Linear Algebra: there is a problem -- let a = (1,0, - 1) t, matrix A = AAT, n be a positive integer, find the determinant of ae-a ^ n? How do you know the eigenvalues of a are 2,0,0

Linear Algebra: there is a problem -- let a = (1,0, - 1) t, matrix A = AAT, n be a positive integer, find the determinant of ae-a ^ n? How do you know the eigenvalues of a are 2,0,0

Matrix A = AAT, then R (a) = 1, then a ^ 2 = aatat = Kaat & nbsp;, (k = ATA)
So a ^ n = k ^ (n-1) a
K = ATA = 2, a ^ n = 2 ^ (n-1) a
aE-A^n=aE-2^(n-1)A
Your problem is how to know the eigenvalue of a is 2,0,0. I'll give you a detailed calculation
I hope it will help you

If the three eigenvalues of invertible matrix A are 1,2,3 respectively, then the three eigenvalues of (2a) ^ - 1 are equal to? The answer is 1,1 / 4,1 / 6?

The three eigenvalues of a are 1,2,3, then the eigenvalues of 2a are 2,4,6, and the eigenvalues of (2a) ^ - 1 are 1 / 2,1 / 4,1 / 6

If λ = 2 is an eigenvalue of invertible matrix A, then the eigenvalue of a-2a ^ - 1 is

2 - 2* (1/2) = 1.

Let a be an invertible matrix of order n, and it is known that a has an eigenvalue of 2, then the inverse of (2a) must have an eigenvalue of?

The eigenvalue of a is a
∴Ax=ax
Multiply a ^ (- 1) twice to get:
x=aA^(-1)x
∴A^(-1)x=(1/a)x,
1 / a of the inverse matrix of a
If the eigenvalue of ∵ A is 2, then the eigenvalue of 2a is 2 * 2 = 4,
One eigenvalue of the inverse matrix of (2a) is 1 / 4

Let a, B, a + B, A-1 + B-1 be invertible matrices of order n, then (A-1 + B-1) - 1 equals ()
A. A-1+B-1B. A+BC. A（A+B）-1BD. （A+B）-1

(1) For option a. ∵ (A-1 + B-1) · (A-1 + B-1) = 2E + a-1b-1 + b-1a-1 ≠ e, ∵ option a is wrong; (2) for option B. ∵ (A-1 + B-1) (a + b) = 2E + a-1b + B-1a ≠ e, ∵ option B is wrong; (3) for option C. ∵ (A-1 + B-1) [a (a + b) - 1B] = (E + B-1a) (a

Matrix a multiplied by matrix B is equal to zero matrix, and matrix A is invertible. Can we judge that matrix B is zero matrix? Why?

Yes
B = 0 is obtained by multiplying a ^ - 1 on both sides of the equation

A and B are n-order matrices, | B | is not equal to 0, the inverse matrix of a + e = transpose of B + e, it is proved that a is invertible

(B + e) transpose = B transpose + e transpose = B transpose + e
And (a + e) ^ (- 1) = (B + e) transpose
So (B + e) transpose (a + e) = (b transpose + e) (a + e) = e, B transpose a + B transpose + A + e = e, (b transpose + e) a = - B transpose, | B + e | a | = | - B|
Because | B | is not equal to 0, so | - B | is not equal to 0, and | a | is not equal to 0
So a is reversible

If a is invertible, what is the solution X of the matrix equation AX = B, XA = B?
Xiaobai began to study linear algebra,
X=A^(-1)B
X=BA^(-1)
I've worked that out, too
Ax = B, XA = B is a dun sign in the middle

The matrix equation AX = B,
Because a is reversible, that is: A ^ (- 1)
Multiply a ^ (- 1) to the left on both sides
A^(-1)AX=A^(-1)B
X=A^(-1)B
Here a ^ (- 1) is equivalent to the reciprocal of a previous number
It's just that there are left multiplication and right multiplication
If a is on the left, multiply left; if a is on the right, multiply right
And XA = B is right multiplied
yes:
X=BA^(-1)

If a times b equals an invertible matrix, then both a and B are invertible matrices?

Because AB is reversible, so | a | B | = | ab | ≠ 0, that is | a | ≠ 0, | B | ≠ 0. Of course, both a and B are invertible matrices

Let 2 be an eigenvalue of matrix A, then matrix 3A must have an eigenvalue?

2 is an eigenvalue of A. according to the definition, | 2e-a | = 0.3 | 2e-a | = 0 | 6e-3a | = 0. According to the definition, 6 is an eigenvalue of matrix 3a