Answer to Question #214070 in Linear Algebra for Simphiwe Dlamini

Question #214070

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


1
Expert's answer
2021-07-07T07:08:49-0400

Let u be

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


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