When AB BA, prove rank A B rank A rank B rank AB

Updated on amusement 2024-06-02
9 answers
  1. Anonymous users2024-02-11

    Let the null space of the four matrices a, b, a+b, and ab.

    They are A, B, C, D

    Since ab=ba, a and b are included in d

    And it is easy to know that a is included in b and c

    The formula is given by dimensionality: dim(a)+dim(b)=dim(a-b)+dim(a-b).

    Combining the above two conditions, there is dim(a)+dim(b)<=dim(c)+dim(d).

    Substituting all four formulas such as dim(a)=n-r(a) is the formula to be proved.

  2. Anonymous users2024-02-10

    For the post-course exercises in the Advanced Algebra textbook, you can refer to the proof method in its exercise set.

  3. Anonymous users2024-02-09

    Let the zero spaces of the four matrices a, b, a+b, and ab be a, b, c, and d respectively since ab=ba, a and b are included in d

    And it is easy to know that a is included in b and c

    The formula is derived from the number of Virvi silver: dim(a)+dim(b)=dim(a-b)+dim(a-b).

    Combining the above two conditions, there is Shen Song dim(a)+dim(b).

  4. Anonymous users2024-02-08

    Summary. First, make the vector Qi and the matrix a, a1, a2 ,..an-1 to form a whole, and they can be represented by a hyperplane, on which any point can be uniquely represented as a linear combination:

    x=k0in+k1a1+k2a2+k3a3+..kn-1an-1.Since a is a matrix of several orders, it can also be represented as a point on a hyperplane, i.e., a=xo=konin+k1a1+k2a2+k3a3+.

    kn-1an-1.According to the definition that any point on the hyperplane can be uniquely represented as a linear combination, it can be obtained: a=konin+k1a1+k2a2+k3a3+.

    kn-1an-1, i.e. a can be written as in, a; A, Chuan, A4"linear combinations", certified.

    2.If a b, prove: (1) rank(a)=rank(b); (211a1=1b1;(3)+r(a)=tr(b);(4)a-~b-1;(5

    This topic is incomplete, dear.

    Can you type here, dear? We can only see the first one.

    First, let the Xiang Dan measure Qi and the matrix a,a1,a2 ,..an-1 to form a whole, and they can be represented by a hyperplane, and any point on the hyperplane can be uniquely represented as a linear combination: x=k0in+k1a1+k2a2+k3a3+.

    kn-1an-1.Since a is a matrix of several orders, it can also be represented as a point on a hyperplane, i.e., a=xo=konin+k1a1+k2a2+k3a3+.kn-1an-1.

    According to the definition that any point on the hyperplane can be uniquely represented as a linear combination, it can be obtained: a=konin+k1a1+k2a2+k3a3+.kn-1an-1, i.e. a can be written as in, a; A, Chuan, A4"The linear combination of "volt argument fiber, proven to be complete."

    The first one can be answered like this.

    What I've learned so far doesn't involve hyperplanes, so I don't understand <> are you new to it?

    This is relatively easy to learn.

    Reasoning is good.

    I'm just a freshman.

    <>Is this an assignment from the teacher?

    This service on our side can only see the first one**.

    Other services can be sent**.

    We can only see the first one here.

  5. Anonymous users2024-02-07

    Let the column vector groups of a and b be a1 and b1, respectively

    Then the column vector of a+b can be linearly represented by the lease a1,b1.

    So r(a+b).

  6. Anonymous users2024-02-06

    1.There seems to be some problem with this relationship you ga, if the spine god a=0, or c=0, then abc=0, obviously rank(abc)=0 is not necessarily equal to rank(b);

    2.If we add the condition, a is a full-rank square of the m order, and c is a full-rank square of the n order, then the conclusion is valid, because the full-rank square matrix can be regarded as the product of a series of elementary matrices, and the function of the elementary matrix is equivalent to making a primary row and column transformation, without changing the rank of the acted matrix.

    To sum up, if you want to make the theory of the loss of the cherry blossom group that you give or only be valid, you must add two conditions, that is, A is a full-rank phalanx of order m, and C is a full-rank square of order N.

    Hope it helps!

    Don't forget!

  7. Anonymous users2024-02-05

    Because a 2=e so a 2-e=0 so (a-e)(a+e)=0

    So r(a-e)+r(a+e)=r(e-a+a+e)=r(2e)=n

    Therefore, the key is Huicha, and in summary, rank(a+e)+rank(a-e)=n

  8. Anonymous users2024-02-04

    Same as solving the chaos. Because rank(ba)=rank(a), b is reversible to accompany the sail.

    bax=0 is multiplied by b (-1) at the same time.

    b^(-1)bax =b^(-1)0

    eax=0ax=0

    So bax=0 and ax=0 are the same as the hail.

  9. Anonymous users2024-02-03

    abc 00 b) -

    abc ab

    0 b) - Rent only.

    0 abbc b)

    Understand that there is no evidence upstairs that there is a problem with hail changes.

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