Prove/Disprove $\left\langle u,v\right\rangle=5\,u_{2}\,v_{2}+8\,u_{1}\,v_{2}+8\,v_{1}\,u_{2}+3\,u_{1}\,v_{1}$ is a inner product

Question

$$\left\langle u,v\right\rangle=5\,u_{2}\,v_{2}+8\,u_{1}\,v_{2}+8\,v_{1}\,u_{2}+3\,u_{1}\,v_{1}$$

1.$$\langle w+u,x\rangle=5(w_2+u_2)v_2+8(w_1+u_1)v_2+8(w_2+u_2)v_1+3((w_2+u_2)v_1=\\=5(x_2v_2+u_2v_2)+8(w_1v_2+u_1v_2)+8(w_2v_1+u_2v_1)+3(w_2v_1+u_2v_1)=\\=5(w_2v_2)+5(u_2v_2)+8(w_1v_2)+8(u_1v_2)+8(w_2v_1)+8(u_2v_1)+3(w_2v_1)+3(u_2v_1)=\\=5(w_2v_2)+8(w_1v_2)+8(w_2v_1)+3(w_2v_1)+5(u_2v_2)+8(u_1v_2)+8(u_2v_1)+3(u_2v_1)=\\=\langle w,v\rangle+\langle u,v\rangle$$

2.$$\left\langle \alpha u,v\right\rangle= 5 \alpha\,u_{2}\,v_{2}+8 \alpha\,u_{1}\,v_{2}+8\,v_{1} \alpha\,u_{2}+3 \alpha\,u_{1}\,v_{1}=\\= \alpha(5\,u_{2}\,v_{2}+8\,u_{1}\,v_{2}+8\,v_{1}\,u_{2}+3\,u_{1}\,v_{1})=\alpha\left\langle u,v\right\rangle$$

3.$$\left\langle v,v\right\rangle=5\,v_{2}\,v_{2}+8\,v_{1}\,v_{2}+8\,v_{1}\,v_{2}+3\,v_{1}\,v_{1}=5(v_2)^2+16v_1v_2+3(v_1)^2$$

For some $v_1$ and $v_2$ ,$\left\langle v,v\right\rangle\neq 0$ but how can I prove it?

Answer 1

Well, continue:

$$\langle v,v\rangle=3v_1^2+16v_1v_2+5v_2^2=3(v_1+v_2)^2+10v_1v_2+2v_2^2=$$

$$3(v_1+v_2)^2+2\left(v_2+\frac52v_1\right)^2-\frac{25}2v_2^2$$

Can the above be non-positive? Yes, it can. For example, with $\;v=\left(-1,1\right)\;$

Answer 2

The standard way to continue from $3v_1^2+16v_1v_2+5v_2^2$ is by completing the square. You have $3v_1^2+16v_1v_2+5v_2^2=3(v_1^2+\frac{16}3v_2^2+\frac53v_2^2)$ and \begin{align*} v_1^2+\frac{16}3v_2^2+\frac53v_2^2 &=(v_1+\frac83v_2)^2+(\frac53-\frac{8^2}{3^2})v_2^2\\ &=(v_1+\frac83v_2)^2+\frac{5\cdot3-8^2}{3^2}v_2^2\\ &=(v_1+\frac83v_2)^2-\frac{49}9v_2^2 \end{align*} If you choose $v_{1,2}$ in such way that $v_2=1$ and $v_1+\frac83v_2=0$, then you get $\langle v_1,v_2\rangle=0$.

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If you have already learned about relation between (real) inner products and symmetric matrices, then this is the same as asking whether the symmetric matrix $$ \begin{pmatrix} 5 & 8 \\ 8 & 3 \end{pmatrix} $$ is positive definite.

If you check Sylvester's criterion, it suffices to check two determinants. And you will find out that the second of them is negative: $$ \begin{vmatrix} 5 & 8 \\ 8 & 3 \end{vmatrix}=5\cdot3-8\cdot8<0. $$