Ask for help for linear algebra problems, ask for help for linear algebra problems

Forget what I said about the Jondan standard type, I didn't think about the condition that "A is an upper triangle formation".

I carefully read the answer written by the man, and I think the result is correct, but the process is indeed a bit abstract, and I guess the person is a master.

I try to write my thoughts in layman's terms:

The conditions that are now known are:

1) The main diagonal element of a is 1,-1,..1

2) A is the upper triangle formation.

3）a^2＝i

The problem is to find out all the eligible A's

That's right, right?

My idea is to use mathematical induction, i.e., to first admit that n-1 matrices must satisfy such a form, and on this basis, to derive that n-order matrices must also be in this form.

Procedure: 1) When a is a matrix of order 2, obviously, for arbitrary. 1 a12

All of them meet the conditions.

2) When a is a matrix of order 3, we assume a

1 a12 a13

0 -1 a23

Then use the condition "a 2 i" to determine a23 = 0, so a

1 a12 a13

3) So we wonder if an should have a form like this:

1 a12 a13 ..a1n

In fact, it is, and now to prove this proposition.

4) If the proposition is true when a is an n-1 matrix, then a(n-1).

1 a12 a13 ..a1n-1

and [a(n-1)] 2=i

a(n-1) b

where b is a column vector to be determined.

From [a(n)] 2=i, and assuming [a(n-1)] 2=i, b must be equal to 0 except for the first element.

This is not difficult to verify, I have calculated, it is true, it is not easy to play the formula, I will not write it, I'm sorry)

Therefore, each additional order of a is equivalent to filling in the left and bottom of the matrix with a row and a column like this: *m

where m is any number.

Therefore all the a's are:

1 a12 a13 ..a1n

There's an obvious mistake upstairs.

Diagonal elements alternate 1 and -1 from top left to bottom right.

The topic seems to be annoying, but it is actually very simple, and oral arithmetic is fine, but it is troublesome to write.

Don't be lazy, do it yourself, and you will know the law after counting the 2nd and 3rd levels.

Finally, when finding all an, rank(an-e)=n-1 is a constraint on n, rather than considering any complex matrix knowledge, don't be intimidated.

The whole problem does not require much matrix knowledge, just general operations, and does not use the Jordanian standard type at all.

It's not worth solving such a simple problem with 100 points. It is recommended to close it.

Solution: coefficient determinant =

So when ≠0 and ≠1, the system of equations has a unique solution.

When =0, the augmentation matrix =

r3-r1-r2, r1+r2

r1*(1/3),r2*(-1)

At this point, the system of equations has an infinite number of solutions. The general solution is: c(-1,1,1).'.

When =1, the augmentation matrix =

r3-r1-2r2

At this point, there is no solution to the system of equations.

A correct. b,c should be the non-zero element of the second row farther to the right than the first row! The second row of non-zero elements of b should start with the third column, and the second row of non-zero elements of c should start from the second column!

dSimilarly, the non-zero elements in the third row should start in the fourth column.

In linear algebra, the row step matrix refers to:

All non-zero rows (the rows of the matrix have at least one non-zero element) are on top of all all-zero rows. That is, all zero rows are at the bottom of the matrix.

The first coefficient of the non-zero row, also known as the principal element, is the first non-zero element on the far left (which in some places requires the first coefficient to be 1), which is strictly further to the right than the first coefficient of the upper row.

In the column where the first coefficient is located, the elements below the first coefficient are zeros.

From the inscription, we know that (1,1,-2)t is the special solution of the linear equation, and (2,1,-1) is the basic solution system of the corresponding secondary equation, so there is, a1+a2-2a3= (Equation 1), 2a1+a2-a3=0(Equation 2), so there is -a2=a1-2a3 (Equation 3), and Equation 2 and Equation 3 are brought into the system of quaternary linear equations, and y1a1+y2a2+y3a3+y4(a1-2a3)=2a1-a3 are obtained, and (y1+y4-2) a1+y2a2+(y3-2y4+1)a3=0, so the system of equations can be obtained,,y1+y4-2=0,y2=0,y3-2y4+1=0, and the solution of this second-linear equation is (2,0,-1,0)t+k(-1,0,2,1)t,

Wrong. Quite simply, take a set of linearly independent vectors, add zero vectors, and then linearly correlate. So the statement is wrong.

To sum it up for you:

When the vector group as a whole is linearly independent, the local must be linearly independent. However, local linearity is not corrosive, and the overall linear independence cannot be introduced.

When there is a local linear correlation, the whole must be linearly correlated; However, the overall linear correlation cannot be deduced.

a。Matrix equivalence means that the rank of the matrix is the same, because matrix equivalence is defined as: if there is an invertible matrix p,q such that paq=b, then a,b is called equivalence.

According to the importance property of the matrix: r(paq)=r(a), where p,q is reversible, r(a)=r(b)

a|=0, then r(a).