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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?

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    $\begingroup$ 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. $\endgroup$ – xbh Feb 14 at 3:38
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    $\begingroup$ For the positive definiteness, use $A_{1,1}, A_{2,2}, \det A$ by taking specific $X, Y$'s. $\endgroup$ – xbh Feb 14 at 3:40
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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.

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    $\begingroup$ I think the converse was done by the OP. $\endgroup$ – xbh Feb 14 at 3:59
  • $\begingroup$ @Theo Bendit very good! Thank you! $\endgroup$ – Ricardo Freire Feb 14 at 12:18

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