from S.L Linear Algebra:
Let $M$ be a square $n \times n$ matrix which is equal to its
transpose. If $X$, $Y$ are column $n$-vectors, then:$$X^TMY $$
is a $1 \times 1$ matrix, which we identify with a number. Show that
the map:$$(X, Y) \mapsto X^TMY$$
satisfies the three properties SP 1, SP 2, SP 3. Give an
example of a $2 \times 2$ matrix $M$ such that the product is not positive
definite.
Let's observe scalar product properties mentioned:
SP 1. We have $\langle v, w \rangle\ = \langle w, v \rangle$ for all $v, w \in V$.
SP 2. If $u$, $v$, $w$ are elements of $V$, then $\langle u, v + w \rangle = \langle u, v \rangle + \langle u, w \rangle$.
SP 3. If $x \in K$, then $\langle xu, v \rangle = x\langle u, v \rangle$ and $\langle u, xv \rangle=x\langle u, v \rangle$
I'm not sure if I'm understanding the question properly, but I assume I have to show that linear map $F: \mathbb{K}^n \rightarrow \mathbb{K}^1$ defined by $(X, Y) \mapsto X^TMY$ is a transformation that inherits scalar product properties.
Thus for SP 1, we would have to show something like:
$$F(X, Y)=X^TMY=F(Y, X)=Y^TMX$$
which is not always true, unless $M$ is an identity matrix, in which case our transformation is a scalar product itself, and every SP rule is obviously satisfied.
But what if $M$ isn't an identity matrix? (In fact what's the purpose of $M$ in this case? If it's a matrix associated with our linear map, shouldn't it have dimensions $1 \times n$?)
For SP 2:
If we had $X, Y, Z$ as column $n$-vectors such that
$$(X, Y + Z) \mapsto X^TM(Y+Z)$$
which, due to distributive property of matrices it is equivalent to:
$$F(X, Y + Z) = X^TMY+ X^TMZ$$
which is sufficient to show that SP2 is satisfied by $F$.
For SP 3, we show that for some $x \in K$:
$$F(Xx, Y)=F(X, Yx)=xF(X, Y)$$
we see that:
$$F(Xx, Y)=(Xx)^TMY = x(X^TMY)$$
which is sufficient to prove SP 3.
For the final request – "give an example of a $2 \times 2$ matrix $M$ such that the product is not positive definite", I believe the simplest answer would be a $2×2$ matrix $(id) * -1$ where $id$ is an identity matrix. Since for any two vectors $X, Y$:
$$F(X,Y) \geq 0$$
is not always true.
Question:
I'm having a difficulty for proving SP 1 in cases where $M$ is not an identity matrix, did I understand a question incorrectly?
For SP 2 and SP 3, I believe I have delivered sufficient generalized information, but I'm not completely certain.
For the final request, I've given a simplest answer I could think of.
How do I show that a linear map $(X, Y) \mapsto X^TMY$ inherits scalar product properties for every $M: \mathbb{F}^n \rightarrow \mathbb{F}^n$? Are my other answers correct?
Thank you!
Best Answer
As Andreas noted in the comments, for the SP1 property you can notice $X^T M Y = (X^T M Y)^T \in \mathbb{K}$. Your proof for the other properties is correct.
About the final request. Let $\mathbb{1}$ be the identity matrix. $-\mathbb{1}$ is not positive definite, but not because $\exists X,Y \in \mathbb{K}^n$ such that $X^T (-\mathbb{1}) Y < 0$: a (real) matrix M is positive (semi)definite if $\forall X \in \mathbb{R}^n X^T M X \geq 0$, so you have to find $X \in \mathbb{R}^2$ such that $X^T (-\mathbb{1}) X < 0$...
An example that shows that this is the "right" definition: for the identity matrix $\mathbb{1}$ (that we want positive definite!) $\exists X,Y \in \mathbb{R}^n$ such that $X^T \mathbb{1} Y < 0$.
$F$ is a map from $\mathbb{K^n} \times \mathbb{K^n}$ to $\mathbb{K}$ and is called "bilinear" because is linear in both variables $X$ and $Y$. If you fix an $X \in \mathbb{K^n}$, then $Y \rightarrow X^T M Y$ is a linear map with associated matrix $X^T M$.