Answer to Question #214070 in Linear Algebra for Simphiwe Dlamini

Question

Suppose u, v $\in$ V. Prove that ||au+bv|| = ||bu+av|| for all a, b $\in$ R if and only if ||u|| = ||v||.

Answer 1

Let u be

$\vec{u}=x_1i+y_1j\\ and\\ \vec{v}=x_2i+y_2j\\ a\vec{u}=ax_1i+ay_1j\\ b\vec{v}=bx_2i+by_2j\\ a\vec{u}+b\vec{v}=bx_2i+by_2j +ax_1i+ay_1j\\ a\vec{u}+b\vec{v}=(ax_1+bx_2)i+(ay_1+by_2)j\\ \|a\vec{u}+b\vec{v}\|=\\\sqrt{(ax_1+bx_2)^2+(ay_1+by_2)^2}\\ b\vec{u}=bx_1i+by_1j\\ a\vec{v}=ax_2i+ay_2j\\ b\vec{u}+a\vec{v}=(bx_1+ax_2)i+(by_1+ay_2)j\\ \|b\vec{u}+a\vec{v}\|=\sqrt{(bx_1+ax_2)^2+(by_1+ay_2)^2}\\ \|b\vec{u}+a\vec{v}\|=\|a\vec{u}+b\vec{v}\|\\ \sqrt{(ax_1+bx_2)^2+(ay_1+by_2)^2}=\\\sqrt{(bx_1+ax_2)^2+(by_1+ay_2)^2}\\ (ax_1+bx_2)^2+(ay_1+by_2)^2=\\(bx_1+ax_2)^2+(by_1+ay_2)^2\\ (ax_1+bx_2)^2-(bx_1+ax_2)^2=\\(by_1+ay_2)^2-(ay_1+by_2)^2\\ (ax_1+bx_2+bx_1+ax_2)(ax_1+bx_2-bx_1-ax_2)\\ =(by_1+ay_2+ay_1+by_2)(by_1+ay_2-ay_1-by_2)\\ ((a+b)x_1+(a+b)x_2)((a-b)x_1-(a-b)x_2)=\\((a+b)y_1+(a+b)y_2)((b-a)y_1-(b-a)y_2)\\ (a^2-b^2)(x_1^2-x_2^2)=-(a^2-b^2)(y_1^2-y_2^2)\\ (x_1^2-x_2^2)=-(y_1^2-y_2^2)\\ (x_1^2-x_2^2)=-y_1^2+y_2^2\\ (x_1^2+y_1^2)=(x_2^2+y_2^2)\\ \sqrt{(x_1^2+y_1^2)}=\sqrt{(x_2^2+y_2^2)}\\ \therefore\\ \|\vec{u_1}\|=\|\vec{v_1}\|$