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.

  • $\begingroup$ If I understand correctly, $$f_A(Y, X) = X^TAY = X^TA^TY = Y^TAX = f_A(X, Y)$$ $\endgroup$
    – PSPACEhard
    Sep 5, 2016 at 12:58
  • $\begingroup$ @NP-hard Why $X^TA^TY=Y^TAX$? $\endgroup$
    – user42912
    Sep 5, 2016 at 13:00
  • 2
    $\begingroup$ $X^TA^TY$ is a real number, so its transpose equals itself. $\endgroup$
    – PSPACEhard
    Sep 5, 2016 at 13:01
  • $\begingroup$ You should also just write it out in terms of the entries of $A$, $X$, and $Y$. $\endgroup$ Sep 5, 2016 at 13:04
  • 1
    $\begingroup$ because the transpose of a real number is the same number $\endgroup$ Sep 6, 2016 at 2:40

1 Answer 1


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 $g\left(\frac{x_1}{x_2}\right)$.


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