# Show that $f_A$ is an inner product

Let $$A$$ be a $$2 \times 2$$ matrix with real entries. For $$X, Y$$ in $$R^{2 \times 1}$$ let

$$f_A(X, Y) = Y^tAX$$.

Show that $$f_A$$ is an inner product on $$R^{2 \times 1}$$ if and only if $$A = A^t$$, $$A_{11} > 0$$, $$A_{22} > 0$$, and $$det A > 0$$.

I was able to solve the first part, assuming the internal product. How to do reciprocal? I could only verify that $$f_A(cX+Z, Y) = cf_A(X, Y)+f_A(Z, Y)$$. But I could not complete the other two conditions, for example

$$f_A(X, Y) = Y^tAX= Y^tA^tX= \{X^tAY\}^t$$, how to conclude that it is equal to $${ f_A(Y, X) }$$ ( in this case there is no conjugation bar because everything is real).

Any tips?

• You should note that $f(\cdot, \cdot)$ is an $1 \times 1$ matrix, i.e. simply a number. Then its transpose is of course itself. – xbh Feb 14 at 3:38
• For the positive definiteness, use $A_{1,1}, A_{2,2}, \det A$ by taking specific $X, Y$'s. – xbh Feb 14 at 3:40

You're basically there with (conjugate) symmetry: note that $$X^\top A Y = f(Y, X)$$. It's also a scalar (well, a $$1 \times 1$$ matrix), so the transpose is superfluous.
As for the positive-definiteness, this is where we need to start making use of the individual entries of $$A$$. Suppose $$X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$. Then, $$f_A(X, X) = \begin{bmatrix} x_1 & x_2 \end{bmatrix} \begin{bmatrix} A_{11} & A_{12} \\ A_{12} & A_{22} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = A_{11}x_1^2 + A_{22} x_2^2 + 2A_{12}x_1x_2$$ We want this to be greater than or equal to $$0$$. Start by completing the square (remember $$A_{11}, A_{22} > 0$$): \begin{align*} f_A(X, X) &= A_{11}\left(x_1^2 + 2\frac{A_{12}}{A_{11}} x_1x_2 \right) + A_{22}x_2^2 \\ &= A_{11}\left(x_1^2 + 2\frac{A_{12}}{A_{11}} x_1x_2 + \left(\frac{A_{12}}{A_{11}} x_2\right)^2\right) + A_{22}x_2^2 - \frac{A_{12}^2}{A_{11}}x_2^2 \\ &= A_{11}\left(x_1 +\frac{A_{12}}{A_{11}}x_2\right)^2 + \frac{A_{22}A_{11} - A_{12}^2}{A_{11}}x_2^2 \\ &= A_{11}\left(x_1 +\frac{A_{12}}{A_{11}}x_2\right)^2 + \frac{\det A}{A_{11}}x_2^2. \end{align*} Since $$A_{11} > 0$$ and $$\det A > 0$$, it follows that $$f_A(X, X) \ge 0$$. Further, if $$f_A(X, X) = 0$$, then both the squares must be $$0$$, in other words, $$x_2 = 0$$ and $$x_1 + \frac{A_{12}}{A_{11}} x_2 = x_1 = 0$$. That is, $$X = 0$$, completing the proof.