## Linear Algebra Question

(1 post)
• Started 2 years ago by lee

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1. lee
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A student asked me the following questions on Linear Algebra:

Question 1. Show that $||x+y||=||x||+||y||$ if and only if $y=ax$ for some $a\in\mathbb R$, i.e. $x$ and $y$ are linearly dependent.

Proof: "If" part is trivial, i.e. if $x$ and $y$ are linearly dependent, the equality clearly holds. To show "only if" part, assume that $||x+y||=||x||+||y||$. Then $(||x+y||)^2=(||x||+||y||)^2$. From this we obtain $\langle x,y\rangle=||x||||y||$. On the other hand, for any vectors $x$ and $y$ in $n$-space, there exists uniquely a real number $\theta\in[0,2\pi)$ such that $\frac{\langle x,y\rangle}{||x||||y||}=\cos\theta$. Since $\cos\theta=1$, $\theta=0$ i.e. $x$ and $y$ are linearly dependent.

Question 2. In an inner product space, the identity $$||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$$ holds. Draw a diagram to explain why it is called parallelogram law and give an example that the identity does not hold in a normed space.

Solution: For the first part see the Wikipedia entry on "Parallelogram Law" here. For the second part, consider a real vector space $V$ with norm $||\cdot||$ defined by $$||x||=\left\{\begin{array}{ccc}1 & \mbox{if} & x\ne 0,\\ 0& \mbox{if} & x=0.\end{array}\right.$$ Let $x=y\ne 0$. Then $||x+y||^2+||x-y||^2=1$ while $2(||x||^2+||y||^2)=4$. Hence Parallelogram Law does not hold in the normed space.

Posted 2 years ago #