How do I generate vectors that are not related to multiple matrices?

Revise it, in fact, it is also caused by my failure to read the meaning of the topic.

Let the lines of m r1, r2, .,rn (let's say n rows).

Find rank(a) and rank(a

m) The latter one is to write the two matrices vertically.

If the two ranks are equal, then all rows of m can be linearly tabled by each row of a, i.e., all row vectors related to m are related to a, i.e., there are no eligible n vectors.

If the two ranks are not equal, if the position of the rows of m is not changed when finding the ranks, you can see which rows of m cause the two ranks to be unequal in the final transformation matrix, and these rows cannot be linearly expressed by the rows of a, which are denoted as ra1, ra2, ra3, .,rai, (A1 to AI are just numbers).

Remember the set a=, (if the ranks are equal, then a is an empty set).

Then compare rank(b) and rank(b

m), to obtain the set b=

Continuing the comparison, we can get the set c= and the set d= respectively

a, b, c, d represent the number of rows in m that are not related to a, b, c, d, respectively, if they are not related to a, b, c, d respectively.

Find the intersection of a, b, c, and d, denoted as j

If it is an empty set, it means that the n-vector that meets the criteria does not exist.

If it is not an empty set, the elements contained in j=, j are related to m and not to a, b, c, d.

Take n as the linear combination of all rows in m, n=k1r1+k2r2+.knrn, as long as the coefficient of the corresponding element (or several) j is not 0, then the n selected in this way is not related to ABCD, but to m.

For example, find j=

As long as n=r1, or n=3r1+4r2+7r4, or n=r1-3r3-8r5 are eligible.

In summary, find the rows in m that are not related to a, b, c, d, and take n as long as these parts are included, and the conditions are met.

I'll give you a specific explanation.

The a, b, c, d, and m columns are equal by the inscription.

Arrange a, b, c, and d longitudinally into a matrix f=abc

d, then write g = f

m is about to arrange f and m longitudinally.

Then any row vector of r(f)=r(g)<=>m can be linearly represented by a group of row vectors of f.

In this way, the linear combination of some row vectors of c as m can also be linearly represented by f, i.e., the group of row vectors of a, b, c, d, so that the c that needs to be found does not exist.

r(f)=t)

Then the expression we require for c is c=k1a1+k2a2+·· ktat+·· knan

Among them, k1, k2··· kn is any real number, and k1, k2, ·· kt is not all 0 if ** doesn't make it clear, tell me.

Specifically, it can be roughly judged as follows: first find the rank of the original matrix, then add the new vector to the original matrix, and then find the rank of the matrix. If these two ranks are equal, they are linearly related, indicating that they can be generated.

Otherwise, the rank will increase by 1, which is linearly independent and cannot be generated.

The second problem is very simple, just use these as the base, and it's easy to generate.

I sent you a message for the specific MATLAB algorithm.

Re-ab can be diagonalized in the case of must be different if a

b (a is not equal to b) are all similar to the same diagonal array c, and if their eigenvectors are the same, then the invertible matrix p used for diagonalization must be the same, i.e., p (-1) ap = c = p (-1) bp, left times p and right times p (-1). then a=b

Contradictions, so the two different matrices are similar, their eigenvectors are unequal, and when they cannot be diagonalized, they are generally different, but not necessarily different. In short, if you pass the similarity, you can't determine that the eigenvalues are the same, and this test is generally a very common sense judgment, just remember it.

Then AB can be diagonalized in the case that must be different, if AB (A is not equal to B) are similar to the same diagonal array C, if theirs.

Eigenvectors. In the same case, it is diagonalized.

Invertible matrix. p must be the same, i.e., p (-1) ap = c = p (-1) bp, left times p and right times p (-1).In short, it cannot be judged by similarity.

Eigenvalue. The same test is generally used as a very common sense judgment, just memorize it.