anonymous
  • anonymous
[linear algebra] <(x1,x2),(y1,y2)> = x1y1 + tx2y2 For which valor of t, it is an in internal (scalar) product?
Mathematics
schrodinger
  • schrodinger
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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?

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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.

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