# Doubt in the question 8 section 8.1 (Hoffman and Kunze linear algebra book)

I'm studying Hoffman and Kunze's linear algebra book and I'm having troubles how to prove this exercise on page 276: I didn't understand why $f_A(X,Y)=f_A(Y,X)$. Even if I assume $A=A^t$ I can't get the symmetry of the real inner product.

• If I understand correctly, $$f_A(Y, X) = X^TAY = X^TA^TY = Y^TAX = f_A(X, Y)$$ Sep 5, 2016 at 12:58
• @NP-hard Why $X^TA^TY=Y^TAX$? Sep 5, 2016 at 13:00
• $X^TA^TY$ is a real number, so its transpose equals itself. Sep 5, 2016 at 13:01
• You should also just write it out in terms of the entries of $A$, $X$, and $Y$. Sep 5, 2016 at 13:04
• because the transpose of a real number is the same number Sep 6, 2016 at 2:40

Hint: As explained in the comments above, we can show that $A$ must be symmetric by noting that $f_A(X,Y) = [f_A(X,Y)]^T$.
As for the other conditions: write $X = (x_1,x_2)^T$. Note that if we assume that $A$ is symmetric, we have $$f_A(X,X) = A_{11}x_{1}^2 + 2A_{12} x_1x_2 + A_{22}x_2^2 = x_2^2 \left( A_{11}\left(\frac{x_1}{x_2}\right)^2 + 2A_{12} \left(\frac{x_1}{x_2}\right) + A_{22}\right)$$ Now, under which conditions on the elements do we have:
• $f_A(X,X) \geq 0$ for all $X$
• $f_A(X,X) = 0$ if and only if $X = 0$
It helps to consider the discriminant of the quadratic function $g\left(\frac{x_1}{x_2}\right)$.