rt. It is proved that if matrix A is commutative with all n-order matrices, then a must be a quantity matrix, that is, a = AE

Let a = AIJ denote the elements of row I and column J as 1 by eij, while the other elements are zero. Since a can be exchanged with any matrix, it must be exchanged with E. from aeij = eija, we get Aji = AIJ, I = J = 1,2,3,... N and AIJ = 0
I is not equal to J
So a is a quantity matrix

How to prove that if a real matrix of order n degenerates, then the transpose of a times a is a positive semidefinite matrix

Because the real matrix A of order n is degenerate,
So IAI = 0
Thus ia'i = IAI = 0
So iaa'i = IAI * ia'i = 0
Therefore, the eigenvalues of AA 'are all greater than or equal to 0, and at least one eigenvalue is equal to 0
So for any real nonzero column vector x,
Then x '(AA') x > = 0
So AA 'is a positive definite matrix

How to prove that a real matrix of order n is nondegenerate, then the transpose of a times a is a positive definite matrix

You can study all the ordinal principal subformulas of AA ', which are greater than 0 (for example, B11 = ∑ a1i & # 178;)
This is because a is not degenerate (I understand it as | a | ≠ 0), so all its ordinal principal subexpressions cannot be zero

Let a be a matrix of order n (n & gt; = 3). If a ≠ 0 but a ^ 3 = 0, we try to prove that a is not diagonalizable

To the contrary, if a can be diagonalized, then a ^ 3 = PD ^ 3P ^ {- 1} = 0 = > d ^ 3 = 0 = > d = a = 0 after diagonalizing a = PDP ^ {- 1}

A is a matrix of order 3, | A-E | = | a-2e | = | a-3e | = 0, find | a * - E|
|E-A|=(-1)^3*|A-E|=0
Similarly, | 2e-a | = | 3e-a | = | e-A | = 0

Because | A-E | = 0
So | e-A | = (- 1) ^ 3 * | A-E | = 0
Similarly, | 2e-a | = | 3e-a | = | e-A | = 0
From this we can know that the three eigenvalues of matrix A are 1,2,3
So matrix A is invertible, and | a | = 1 × 2 × 3 = 6
AA*=|A|E
So a * = | a | a ^ (- 1) [a ^ (- 1) denotes the inverse matrix of a]
The eigenvalues of a are 1,2,3
So the eigenvalues of a ^ (- 1) are 1,1 / 2,1 / 3
So the eigenvalues of a * are 6,3,2, because a * = |a|a ^ (- 1)
Therefore, we know that there exists an invertible matrix P and its inverse matrix Q [q = P ^ (- 1),], such that the result of PA * q is a diagonal matrix D, that is
PA * q = D, and the diagonal elements of D are 6,3,2
So | a * - e | = | P | a * - e | Q | = | PA * q-peq | = | D-E | because P and Q are inverse matrices of each other | P | * | Q | = 1, PEQ = E
The result of D-E is a diagonal matrix with diagonal elements of 5,2,1
So | a * - e | = | D-E | = 5 × 2 × 1 = 10
For matrix e-A, it is equivalent to multiplying every row of matrix A-E by - 1
When calculating the determinant, if a row (column) has a common factor K, it can be said that K is outside the determinant
So when we calculate | e-A | we put forward the common factor - 1 for each line, and we get | A-E | in total, three lines
So | e-A | = (- 1) ^ 3 * | A-E | = 0

Let n-order real square matrix a satisfy a ^ 2-4a + 3E = 0, and prove that B = (2e-a) ^ t (2e-a) is a positive definite matrix

Because a ^ 2-4a + 3E = 0
So a (a-2e) - 2 (a-2e) - E = 0
So (a-2e) (a-2e) = E
So a-2e is reversible
So 2e-a is reversible
So B = (2e-a) ^ t (2e-a) is a positive definite matrix
--Positive definite contract in identity matrix

A is a third-order matrix, if | a + e | = 0, | a + 2e | = 0, | a + 3E | = 0, then | a + 4E | =?
Why?

If | a | = 0, then rank a

A is a matrix of order n, and a ^ 2-A = 2E. It is proved that a can be diagonalized
This is a kind of matrix diagonalization problem ~ please know a little proof~

Obviously, because minimal polynomials have no multiple roots

Proof: let a be a matrix of order n, and a ^ 2-A = 2E. Prove that a can be diagonalized

This problem can be solved in different ways at different stages
If we have learned the Jordan canonical form and the minimum polynomial of a matrix, we can use:
The sufficient and necessary condition for diagonalization of a matrix is that its minimum polynomial has no multiple roots (i.e. Jordan blocks are all of order 1)
From a & # 178; - a = 2E, we know that X & # 178; - X-2 = (X-2) (x + 1) is a zeroing polynomial of A
Note that the polynomial has no multiple roots, and the minimum polynomial must be a factor of the zeroing polynomial
So a can be diagonalized
If you are not learning Jordan Standard type, you can use:
The sufficient and necessary condition for diagonalization of a matrix is that the geometric multiplicity of any eigenvalue is equal to the algebraic multiplicity
Here, the geometric multiplicity of eigenvalue λ is the dimension of solution space AX = λ X,
Algebraic multiplicity refers to the multiplicity of the root of the characteristic polynomial of a (provable geometric multiplicity ≤ algebraic multiplicity)
Because eigenvectors belonging to different eigenvalues are linearly independent, the above conditions are equivalent to finding n linearly independent eigenvectors
From a & # 178; - a = 2E, we know (a + e) (a-2e) = 0
Then R (a + e) + R (a-2e) - n ≤ R ((a + e) (a-2e)) = 0, i.e. R (a + e) + R (a-2e) ≤ n
-The geometric multiplicity of 1 as the eigenvalue of a is N-R (a + e), while the geometric multiplicity of 2 is N-R (a-2e)
So the algebraic multiplicity of n ≥ - 1 + the algebraic multiplicity of 2
Geometric multiplicity of ≥ - 1 + 2
= n-r(A+E)+n-r(A-2E)
≥ n,
It is known that a has no eigenvalues other than - 1 and 2, and the geometric multiplicity of - 1 and 2 = algebraic multiplicity, so a can be diagonalized

It is known that matrix A of order n satisfies a ^ 2 (a-2e) = 3a-11e. It is proved that a + 2E is invertible and (a + 2e) ^ - 1 is obtained

Because a ^ 2 (a-2e) = 3a-11e
So a ^ 3-2a ^ 2-3a + 11e = 0
So a ^ 2 (a + 2e) - 4a (a + 2e) + 5 (a + 2e) + e = 0
So (a ^ 2-4a + 5e) (a + 2e) = E
So a + 2E is reversible, and (a + 2e) ^ - 1 = a ^ 2-4a + 5E

rt. It is proved that if matrix A is commutative with all n-order matrices, then a must be a quantity matrix, that is, a = AE