Solving a problem about similar matrices, about matrix similarity?

Why is it necessary to have the same eigenvalue as a to be similar to a?"

This conclusion is generally incorrect, as the counterexample given in the [commentary] shows that the same eigenvalues are not necessarily similar.

However, if you add one more condition, you can do it

If a and b are similar to the diagonal matrix and their eigenvalues are the same, then a and b are similar.

If a is similar to the diagonal matrix, it can be proved that the numbers on the main diagonal elements of the diagonal matrix are all eigenvalues of a.

So, a,b are similar to the same diagonal matrix.

i.e. there is p -1ap = q -1bq

So there is (pq -1) -1 a (pq -1) = b i.e. a is similar to b.

Option B takes advantage of the similarity of matrices.

1. The similarity matrix has equal traces (i.e., the sum of the elements on the main diagonal is equal) to obtain a=02, and the similar matrix has an equal determinant, which can be obtained.

b=2。

Let these three eigenvectors form the matrix p, then p (-1)ap=diag(1,0,-1) then a=pdiag(1,0,-1)p (-1) a 11=(p p (-1)) 11=p 11p (-1)=p p (-1)=a itself.

<>b is a diagonal matrix.

The eigenvalue is b. Because a and b are similar, the eigenvalues of a and b are the same.

So 1 is a eigenvalue of a, substituting 1 into the eigenpolynomial.

f(λ)=|λe-a|=0, we can get 2(1-a)+4=0, and find a=3. Only the c option is a=3, and the answer is c.

Of course, you can also continue to ask for b. Replace a=3 generations back to |λe-a|, f( )=( -2)( 1) , so the eigenvalue of matrix a is , and the comparison with matrix b yields b=1, which is consistent with the answer.

If the matrices are similar, the eigenvalues are the same.

Because the eigenvalues are the same, there are:

The determinants are the same (deta = product of eigenvalues);

The traces are also the same (traces = sum of eigenvalues).

So the matrix is similar to find the parameters.

Use |a-into e|

Or the elementary row transforms the upper (lower) triangle.

Finding the eigenvalues [Recommended|.]a-into e丨, the elementary row transformation will change the value of the determinant] the left and right eigenvalues correspond one-to-one, find a, b

Since a is an upper triangle matrix, the elements on the main diagonal are eigenvalues of a. Let the eigenvalues of a be 1, 2,..n。

Because the main diagonal elements are equal, 1= 2=.n=λ。

Since a(ij)≠0(i j), r(a) 1, r( e-a) 1, so the number of solution vectors contained in the system of equations( e-a)x=0 in the basic solution system s=n-r( e-a) n-1, i.e., the n-fold eigenvalue of a 1=.n corresponds to a maximum of n-1 linearly independent eigenvectors.

So a is not diagonal (i.e., a is not similar to a diagonal array).

Certification. Solution: A is an abstract matrix, so the key to solving the problem lies in the rank of the matrix, and the sufficient and necessary conditions for the matrix to be diagonalized (the sufficient condition for the diagonalization of the n-order square matrix is that n linear independent eigenvectors can be found).

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Matrix A can be similar to the diagonal array of the rolling shed. That is, similar to diagonal elements are matrices with eigenvalues. Since the matrix is similar and transitive, if the eigenvalues of these matrices are the same, that is, the major difference can be similar to the same diagonal array, it must be similar to each other.

We don't seem to learn this in Hebei High School.

Question 1: So a(tse)0d=c0b, so ad=bc, so it is true. Question 2:

a=jus0（7+） b=m asq07 。［a］0c..〘b!

0.*7) d6 is the solution of the matrix.

Question 3 t=

1 k 01 k 0

The 1 k0 similarity matrix has the same eigenvalues and therefore the traces are the same, then.

1+k+0=2+1+0

The solution of the 4th problem of k=2 has the same eigenvalue, so the eigenvalues of a and b are both 2, b, 0, then |2i-a|=|bi-a|=|a|=0

i.e. the determinant. 1 -a -1

a 2-b -a

1 -a 1

0b-1 -a -1

a 0 -a

1 -a b-1

0 respectively.

2a(-2a)=0

a(-ab-ab)=0

The solution is a=0, and b is an arbitrary constant and chooses b