3
$\begingroup$

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!

$\endgroup$
2
  • 3
    $\begingroup$ For SP 1, you may note that since $X^{T} M Y$ is a $1 \times 1$ matrix, it equals its transpose $(X^{T} M Y)^{T} = Y^{T} M^{T} (X^{T})^{T} = Y^{T} M X.$ $\endgroup$ Mar 15, 2019 at 14:49
  • $\begingroup$ @AndreasCaranti Thank you for the information! That's sufficient to show the SP 1 property. $\endgroup$
    – ShellRox
    Mar 15, 2019 at 15:18

1 Answer 1

1
$\begingroup$

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 $X^T \mathbb{1} Y < 0$.

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×𝑛?)

$F$ 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 $X^T M$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .