Solve a matrix proof problem, a proof question about matrices

A is a sufficient and necessary condition for an orthogonal matrix is aa'=e.

Take the determinant on both sides to get |a||a'| = |e|.

a'It is a transpose of A. e is the identity matrix.

So |a'| = |a|, e|= 1 so |a|^2 = 1.

When|a|= -1.

a+e| = |a+aa'| = |a(e+a')| = |a||e+a'| = |a||(e+a)'| = -|e+a|.

So |a+e| = 0.

So -1 is a eigenvalue of a.

When|a|= 1 and a is of odd order, a-e| = |a-aa'| = |a(e-a')| = |a||e-a'| = |(e-a)'| = |e-a|

a-e)| = (-1)^n|a-e| = -|a-e|.

So |a-e| = 0.

So -1 is a eigenvalue of a.

Satisfied

1)ab+a+b=e

Add the identity matrix e on both sides at the same time

ab+a+b+e=2e

b+e)(a+e)=2e

b+e)[(a+e)/2]=e

Therefore, b+e is an invertible matrix.

2) From (1) it is known that (b+e)[(a+e) 2]=e, so the inverse matrix of b+e is (a+e) 2

Uniqueness:

If there are two forms.

i.e. a = b + c b symmetrical c antisymmetric.

a = f + g f symmetrical g antisymmetric.

So there's a'Represents a transpose.

a' = b' + c' = b - c

a' = f' + g' = f - g

From f + g = b + c

f - g = b - c

The addition of the two formulas gives 2f=2b, f=b

Further we get g = c

So it was proven.

Proof: Take x= i=(0,..1,..0) t, the i-th component is 1, and the other components are 0

It is known that x tax = aii = 0, i=1,2 ,..n.

Take x= ij=(0,..1,..1,..0) t, the ith and jth components are 1, and the other components are 0

It is known that x tax = 2aij = 0, i,j=1,2 ,..n, i≠j.

In summary, aij = 0, i,j=1,2,..n is a = 0

1.Because if both a and b are nth-order orthogonal matrices.

So aa' =a'a = e, bb'Deficit blouse = b'b = e, so (ab).'(ab) =b'a'ab = b'b = e so ab is an orthogonal matrix.

2.Because (a+a')'a'+(a')'a' +a = a+a'

So aa'The pin cavity is a symmetrical matrix.

Because (a+a')'Stuffiness = a'-(a')'a' -a = a+a' )

So aa'is an antisymmetric matrix.