Do two similar matrices AB have the same rank in linear algebra

Do two similar matrices AB have the same rank in linear algebra

If a is similar to B, then there is an invertible matrix P such that P ^ (- 1) AP = B
There is a conclusion: when P and Q are reversible, R (a) = R (PA) = R (AQ)
[this is because the invertible matrix can be expressed as the product of the elementary matrix, and the elementary matrix does not change the rank of the matrix]
So r (b) = R (P ^ (- 1) AP) = R (a)

Matrix a multiplied by matrix B must satisfy__________ condition

① The number of columns of the left matrix A is equal to the number of rows of the right matrix B
② Let C = AB, then
The elements of the i-th row and j-th column in matrix C are equal to the sum of the products of the i-th row elements of the left matrix A and the corresponding elements of the j-th column of the right matrix B;
The number of rows of matrix C is equal to the number of rows of matrix A on the left, and the number of columns of matrix C is equal to the number of columns of matrix B on the right

Commutative conditions of matrices
linear algebra

The same matrix is a typical case
The question of the owner is not clear. Under what conditions is the exchange? + -? * /?

If two matrices are multiplied by each other to form a 0 matrix, and one of them is a diagonal matrix, then is the other necessarily a 0 matrix

Of course not
for instance
diag{1,0,1,0} * diag{0,1,0,1} = 0

Is the method of transforming a matrix into diagonal canonical form the same as Jordan canonical form? Do we use the formula p ^ (- 1) AP to solve it?

It is the same only when all eigenvalues are single root, otherwise it is different
The diagonal standard form only needs to obtain its eigenvalues, and then arrange the eigenvalues on the diagonal. Its transformation matrix P can be obtained by AP = Pb, or by the corresponding eigenvector arrangement
Jordan canonical form needs to find the roots of its minimum polynomials. These roots are arranged in the form of multiplicity and Jordan canonical form. The transformation matrix P can be obtained by AP = Pb or by generalized eigenvector

A is a matrix of order 3. Judge whether a is similar to a diagonal matrix,
If there is only one linearly independent eigenvector of a, is it similar to a diagonal matrix? Is it that only three linearly independent eigenvectors are similar to a diagonal matrix

A square matrix of order n is similar to a diagonal matrix if and only if a has n linearly independent eigenvectors
If the number of linearly independent eigenvectors is less than N, then a is not similar to a diagonal matrix

The diagonal matrix similar to the matrix A = 1 2 {} is? 6 3
The matrix is a = {1 2, the second line 6 3)

The elements on the main diagonal of a diagonal matrix are the eigenvalues of A
The eigenvalue of a is
-1.6056
five point six zero five six
There is something wrong with this question

Matrix A finds the invertible matrix P such that P ^ - 1AP is a diagonal matrix and writes this diagonal matrix

|A-λE| =
-1-λ 3 3
3 -1-λ 3
3 3 -1-λ
=
5-λ 3 3
5-λ -1-λ 3
5-λ 3 -1-λ
=
5-λ 3 3
0 -4-λ 0
0 0 -4-λ
= (5-λ)(-4-λ)^2.
The eigenvalues of a are 5, - 4, - 4
The basic solution system of (a-5e) x = 0 is: A1 = (1,1,1) ^ t
The basic solution system of (a + 4e) x = 0 is: A2 = (1, - 1,0) ^ t, A3 = (1,0, - 1) ^ t
Let P = (A1, A2, A3), then p is reversible and P ^ - 1AP = diag (5, - 4, - 4)

Which of the following matrices can be diagonalized? For the diagonalized matrix A, find an invertible matrix P so that P ^ - 1AP becomes a diagonal matrix
2 0 -2
0 3 0
0 0 3

|A-λE|=(2-λ)(3-λ)^2.
So the eigenvalues of a are 2,3,3
The basic solution system of (a-2e) x = 0 is A1 = (1,0,0) '
The basic solution system of (a-3e) x = 0 is A2 = (0,1,0) ', A3 = (- 2,0,1)'
Let P = (A1, A2, A3), then p is an invertible matrix,
And P ^ - 1AP = diag (2,3,3)

Let p ^ - 1AP be a diagonal matrix. I have found that the eigenvalues of a are 0,5
A=1 2
2 4

For each eigenvalue λ, the basic solution system of (a - λ E) x = 0 is obtained, and P
The basic solution system of AX = 0 is A1 = (- 2,1) '
The basic solution system of (a-5e) x = 0 is A2 = (1,2) '
Let P = (A1, A2)=
-2 1
1 2
Then p is reversible and P ^ - 1AP = diag (0,5)