# Showing scalar product properties on certain matrix multiplication

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 $$2x2$$ 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!

• For SP 1, you may note that since $X^{T} M Y$ is a $1 \times 1$ matrix, it equals its transpose $(X^{T} M Y)^{T} = Y^{T} M^{T} (X^{T})^{T} = Y^{T} M X.$ Mar 15 '19 at 14:49
• @AndreasCaranti Thank you for the information! That's sufficient to show the SP 1 property. Mar 15 '19 at 15:18

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 $$X^T M$$.