1
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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

2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.