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My analysis instructor has posed an exercise to me in the following format:
"For [tex]{\bf{u}} = (u_1,u_2), {\bf{v}} = (v_1,v_2) \in R^2[/tex] define [tex]\left\langle{\bf{u}},{\bf{v}}\right\rangle = 3u_1u_2 - u_1v_2 - u_2v_1 + \frac{1}{2}u_2v_2..[/tex]
Show that this is an inner product on ##R^2##."
Not sure what the dots at the end represent, but I'm just posting it exactly as it appears. The textbook we're using (Advanced Calculus, Edwards) gives the criteria for an inner product as
1)[tex]\left\langle{\bf{x}},{\bf{x}}\right\rangle > 0, {\bf{x}} \neq 0 [/tex]
2) [tex]\left\langle{\bf{x}},{\bf{y}}\right\rangle = \left\langle{\bf{y}},{\bf{x}}\right\rangle[/tex]
3) [tex]\left\langle{a}{\bf{x}} + b{\bf{y}},{\bf{z}}\right\rangle = a\left\langle{\bf{x}},{\bf{z}}\right\rangle + b\left\langle{\bf{y}},{\bf{z}}\right\rangle[/tex]
I assume proving these would be sufficient, since this course doesn't deal with complex space. Thus, my attempt at a solution is:
1) [tex]\left\langle{\bf{u}},{\bf{u}}\right\rangle = 3u_1u_2 - u_1u_2 - u_2u_1 + \frac{1}{2}{u_2}^2 = u_1u_2 + \frac{1}{2}{u_2}^2[/tex]
Which is non-zero if ##{\bf{u}}## is non-zero. Next,
2) [tex]\left\langle{\bf{v}},{\bf{u}}\right\rangle = 3v_1v_2 - v_1u_2 - v_2u_1 + \frac{1}{2}v_2u_2 = \left\langle{\bf{u}},{\bf{v}}\right\rangle[/tex]
Proving property 3 seems to be giving me trouble. I believe I should begin by letting: [tex]{\bf{w}} = (w_1,w_2)[/tex]
Then, [tex]\left\langle{a}{\bf{u}} + b{\bf{v}},{\bf{w}}\right\rangle = 3(au_1 + bv_1)(au_2 + bv_2) - (au_1 + bv_1)w_2 - (au_2 + bv_2)w_1 + \frac{1}{2}(au_2 + bv_2)w_2[/tex]
The first term seems to pose a problem in rewriting in the desired form because I'm getting a quadratic-like form with the constants. Other than that, that's the only way I see how to show this. Any thoughts? Much appreciated.
"For [tex]{\bf{u}} = (u_1,u_2), {\bf{v}} = (v_1,v_2) \in R^2[/tex] define [tex]\left\langle{\bf{u}},{\bf{v}}\right\rangle = 3u_1u_2 - u_1v_2 - u_2v_1 + \frac{1}{2}u_2v_2..[/tex]
Show that this is an inner product on ##R^2##."
Not sure what the dots at the end represent, but I'm just posting it exactly as it appears. The textbook we're using (Advanced Calculus, Edwards) gives the criteria for an inner product as
1)[tex]\left\langle{\bf{x}},{\bf{x}}\right\rangle > 0, {\bf{x}} \neq 0 [/tex]
2) [tex]\left\langle{\bf{x}},{\bf{y}}\right\rangle = \left\langle{\bf{y}},{\bf{x}}\right\rangle[/tex]
3) [tex]\left\langle{a}{\bf{x}} + b{\bf{y}},{\bf{z}}\right\rangle = a\left\langle{\bf{x}},{\bf{z}}\right\rangle + b\left\langle{\bf{y}},{\bf{z}}\right\rangle[/tex]
I assume proving these would be sufficient, since this course doesn't deal with complex space. Thus, my attempt at a solution is:
1) [tex]\left\langle{\bf{u}},{\bf{u}}\right\rangle = 3u_1u_2 - u_1u_2 - u_2u_1 + \frac{1}{2}{u_2}^2 = u_1u_2 + \frac{1}{2}{u_2}^2[/tex]
Which is non-zero if ##{\bf{u}}## is non-zero. Next,
2) [tex]\left\langle{\bf{v}},{\bf{u}}\right\rangle = 3v_1v_2 - v_1u_2 - v_2u_1 + \frac{1}{2}v_2u_2 = \left\langle{\bf{u}},{\bf{v}}\right\rangle[/tex]
Proving property 3 seems to be giving me trouble. I believe I should begin by letting: [tex]{\bf{w}} = (w_1,w_2)[/tex]
Then, [tex]\left\langle{a}{\bf{u}} + b{\bf{v}},{\bf{w}}\right\rangle = 3(au_1 + bv_1)(au_2 + bv_2) - (au_1 + bv_1)w_2 - (au_2 + bv_2)w_1 + \frac{1}{2}(au_2 + bv_2)w_2[/tex]
The first term seems to pose a problem in rewriting in the desired form because I'm getting a quadratic-like form with the constants. Other than that, that's the only way I see how to show this. Any thoughts? Much appreciated.