Prove that if (x+y).(x-y)=0 then ||x||=||y|| where x, y are vectors in |Rn | LINEAR ALGEBRA GE

Q) Prove that if (x+y).(x-y)=0 then ||x||=||y|| where x,y are vectors in |Rn

Ans. 

To prove that if $(𝑥+𝑦)\cdot (𝑥-𝑦)=0$, then $\mathrm{\parallel }𝑥\mathrm{\parallel }=\mathrm{\parallel }𝑦\mathrm{\parallel }$ where $𝑥$ and $𝑦$ are vectors in ${\mathbb{𝑅}}^{𝑛}$, let's use the given dot product equation and properties of vectors:

Given: $(𝑥+𝑦)\cdot (𝑥-𝑦)=0$

Expanding the dot product: $(𝑥+𝑦)\cdot (𝑥-𝑦)=𝑥\cdot 𝑥-𝑥\cdot 𝑦+𝑦\cdot 𝑥-𝑦\cdot 𝑦$ $=\mathrm{\parallel }𝑥{\mathrm{\parallel }}^2-\mathrm{\parallel }𝑦{\mathrm{\parallel }}^2$

Given that $(𝑥+𝑦)\cdot (𝑥-𝑦)=0$, we have: $\mathrm{\parallel }𝑥{\mathrm{\parallel }}^2-\mathrm{\parallel }𝑦{\mathrm{\parallel }}^2=0$

This implies: $\mathrm{\parallel }𝑥{\mathrm{\parallel }}^2=\mathrm{\parallel }𝑦{\mathrm{\parallel }}^2$

Taking the square root of both sides (assuming $\mathrm{\parallel }𝑥\mathrm{\parallel }$ and $\mathrm{\parallel }𝑦\mathrm{\parallel }$ are non-negative): $\mathrm{\parallel }𝑥\mathrm{\parallel }=\mathrm{\parallel }𝑦\mathrm{\parallel }$

Thus, we've proven that if $(𝑥+𝑦)\cdot (𝑥-𝑦)=0$, then $\mathrm{\parallel }𝑥\mathrm{\parallel }=\mathrm{\parallel }𝑦\mathrm{\parallel }$ for vectors $𝑥,𝑦$ in ${\mathbb{𝑅}}^{𝑛}$

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