow V _ {2, N_2} \ rightarrow 0. \ & \ cdots \ & V _ {q, 1} \ rightarrow V _ {q, 2} \ rightarrow \ cdots \ rightarrow V _ {q, n_q} \ rightarrow 0. \ end {array} \]

Because $ V _ {I, 1} \ In N (v) $, you can set $ V _ {I, 1} = nw_ I $, so we can get a group of longer tracks (that is, add one above)

\ [\ Begin {array} {L} & W_1 \ rightarrow V _ {1, 1} \ rightarrow V _ {1, 2} \ rightarrow \ cdots \ rightarrow V _ {1, n_1} \ rightarrow 0. \ & W_2 \ rightarrow V _ {2, 1} \ rightarrow V _ {2} \ rightarrow \ cdots \ rightarrow V _ {2, N_2} \ rightarrow 0. \ & \ cdots \ & w_q \ rightarrow V _ {q, 1} \ rightarrow V _ {q, 2} \ rightarrow \ cdots \ rightarrow V _ {q, n_q} \ rightarrow 0. \ end {array} \]

Are the vectors contained in these new tracks a group of bases of $ V $? The answer is that we need to add some tracks with a length of 1 in $ V $ but "disappears" in $ N (v) $: note $ \ {v _ {1, n_1}, \ cdots, V _ {q, n_q }\}$ is a linear irrelevant element in $ \ Ker N $, however, $ \ Ker N $ may have other base vectors. Expand them to a group of bases of $ \ Ker N $

\ [\ {V _ {1, n_1}, \ cdots, V _ {q, n_1 }\}\ cup \ {w _ {q + 1}, \ cdots, W _ {k }\}\ quad K = \ dim \ Ker n. \]

In this way, we finally get the following orbital diagram:

\ [\ Begin {array} {r} \ mathbf {W_1 \ rightarrow} V _ {1, 1} \ rightarrow V _ {1, 2} \ rightarrow \ cdots \ rightarrow V _ {1, n_1} \ rightarrow 0. & \\ mathbf {W_2 \ rightarrow} V _ {2, 1} \ rightarrow V _ {2, 2} \ rightarrow \ cdots \ rightarrow V _ {2, N_2} \ rightarrow 0. & \\ cdots \ cdots & \\ mathbf {w_q \ rightarrow} V _ {q, 1} \ rightarrow V _ {q, 2} \ rightarrow \ cdots \ rightarrow V _ {q, n_q} \ rightarrow 0. & \\ mathbf {w _ {q + 1} \ rightarrow} 0. & \\ cdots \ cdots & \\ mathbf {w_k \ rightarrow0 }. \ end {array} \]

You can see that $ W _ {q + 1}, \ ldots, w_k $ are exactly the ones whose length is 1 in $ V $, but in $ N (v) $.

The last step is to verify that these vectors actually constitute a group of bases for $ V $. Obviously, these vectors have a total of $ \ dim N (v) + \ dim \ Ker n = \ dim v $, so as long as they are linear independent.

Assuming wired relationships

\ [\ Cdots + (c_0w_ I + c_1v _ {I, 1} + \ cdots + C _ {n_ I} V _ {I, n_ I }) + \ cdots + \ sum _ {J = q + 1} ^ K d_1__j = 0, \]

We need to show that all coefficients in the above formula are 0. Use $ N $ on the left to get

\ [\ Cdots + (c_0v _ {I, 1} + c_1v _ {I, 2} + \ cdots + C _ {n_i-1} V _ {I, n_ I }) + \ cdots = 0. \]

This is a linear relationship between a group of bases for $ N (v) $, so $ c_0 = \ cdots = C _ {n_i-1} = 0 $, and the remaining linear relationship is

\ [\ Cdots + C _ {n_ I} V _ {I, n_ I} + \ cdots + \ sum _ {J = q + 1} ^ K d_1__j = 0. \]

This is a linear relationship between a group of bases of $ \ Ker N $, therefore, $ C _ {n_ I} = D _ {q + 1} = \ cdots = d_k = 0 $, so that all coefficients are 0, this completes the proof of the existence of the Jordan standard form.

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Proof of decomposition uniqueness:

Finally, we still have proof of the decomposition uniqueness theorem. This part is much simpler. It mainly utilizes a special property of the Jordan block:

\ [J_0 =\begin {pmatrix} 0 & 1 & \\\ ddots & \ ddots \\\ & 0 & 1 \\& & 0 \ end {pmatrix }_{ n \ times n} \]

Is a Jordan block with 0 feature values, then $ j_0 ^ 2 $ is to translate 1 on the diagonal line to the top right, $ j_0 ^ 3 $ is to translate two steps to the top right, and so on, $ j_0 ^ {n-1} $ change

\ [\ Begin {pmatrix} 0 & \ cdots & 1 \ & \ ddots & \ vdots \ & 0 \ end {pmatrix}, \]

Finally, $ j_0 ^ n = 0 $. With this rule, we can calculate $ \ Lambda \ In \ mathbb {c} $ and $ m \ In \ mathbb {z} ^ + $, $ T $ the number of $ M $ level Jordan blocks in the Jordan Standard Form $ J _ {\ lambda, m} $ n_m $: \ [n_m = \ Text {rank} (t-\ Lambda I) ^ {M-1}-2 \ cdot \ Text {rank} (t-\ Lambda I) ^ {m} + \ Text {rank} (t-\ Lambda I) ^ {m + 1 }. \]

The reason is: Take $ \ Lambda = 0 $ as an example. It calculates the number of 0 feature value Jordan blocks, and you will calculate the number of the Jordan blocks of any feature value. Set a Jordan standard for $ T $

\ [T = \ left (\ bigoplus _ {k \ geq1} n_k J _ {0, k} \ right) \ bigoplus _ {\ MU \ ne0} J _ \ Mu, \]

Then $ t ^ m $ is

\ [T ^ {m} = \ left (\ bigoplus _ {k \ geq1} n_k J _ {0, k} ^ {m} \ right) \ bigoplus _ {\ MU \ ne0} J _ {\ Mu} ^ {m }. \]

Note that the second half of $ \ oplus _ {\ MU \ ne0} J ^ m _ \ Mu $ is full-rank for any $ M $, so the rank of this part remains unchanged. In the previous section, all the Jordan blocks with orders less than or equal to $ M $ J _ {0, k} (k \ Leq m) $'s $ M $ power is changed to a 0 matrix. $ J _ {0, m + 1} ^ the rank of M $ is 1; $ J _ {0, m + 2} ^ the rank of M $ is 2... and so on, so

\ [\ Text {rank} t ^ m = N _ {m + 1} \ cdot1 + N _ {m + 2} \ cdot2 + \ cdots + \ Text {rank }\ bigoplus _ {\ MU \ ne0} J _ \ Mu ^ m. \]

Likewise

\ [\ Text {rank} t ^ {m + 1} = N _ {m + 2} \ cdot1 + N _ {M + 3} \ cdot2 + \ cdots + \ Text {rank} \ bigoplus _ {\ MU \ ne0} J _ \ Mu ^ {m + 1 }. \]

Therefore

\ [\ Text {rank} t ^ m-\ Text {rank} t ^ {m + 1} = N _ {m + 1} + N _ {m + 2} + \ cdots, \]

Still the same

\ [\ Text {rank} t ^ M-1}-\ Text {rank} t ^ {m} = N _ {m} + N _ {m + 1} + \ cdots, \]

So

\ [N_m = \ Text {rank} t ^ {M-1}-2 \ cdot \ Text {rank} t ^ {m} + \ Text {rank} t ^ {m + 1 }. \]

Now, you can see that the expression $ n_m $ does not depend on the specific base selection, but only on the similar constants of linear transformation, therefore, the $ T $ standard form is unique in the sense of only one arrangement.

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An interesting question is that given

\ [J _ \ Lambda =\begin {pmatrix} \ Lambda & 1 & \\\& \ ddots & \ ddots &\\& \ Lambda & 1 \\&&& \ Lambda \ end {pmatrix }_{ n \ times n} \]

Calculate the Jordan Standard Form of the $ K $ power $ J _ \ Lambda ^ K $ for a Jordan block with an feature value $ \ Lambda $.

When $ \ Lambda \ ne0 $,

\ [J _ \ Lambda ^ K = \ begin {pmatrix} \ Lambda ^ K & K \ Lambda ^ {k-1} & \ ddots & \ Lambda ^ K & \ ddots & \ ddots \ & \ ddots & K \ Lambda ^ {k-1} \ & \ Lambda ^ k \ end {pmatrix }. \]

(Do you know how to calculate $ J _ \ Lambda ^ K $? Remember this tip: unfold the polynomial $ x ^ K $ at $ \ Lambda $ in taloy: \ [x ^ k = (X-\ lambda) ^ K + A _ {k-1} (X-\ lambda) ^ {k-1} + a_1 (X-\ lambda) + A_0, \] And then substitute $ J _ \ Lambda $ .)

The proof is exactly the same as that of the non-decomposition of Jordan blocks. We found that the rank of $ J _ \ Lambda ^ K-\ Lambda ^ k I $ is $ n-1 $, therefore, the space for solving the equations $ J _ \ Lambda ^ kx = \ Lambda ^ k x $ is one-dimensional, so $ J _ \ Lambda ^ K $ cannot be decomposed, therefore, there is only one Jordan standard form, that is

\ [\ Begin {pmatrix} \ Lambda ^ K & 1 & \\\& \ ddots & \ ddots \\\& \ Lambda ^ K & 1 \\&&&\ lambda ^ k \ end {pmatrix }_{ n \ times n }. \]

The most interesting situation occurs when $ \ Lambda = 0 $. At this time, Jordan splits evenly into the sum of some small Jordan blocks.

$ J_0 $ is a shift operator at this time:

\ [J_0: \ quad v_n \ rightarrow V _ {n-1} \ rightarrow \ cdots \ rightarrow v_1 \ rightarrow 0. \]

There is only one track. However, $ j_0 ^ K $ indicates $ K $ step $ K $:

\ [J_0 ^ K: \ quad \ left \ {\ begin {array} {L} v_n \ rightarrow V _ {n-k} \ rightarrow \ cdots \ rightarrow0, \ V _ {n-1} \ rightarrow V _ {n-1-k} \ rightarrow \ cdots \ rightarrow 0, \ cdots \ V _ {n-k + 1} \ rightarrow V _ {n-2k + 1} \ rightarrow \ cdots \ rightarrow 0. \ end {array} \ right. \]

Therefore, $ j_0 ^ K $ has $ K $ tracks. Each track is a Jordan block, that is, $ K $ is included in the standard form of $ j_0 ^ K $. Set $ n = qk + r$. Here, $0 \ Leq r <K $, $ r$ of the $ K Jordan blocks is $ q + 1 $, $ K-r$ is of the order of $ q $.

For example, we can see what the Jordan standard form of an eight-level 0 feature value Jordan block $ j_0 $, $ j_0 ^ 3 $ looks like? At this time, $ j_0 ^ 3 $ has three tracks $ \ {v_8, V_5, V_2 \}$, $ \ {v_7, V_4, v_1 \}$, $ \ {V_6, v_3 \} $, so the $ j_0 ^ 3 $ Jordan standard has two three-level Jordan blocks and one two-level Jordan blocks.

To sum up, the higher power of the zero-feature-value Jordan block is always split, and is split as evenly as possible. Any power of the non-zero-feature-value Jordan block is not split.

An irrevocable algebraic structure can even "break" in a limited or expanded sense, which is a common and important phenomenon in modern mathematics. For example, if $ F $ is an irrevocable polynomial on the rational number field $ \ mathbb {q} $, $ F $ is a formal expanded domain of $ q $, if $ F $ is approx-able on $ F $, $ F $ is inevitably decomposed into the product of Polynomial with the same number of times: \ [f = f_1f_2 \ cdots f_r, \ quad \ deg F_1 = \ cdots = \ deg f_r. \]

Similarly, there is an ideal factorization in algebraic number theory. In group embedding theory, it cannot be expressed (under guidance and Restriction), and in algebra, it cannot be reduced.

Jordan Standard Form