(7) Let $\alpha = (u_1, u_2, u_3)$ and $\beta = (v_1, v_2, v_3)$ be vectors in $\mathbb{R}^3$. Does the function \begin{equation*} <\alpha, \beta> = u_1v_1 - 2u_2v_2 + u_3v_3 \end{equation*} defines an inner product on $\mathbb{R}^3$? If the answer is not show which of the properties of the definition of inner product are not valid.
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The first property of an inner product is that it is linear in the first argument. This means that for any vectors a, b, and c, and any scalar k, we have <ka + b, c> = k<a, c> + <b, c>. Let's check if this property holds for the given function: <ka + b, c> = (ka1 Show more…
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Q.4.(a): Show that the following function is not an inner product on R3, where u = (U1, U2, U3) and v = (V1, V2, V3) <u,v> = U1V1 + 2U2V2 + U3V3 (b): Let f and g be real-valued continuous functions in the vector space C[a,b]. Show that <f,g> = ∫(a to b) f(t)g(t) dt defines an inner product on [a,b].
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Show that the function does not define an inner product on $R^{3},$ where $u=\left(u_{1}, u_{2}, u_{3}\right)$ and $\mathbf{v}=\left(v_{1}, v_{2}, v_{3}\right)$ $$\langle\mathbf{u}, \mathbf{v}\rangle= 2 u_{1} u_{2}+3 v_{1} v_{2}+u_{3} v_{3}$$
Inner Product Spaces
Show that the function defines an inner product on $R^{3},$ where $\mathbf{u}=\left(u_{1}, u_{2}, u_{3}\right)$ and $\mathbf{v}=\left(v_{1}, v_{2}, v_{3}\right)$ $$\langle\mathbf{u}, \mathbf{v}\rangle= 2 u_{1} v_{1}+3 u_{2} v_{2}+u_{3} v_{3}$$
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