Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\langle x, y \rangle_{A_k} :=x^{t}A_{k}y$ for all $x,y \in \mathbb{R}^{3}$ with $ A_{k} = \left( \begin{array}{rrr} 1 & k & 0 \\ k & 3 & -2 \\ 0 & -2 & 5 \end{array}\right),k \in \mathbb{R}^{3}.$

For which $k \in \mathbb{R}$ is the vector space $(\mathbb{R}^3, \langle , \rangle_{A_k})$ a euclidean space?

I wasn't really sure what they were trying to have me test here... By a euclidean vector space all I understand is a pair $(V, \langle \cdot, \cdot \rangle)$ made up of a real vector space and a scalar product $\langle \cdot, \cdot \rangle$ on $V$.

So does this mean that the task at hand is to examine $\langle x, y \rangle_{A_k}$ and see for which $k$ it will actually be a scalar product? This entails showing (i) bilinearity, (ii) symmetry, and (iii) positive definiteness ?

I haven't encountered scalar products with a matrix before, is it safe to say that a scalar product of the form $x^{t}Ay$ is symmetric $\Leftrightarrow A$ is symmetric? If so, then (ii) is already taken care of...

I don't really know what should be done for (i)...

For (iii) I think there is some theorem... What are the minors along the diagonal called? I think: all minors along the diagonal have positive determinants $\Rightarrow A$ is positive definite... Is the converse also always true? What about with eigenvalues? All eigenvalues are positive $\Leftrightarrow A$ is positive definite, right?

In this case I tried saying: $\det(1) > 0 \forall k \in \mathbb{R}$, $\det \left( \begin{array}{rr} 1 &k \\ k & 3 \end{array} \right) > 0 \Leftrightarrow k < \sqrt{3}$, and $\det \left( \begin{array}{rrr} 1 & k & 0 \\ k &3 &-2 \\ 0 &-2 &5 \end{array} \right) > 0 \Leftrightarrow k< \sqrt{\frac{11}{5}}$.

Since $\sqrt{\frac{11}{5}} < \sqrt{3}$ we take the stricter condition and say for $\{k \in \mathbb{R} | k< \sqrt{\frac{11}{5}} \}$ the vector space $(\mathbb{R}^{3}, \langle , \rangle_{A_k})$ is a euclidean space. Is this correct?

I also have a question about the way I am wording this: should I say "the vector space" when talking about the same pair with respect to other $k$. I mean assuming that what I claim about $k$ is correct, if $k \geq \sqrt{\frac{11}{5}}$ how would one correctly refer to this: $(\mathbb{R}^{3}, \langle , \rangle_{A_k})$ ? It would definitely not be true to say that $\langle , \rangle_{A_k}$ is a scalar product at least(I assume), but what sort of vector space would that pair be if it weren't euclidean?

As always, any help is very much appreciated :)

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

Your interpretation is correct. Indeed, (i) is straightforward, (ii) is always true because the matrix is symmetric. As for the last point, what you're talking about is called Sylvester's criterion and is an equivalence. Also, note that $k^2 < a^2$ implies $|k| < |a|$ (which is two inequalities for $k$ : $-|a|< k < |a|$).

share|improve this answer
    
@Listing : You're right, it's simple but not a definition. –  Joel Cohen Jul 5 '11 at 12:01
    
Ok, I removed the comment :-) –  Listing Jul 5 '11 at 12:06
    
thank you for the explanation! –  ghshtalt Jul 6 '11 at 12:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.