Let a be a real symmetric matrix of order n, and a-3a + 3a-e = 0, prove that a = E

Let a be a real symmetric matrix of order n, and a-3a + 3a-e = 0, prove that a = E

Let λ be the eigenvalue of a, then
λ^3-3λ^2+3λ-1=0
λ=1
So, a is similar to E
There is an invertible matrix P such that
P^(-1)·A·P=E
∴A=P·E·P^(-1)=E

Proof: if a and B are symmetric matrices of order n, then 2a-3b are symmetric matrices and ab-ba are antisymmetric matrices

If a and B are symmetric matrices of order n, then transpose of a = a, transpose of B = B. transpose of (2a -- 3b) = transpose of 2 * a - transpose of 3 * b = transpose of 2A -- 3B 〈 2a-3b is also symmetric matrix. Transpose of (AB -- BA) = transpose of (AB) - (BA) = transpose of B * transpose of a - transpose of a * transpose of B = transpose of Ba -- AB = (AB -- BA) 〉 ab-ba

Let a be an orthogonal matrix and prove that a ^ 2 is also an orthogonal matrix

Definition of orthogonal matrix:
Let a be a square matrix of order n. if a'a = e, then a is called an orthogonal matrix, where a 'denotes the transpose matrix of A
It is proved that a'a = e because a is an orthogonal matrix
By the property of transpose (AB) '= b'a'
So (a ^ 2) '(a ^ 2) = (a'a') (AA) = a '(a'a) a = a'ea = a'a = E
So a is an orthogonal matrix#

Let a be an orthogonal matrix, and prove that a ^ * is also an orthogonal matrix

Because a is an orthogonal matrix, so | a ^ ^ 2 = 1, a ^ - 1 is also an orthogonal matrix, ((a ^ - 1) ^ t (a ^ - 1) = (a ^ t) ^ - 1 (a ^ - 1) = (AA ^ t) ^ - 1 = e ^ - 1 = e), so (a *) ^ ta * = (| a ^ - 1) ^ t (| a ^ - 1) = | a ^ ^ 2 (a ^ - 1) ^ t (a ^ - 1) = e, so a * is also an orthogonal matrix

Let a and B be orthogonal matrices of order n, and | ab|

Proof: because the determinant of orthogonal matrix is positive and negative 1
And then by the AB|

Let a and B be orthogonal matrices of order n, and | a | * | B | = - 1, prove that | a + B | = 0 is different!

Because a and B are orthogonal matrices
So AA ^ t = a ^ TA = e, BB ^ t = B ^ TB = E
And because | a | B | = - 1
So - | a + B|
= - |(A+B)^T|
= - |A^T+B^T|
= |A||A^T+B^T||B|
= |AA^TB+AB^TB|
= |B+A|
= |A+B|
So | a + B | = 0

Let AB be an orthogonal matrix of order n and | a | B | = - 1, prove | a + B | = 0

Because a and B are orthogonal moments, AA ^ t = e, BB ^ t = E
So a ^ t (a + b) B ^ t = B ^ t + A ^ t = (a + b) ^ t
therefore
|A^T(A+B)B^T|=|(A+B)^T|=|A+B|
Namely
|A^T||(A+B)||B^T|=|A+B|
|A||A+B||B|=|A+B|
-|A+B|=|A+B|
|A+B|=0

It is proved that if n is even and | a | = - 1, then | e-A | = 0

Let A. B be an orthogonal matrix of order n and n be an odd number. It is proved that | (a-b) (a + b) | = 0

A and B are orthogonal matrices of order n, and N is odd. It is proved that: (a-b) (a + b) = 0

Finally, we prove that determinant is 0, not matrix product is 0
Counter proof: if A-B and a + B are nonsingular, then (a-b) ^ t (a + b) = a ^ ta-b ^ TA + A ^ tb-b ^ TB = a ^ tb-b ^ TA is nonsingular, but a ^ tb-b ^ TA is odd antisymmetric, and the determinant must be 0, which is contradictory