Let \(A\) be a \(2 \times 2\) matrix with real entries. For \(X\), \(Y\) in \(\RR^{2 \times 1}\) let \[ f_A(X, Y) = Y^t A X. \] Show that \(f_A\) is an inner product on \(\RR^{2 \times 1}\) if and only if \(A = A^t\), \(A_{11} > 0\), \(A_{22} > 0\), and \(\det A > 0\).
Solution
Write \[ A=\begin{pmatrix}a&b\\c&d\end{pmatrix}. \] The form \(f_A(X,Y)=Y^tAX\) is an inner product exactly when \[ \boxed{A=A^t,\qquad a>0,\qquad d>0,\qquad \det A>0.} \]
1. Symmetry of the form is equivalent to symmetry of \(A\).
For all column vectors \(X,Y\), \[ f_A(X,Y)=Y^tAX, \qquad f_A(Y,X)=X^tAY=Y^tA^tX. \] Thus \(f_A(X,Y)=f_A(Y,X)\) for every \(X,Y\) if and only if \(Y^t(A-A^t)X=0\) for every \(X,Y\), which is equivalent to \(A=A^t\). Bilinearity is automatic for every real matrix \(A\).
2. Positive definiteness implies the three numerical inequalities.
Assume \(f_A\) is an inner product. By step 1 write \(A=\begin{pmatrix}a&b\\b&d\end{pmatrix}\). Testing the nonzero standard basis vectors gives \(a>0\) and \(d>0\). For every \(t\in\mathbb R\), \[ 0<f_A((t,1)^t,(t,1)^t)=at^2+2bt+d. \] A real quadratic with positive leading coefficient that is strictly positive for every \(t\) has negative discriminant, so \[ 4b^2-4ad<0, \] i.e. \(ad-b^2=\det A>0\).
3. The stated conditions imply positive definiteness.
Conversely assume \(A=A^t\), \(a>0\), \(d>0\), and \(ad-b^2>0\). For \(X=(x,y)^t\), \[ X^tAX=ax^2+2bxy+dy^2 =a\left(x+\frac ba y\right)^2+\frac{ad-b^2}{a}y^2. \] Both coefficients are positive. Hence this expression is nonnegative and can vanish only when \(y=0\) and then \(x=0\). Therefore \(f_A\) is positive definite. Together with bilinearity and symmetry, it is an inner product [@HK71].