To find a diagonal matrix A, teach A is a square matrix of order n, A 2 A, and prove that A is simil

adjoint matrix of a.

Same. Adjoint matrices of diagonal matrices (denoted m) similar to a.

It must be similar, so you don't need to prove it. (I'm using eigenvalues, all eigenvalues are the same, including multiplies).

The following focuses on the case of diagonal matrices with a.

When a is a full-rank matrix, a*

a|a^(-1).

If you want to make a* similar to m, by similar transitivity, then required.

m is similar to m*.

take m as diag(1,2,3).then m* is diag(6,3,2).The eigenvalues are not the same, so they are not similar (but can be proved to be similar in the case of the second order).

So it's more than a third-order matrix.

a* is similar to m.

Generally not true.

When the nth order matrix a is not a full-rank matrix, let the function r(x) denote the rank of the matrix x, then there is.

r(a*)1, when r(a)n-1.

r(a*)0, when r(a)n-1.

As for why, you can use the definition to represent a*, and pay attention to the relationship between the value of the determinant and the rank of the matrix).

The rank of the similarity matrix is invariant. The diagonal matrix similar to a is still set to mRule.

r(m)r(a)

To m be similar to a*, the rank must be equal, r(a).

r(m)r(a*)

This is clearly true when r(a)0.

When r(a)!=0, it can only be r(a).

1, n=2 is possible. In this case, m and m* are similar, and a* and m are similar from the similar transitivity.

In general, for the second-order situation, it is indeed similar. Except for special cases of the second order, it is generally not similar.

The formula can be formed.

a+e)(a-4e)=0

So a=-e or a=4e

Then the diagonal matrix a is a matrix of identity multiplied by -1 or 4.

Generally, no.

It depends on whether the matrix b is similar to the diagonal matrix or not.

A sufficient and necessary condition for a matrix to be similar to a diagonal matrix is that it has n linearly independent eigenvectors.

Evidence 1:

Since a 2=a knows that x 2-x is a zeroing polynomial of a, so that the minimum polynomial of a has no heavy root, so a is similar to a diagonal matrix.

Method 2: Easy to know r(a)+r(a-e)=n. If r(a)=r, then the basic solution system of ax=0 has n-r solution vectors (linearly independent column vectors of a-e), i.e., a has n-r linearly independent eigenvectors belonging to eigenvalue 0, and similarly, a has r linearly independent eigenvectors belonging to eigenvalue 1. In short, a has n linearly independent eigenvectors, so a is similar to a diagonal matrix.

Proof: Because.

a 2 = a, so.

a(a-e)=0

So r(a) + r(a-e).

n.And because.

n=r(e)=r[a-(a-e)]ax=0

The basic solution system contains.

n-r(a)

solution vectors. a-e)x=0

The basic solution system contains.

n-r(a-e)

solution vectors. Therefore, the number of linearly independent eigenvectors of a belonging to eigenvalue 0,1 is [n-r(a)]+n-r(a-e)].

n So a can be diagonalized, i.e. a is similar to a diagonal matrix.

Chu Chunchang's hands and other lines have to be changed.

r2-r3,r1-r3*a~

0 1-a 1-a^2

0 a-1 1-a

1 1 A r1+r2, exchange r1r3

1 1 a0 a-1 1-a

0 0 2-a-a^2

If r(a)=2, then 2-a-a2=0, and a-1 is not equal to 0, so we get rapid permeation a= -2

Similar to the conditions of the diagonal matrix:1. A sufficient and necessary condition for the similarity of the square matrix to the diagonal matrix is that the square matrix has n linearly independent eigenvectors.

2. If there are several different eigenvectors in the matrix, these eigenvectors are linearly independent.

3. If the eigenvalues of the matrix are different from each other, it is similar to the diagonal matrix.

A diagonal matrix is a matrix in which all elements outside the main diagonal are 0, often written as diag(a1,a2,..an)。Diagonal matrices can be considered to be the simplest type of matrix.

The elements on the diagonal can be 0 or other values, and diagonally moment crypto-hail arrays with equal elements on the diagonal are called quantity matrices. A diagonal matrix with all elements on the diagonal being 1 is called an identity matrix. The operation of the diagonal matrix includes the sum operation, the difference operation, the number multiplication operation, and the product of the same order excitation diagonal matrix, and the result is still a diagonal matrix.

That's the identity matrix. In the multiplication of matrices, there is a kind of matrix that plays a special role, like 1 in the multiplication of numbers, and this matrix is called the identity matrix. It is a phalanx with elements 1 on the diagonal (called the main diagonal) from the upper left corner to the lower right corner of the ruckus.

All other than that, it's 0.

The identity matrix is.

The adjoint matrix of a is the same as the adjoint matrix of a diagonal matrix (denoted m) that is similar to a, so there is no need to prove it. (I'm using eigenvalues, all eigenvalues are the same, including multiplies).

The following focuses on the case of diagonal matrices with a.

When a is a full-rank matrix, a* = a| *a^(-1).

If you want to make a* similar to m, by similar transmissibility, then m is required to be similar to m*.

take m as diag(1,2,3).then m* is diag(6,3,2).The eigenvalues are not the same, so they are not similar (but can be proved to be similar in the case of the second order).

So it is generally not true to say that more than a third-order matrix a* is similar to m.

When the nth order matrix a is not a full-rank matrix, let the function r(x) denote the rank of the matrix x, then there is.

r(a*)1, when r(a)n-1.

r(a*) 0, when r(a) (as for why, you can use the definition to express a*, just note the value of the determinant in relation to the rank of the matrix).

The rank of the similarity matrix is invariant. The diagonal matrix similar to a is still set to mRule.

r(m) =r(a)

To m be similar to a*, the rank must be equal, r(a) = r(m) = r(a*).

This is clearly true when r(a)0.

When r(a)!=0, only r(a) =1, n=2 can be true. In this case, m and m* are similar, and a* and m are similar from the similar transitivity.

In general, for the second-order situation, it is indeed similar. Except for special cases of the second order, it is generally not similar.

If A is similar to B, ADJ(A) is also similar to ADJ(B). Proof is easy, just know adj(p ap)=p adj(a)p.

However, adj(a) and b are not directly related, and even adj(a) and adj(a) are generally not similar, and the eigenvalues are different.

Wrong, matrix a reversible can only be pushed out |a|is not equal to 0, and the sufficient and necessary condition for a similarity and diagonal array is that a has n linear independent eigenvectors.

Matrix A is similar to a diagonal array, then a* is similar to the adjoining matrix of this diagonal array.

Here we use a conclusion: (ab)*b*a*Linear algebra ranges are not commonly used.

a*)^1 = a^-1)*.

Let a = p p -1

then a* =p p -1)* p -1)* p* =p*) 1 *p*

Let p*=q, then there is a* =q -1 *q

A diagonal matrix is a matrix in which all elements outside the main diagonal are 0s. Elements on the diagonal can be 0 or other values. 1. Let m=( ij) be the nth-order phalanx.

All elements with equal two subscripts of m are called diagonal elements of m, and the sequence (ii) (1 i n) is called the principal diagonal of m.

2. An nth-order matrix in which all non-principal diagonal elements are equal to zero is called a diagonal matrix or a diagonal matrix. It is also often written as diag (a1, a2,..an) It is worth mentioning that:

Elements on the diagonal can be 0 or other values. Therefore, a matrix = (ai,j) in n rows and n columns is diagonal if it meets the following properties: ai,j=0 and i ≠j.

A matrix with all zeros on the diagonal is a special diagonal matrix, but it is generally called a zero matrix. 1. Diagonal matrix.

d=[ a, 0, 0]

0, b, 0]

0, 0, c]

and matrix a = [1 2 3].

d*a=[ a, 2*a, 3*a]

4*b, 5*b, 6*b]

7*c, 8*c, 9*c]

a*d=[ a, 2*b, 3*c]

4*a, 5*b, 6*c]

7*a, 8*b, 9*c]

When a=b=c, there is d*a=a*d

When a=b=c=, d*a=a*d= aAt this point d is called the scalar matrix.

When =1, d is the unit matrix i.