Answer to Question #214278 in Linear Algebra for sabelo Zwelakhe Xu

Question #214278

 Suppose u; v € V. Prove that

||au + bv|| = ||bu +

av|| for all a; b € R if and only if ||u|| = ||

v||

.


1
Expert's answer
2021-07-20T12:22:53-0400

u=x1i+y1jandv=x2i+y2jau=ax1i+ay1jbv=bx2i+by2jau+bv=bx2i+by2j+ax1i+ay1jau+bv=(ax1+bx2)i+(ay1+by2)jau+bv=(ax1+bx2)2+(ay1+by2)2bu=bx1i+by1jav=ax2i+ay2jbu+av=(bx1+ax2)i+(by1+ay2)jbu+av=(bx1+ax2)2+(by1+ay2)2bu+av=au+bv(ax1+bx2)2+(ay1+by2)2=(bx1+ax2)2+(by1+ay2)2(ax1+bx2)2+(ay1+by2)2=(bx1+ax2)2+(by1+ay2)2(ax1+bx2)2(bx1+ax2)2=(by1+ay2)2(ay1+by2)2(ax1+bx2+bx1+ax2)(ax1+bx2bx1ax2)=(by1+ay2+ay1+by2)(by1+ay2ay1by2)((a+b)x1+(a+b)x2)((ab)x1(ab)x2)=((a+b)y1+(a+b)y2)((ba)y1(ba)y2)(a2b2)(x12x22)=(a2b2)(y12y22)(x12x22)=(y12y22)(x12x22)=y12+y22(x12+y12)=(x22+y22)(x12+y12)=(x22+y22)u1=v1\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}\|



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