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.
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$\begingroup$ If I understand correctly, $$f_A(Y, X) = X^TAY = X^TA^TY = Y^TAX = f_A(X, Y)$$ $\endgroup$– PSPACEhardCommented Sep 5, 2016 at 12:58
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$\begingroup$ @NP-hard Why $X^TA^TY=Y^TAX$? $\endgroup$– user42912Commented Sep 5, 2016 at 13:00
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2$\begingroup$ $X^TA^TY$ is a real number, so its transpose equals itself. $\endgroup$– PSPACEhardCommented Sep 5, 2016 at 13:01
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$\begingroup$ You should also just write it out in terms of the entries of $A$, $X$, and $Y$. $\endgroup$– Matthew LeingangCommented Sep 5, 2016 at 13:04
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1$\begingroup$ because the transpose of a real number is the same number $\endgroup$– Steven Alexis GregoryCommented Sep 6, 2016 at 2:40
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1 Answer
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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 $g\left(\frac{x_1}{x_2}\right)$.
answered Sep 6, 2016 at 2:16