anonymous
  • anonymous
[linear algebra] <(x1,x2),(y1,y2)> = x1y1 + tx2y2 For which valor of t, it is an in internal (scalar) product?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The internal product (i.e. dot product) of \[\left(\begin{matrix}x1 \\ y1\end{matrix}\right)\] with \[\left(\begin{matrix}x2 \\ y2\end{matrix}\right)\] is \[\left(\begin{matrix}x1+x2 \\ y1+y2\end{matrix}\right)\] So the only value of t which gives this is t=1
helder_edwin
  • helder_edwin
u have to check one by one the properties of an inner product: (assuming an vector space over \(\mathbb{R}\)) \[ \large \langle a,b\rangle=\langle b,a\rangle \] \[ \large \langle a,b+c\rangle=\langle a,b\rangle+\langle a,c\rangle \] \[ \large \langle\alpha a,b\rangle=\langle a,\alpha b\rangle=\alpha\langle a,b\rangle \] \[ \large \langle a,a\rangle\geq0\quad\text{and}\quad \langle a,a\rangle=0\Leftrightarrow a=0 \]
anonymous
  • anonymous
Ok so I guess inner product isn't just another name you're using for dot product then?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

helder_edwin
  • helder_edwin
that is not true. u can write: \[ \large \langle a,b\rangle=(a_1\quad a_2)\begin{pmatrix} 1 & 0\\ 0 & t \end{pmatrix} \binom{b_1}{b_2} \]
helder_edwin
  • helder_edwin
\[ \large =a_1b_1+ta_2b_2 \]
anonymous
  • anonymous
The determinant must be > 0, right? Wich would give the answer t>0. I didn't learn all this procces (I don't really know why, but my teacher didn't explicitly show this transformation of scalar product into matrices product). How I solve by the properties?
helder_edwin
  • helder_edwin
not the determinant!! for example u might have the matrix \[ \large \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix} \] has determinant >0 but the resulting expresion is NOT an inner product.
anonymous
  • anonymous
I see. I'm gonna study more of Linear Algebra, for I don't want to ask you to teach me everything about it hahaha. Thank you very much ;)
helder_edwin
  • helder_edwin
i think it is a terrific idea. i recomend any of Strang's linear algebra books. there are just great. also hoffman & kunze's
anonymous
  • anonymous
This Strang is from Gilbert Strang?
helder_edwin
  • helder_edwin
yes. the one and only
anonymous
  • anonymous
Understood! Thanks again.

Looking for something else?

Not the answer you are looking for? Search for more explanations.