contestada

A square matrix A E Rn is said to be skew-symmetric if A =-A Prove that if A is skew-symmetric, then x'Ax = 0 for all x E R". (Hint The T T T quantity x'Ax is scalar so that (x'Ax) = x'Ax)

Respuesta :

Answer:

You can use the given hint as follows:

Step-by-step explanation:

Let [tex]A[/tex] be a square matrix that is a skew-symmetric matrix. Since the matrix [tex]R={\bf x}^{T}A{\bf x}[/tex] is matrix of size [tex]1\times 1[/tex] then it can be identified with an scalar. It is clear that [tex]R=R^{T}[/tex]. Then applying the properties of transposition we have

[tex]({\bf x}^{T}A{\bf x})^{T}=({\bf x}^{T})A^{T}({\bf x}^{T})^{T}={\bf x}^{T}(-A){\bf x}=-{\bf x}^{T}A{\bf x}[/tex]

Then,

[tex]{\bf x}^{T}A{\bf x}+{\bf x}^{T}A{\bf x}=0[/tex]

[tex]2{\bf x}^{T}A{\bf x}=0[/tex]

Then,

[tex]{\bf x}^{T}A{\bf x}=0[/tex]

For all column vector [tex]{\bf x}[/tex] of size [tex]n\times 1[/tex] .