Let A1 and A2 be n-dimensional column vectors and a be n-order orthogonal matrix. It is proved that: (1) [Aa1, aa2] = [A1, A2] (2) {Aa1} = {A1}

1、=(Aa1)^T*(Aa2)=(a1)^T*A^T*A*a2=(a1)^T*(a2)=.
2. Let a 2 = a 1, where a 1 has a 1 | a 1 | a ^ 2 = | a 1 | a ^ 2

Let a be a matrix of order n, A1, A2,... An be n-dimensional column vector, an! = 0, Aa1 = A2,... Aan = 0, and prove
Let a be a matrix of order n, A1, A2,... An be n-dimensional column vectors, an! = 0, Aa1 = A2,... Aan = 0, and prove that a cannot be similar diagonalized

First, the definition of linear independence is used to verify the linear independence of A1, A2,..., an
Then let x = [A1, A2,..., an], then x is a nonsingular matrix and satisfies x ^ {- 1} AX = J, where
J=
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
Is the Jordan canonical form of the lower triangular form

Let A1, A2,... As be n-dimensional sequence vectors and a be m × n matrix If, as is linearly independent, then Aa1, aa2 Is AAS linearly independent wrong?

It's wrong
When a = 0, Aa1, aa2 And AAS was linear correlation

Let A1, A2,..., an be a base of n-dimensional column vector space R ^ n, and a be any invertible matrix of order n. It is proved that n-dimensional column vector group Aa1, aa2,..., aan
It must be the base of R ^ n

In n-dimensional Euclidean space, any n linearly independent vectors can be used as a set of bases
In this problem, the n column vectors of an invertible matrix are linearly independent, so they can be used as a set of bases

Let A1 and A2 be n-dimensional column vectors and a be n-order orthogonal matrix. It is proved that: (1) [Aa1, aa2] = [A1, A2] (2) {Aa1} = {A1}