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If ||x||$_2$ = 3, ||4x-5y||$_2$ = 14 and ||5x+2y||$_2$ = 8 find ||y||$_2$

I had attempted to write at the vectors in explicit form and in doing so arrived at a point where I would add (||4x-5y||$_2$)$^2$ = (14)$^2$ and (2)(||5x+2y||$_2$)$^2$= 8$^2$(2) to remove <x,y>. From there I would subtract out 66||x||$_2$ by adding on (-66)(||x||$_2$)$^2$ = (-66)(9). Now I get a negative answer of -270/33. Is it possible to get a negative answer? I was under the impression that Euclidean norms acted like absolute value signs. Any advice and explanation would be appreciated.

Nuemann12
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  • What space are $x,y$ in? – Stuck Nov 17 '20 at 02:43
  • It is in the real numbers – Nuemann12 Nov 17 '20 at 02:48
  • Are you sure? Because the 2-norm of a real number is just the absolute value. Seems kind of silly to write 2-norm in that case. – Stuck Nov 17 '20 at 02:52
  • I am practicing for a class that covers the study of Real Numbers. I am assuming that it would be the Real Numbers, but I am treating this numbers by the standard definition of Euclidean norm anyway – Nuemann12 Nov 17 '20 at 02:56
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    My advice to you is to verify that the space is $\mathbb{R}$ (the real numbers) and not $\mathbb{R}^n$ for some natural number $n$. To answer one of your questions though, the $2$-norm cannot be negative. – Stuck Nov 17 '20 at 03:02
  • It is R$^n$ and I had been unwittingly treating it as such. Is there any reason why I keep getting negative answers? Can I simply negate it and say is 270/33 if its like an absolute value sign? – Nuemann12 Nov 17 '20 at 03:08
  • You cannot just negate the negative answer. You're getting a negative because you're making a mistake somewhere. – Stuck Nov 17 '20 at 03:29

2 Answers2

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Assume $\textbf{x},\textbf{y} \in \mathbb{R}^n$. Recall the definition of the Euclidean norm

$$||\textbf{x}||_2:=\sqrt{\textbf{x}\cdot \textbf{x}}.$$ So the equations you were given really read

\begin{align*} ||\textbf{x}||_2 &=\sqrt{\textbf{x}\cdot \textbf{x}}=3,\\ ||4\textbf{x}-5\textbf{y}||_2 &=\sqrt{(4\textbf{x}-5\textbf{y})\cdot (4\textbf{x}-5\textbf{y})}=14,\\ ||5\textbf{x}+2\textbf{y}||_2 &=\sqrt{(5\textbf{x}+2\textbf{y})\cdot (5\textbf{x}+2\textbf{y})}=8. \end{align*} Squaring both sides to remove the radical gives \begin{align*} \textbf{x}\cdot \textbf{x} &=9,\\ (4\textbf{x}-5\textbf{y})\cdot (4\textbf{x}-5\textbf{y})&=196,\\ (5\textbf{x}+2\textbf{y})\cdot (5\textbf{x}+2\textbf{y})&=64. \end{align*}

Using properties of the dot product, can you solve this system of equations for $\textbf{y}\cdot \textbf{y}$?

Stuck
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  • Combining the last 2 equations while first multiplying the bottom equation by 2 I get that, 66x$^2$+33y$^2$=324 . Substituting x$^2$ i get 66*9+33y$^2$=324 and there is my issue. Still confused where my issue is – Nuemann12 Nov 17 '20 at 04:13
  • I obtain the same thing now that I actually finish the problem (to be honest I didn't really understand your remarks about the steps you took to solve the problem in the OP -- I just looked at the problem). But there are no mistakes here. So either the answer is there is no solution or you miswrote the problem. – Stuck Nov 17 '20 at 04:43
  • Sadly I have checked the question countless times and it is the right question. I may have to sleep on it – Nuemann12 Nov 17 '20 at 05:06
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So, the answer is that there is no solution as seen by the fact that we get a negative solution which is not possible

Nuemann12
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