Question

Help with Isometry

Answer 1

I have this matrix
\[\left[\begin{matrix}0 & 1 &-1 \\ 1 & -1 &0 \\ 0&0&1\end{matrix}\right]\]
and i need to determine if the matrix i isometry.
Can someone help me?

Answer 2

@RFJ-86 back for more? :D

Answer 3

I will add your name on my assignment.. :)

Answer 4

Terence Jason Wong, and no, Terence is spelled with only one r, thank you very much.
Now... to business....
(Note: I have no idea what an isometry is, but I guess we'll find out... unless you tell me now XD )

Answer 5

I the problem it said "Determine if the matrix i isometry (length preserving)"

Answer 6

And something tell me that matrix is length preserving if
\[\left| \left| Ax \right| \right|=\left| \left| x \right| \right|\]
for any x in R^n

Answer 7

do we have a definition for ||x|| ?

Answer 8

Nop...
Problem B i)

Answer 9

I need to look up definitions :/

Answer 10

\[Ax = \left[\begin{matrix}0 & 1&-1 \\ 1 & -1&0 \\ 0&0&1\end{matrix}\right]\left(\begin{matrix}x \\ y\\z\end{matrix}\right)=\left(\begin{matrix}y \\ x-y\\-x+z\end{matrix}\right) => \left| \left| Ax \right| \right|= \sqrt{(1-(-1))^2+(1-(-1))^2+1^2}\]

Answer 11

ahh okay understood :)

Answer 12

<realisation moment>

Answer 13

So if I let x be <1 1 1> I skal show the map not isometry.. Agree?

Answer 14

uhh... I think you just try the elements of the basis first. <1,0,0> , <0,1,0> and <0,0,1>

Answer 15

I think that should suffice.

Answer 16

Because\[\left| \left| Ax \right| \right|= \sqrt{(1-(-1))^2+(1-(-1))^2+1^2}=\sqrt{9}\]
\[\left| \left| x \right| \right|= \sqrt{1^2+1^2+1^2}=\sqrt{3}\]
so
\[\left| \left| Ax \right| \right|\neq \left| \left| x \right| \right|\]

Answer 17

@terenzreignz Do you think this is the answer? if \[x=\left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)\]

Answer 18

Maybe you ought to calculate the product of your matrix and that vector again, something seems off...