Question

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

Answer 1

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

Answer 2

u have to check one by one the properties of an inner product: (assuming an vector space over \(\mathbb{R}\))
\[ \large \langle a,b\rangle=\langle b,a\rangle \]
\[ \large \langle a,b+c\rangle=\langle a,b\rangle+\langle a,c\rangle \]
\[ \large \langle\alpha a,b\rangle=\langle a,\alpha b\rangle=\alpha\langle a,b\rangle \]
\[ \large \langle a,a\rangle\geq0\quad\text{and}\quad \langle a,a\rangle=0\Leftrightarrow a=0 \]

Answer 3

Ok so I guess inner product isn't just another name you're using for dot product then?

Answer 4

that is not true. u can write:
\[ \large \langle a,b\rangle=(a_1\quad a_2)\begin{pmatrix}
1 & 0\\ 0 & t
\end{pmatrix}
\binom{b_1}{b_2} \]

Answer 5

\[ \large =a_1b_1+ta_2b_2 \]

Answer 6

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?

Answer 7

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.

Answer 8

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 ;)

Answer 9

i think it is a terrific idea.
i recomend any of Strang's linear algebra books. there are just great.
also hoffman & kunze's

Answer 10

This Strang is from Gilbert Strang?

Answer 11

yes. the one and only

Answer 12

Understood! Thanks again.