# Show that if `A` is an `nxxn` symmetric matrix, that is, `A^T = A`, then ⟨Ax; y⟩ = ⟨x;Ay⟩ for all x, y in `RR^n` where ⟨ ; ⟩ denotes the standard inner product on `RR^n.`

degeneratecircle | Certified Educator

To prove this, we'll express the standard inner product as a matrix multiplication. Writing our vectors `vecx` and `vecy` as `nxx1` matrices, and renaming them `X` and `Y` respectively, we have

`X=[[x_1],[x_2],[.],[.],[.],[x_n]]` and `Y=[[y_1],[y_2],[.],[.],[.],[y_n]].`

Then `<vecx;vecy> =x_1y_1+x_2y_2+...+x_ny_n=X^TY.` (see the remark at the bottom)

Now using this, the fact that `(AX)^T=X^TA^T` , and associativity of matrix multiplication, we get

`<Avecx;vecy> = (AX)^TY=X^TA^TY`

`=X^T(AY)= <vecx;Avecy>,` where the third equality is true because `A=A^T.` This proves the theorem.

Remark: To be pedantic, the right side of this is a `1xx1` matrix whose element is ` ``x_1y_1+x_2y_2+...+x_ny_n.` Since there is a one to one correspondence between `1xx1` matrices and real numbers, this distinction isn't important.