Question

Help.. :)

Answer 1

I find the matrix of the map to:

Answer 2

And i found the map to be invertible because rank=3=n

Answer 3

But how do I determine if the map are isometry?

Answer 4

A linear map with matrix A is an Isometry if :
\[\Large\|Ax\|=\|x\|,\quad \forall x\in \mathbb R^n\]

Answer 5

Will you take me step-by-step? Please :)

Answer 6

OK !
First, we have :
\[Ax=\begin{pmatrix}1&1&1\\1&0&1\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}\Rightarrow \|Ax\|=\sqrt{(x+y+z)^2+(y+z)^2+z^2}\]
When we take :
\[x=\begin{pmatrix}1\\1\\1\end{pmatrix}\] then :
\[\|Ax\|=\sqrt{3^2+2^2+1^2}=\sqrt{14}\\
\|x\|=\sqrt{1^2+1^2+1^2}=\sqrt3\ne \|Ax\|\]
So, this map is not an isometry

Answer 7

Great.. Thank you..

Answer 8

Are my first two results correct? Can you see that?

Answer 9

To find the matrix or the map in the standard bases we have to find :
\[\Large Ae_1,~Ae_2~~\text{ and }~~Ae_3 \]
Where :
\[\Large e_1=\begin{pmatrix}1\\0\\0\end{pmatrix}~~~,~~e_2=\begin{pmatrix}0\\1\\0\end{pmatrix}~~~\text{and}~~~e_3=\begin{pmatrix}0\\0\\1\end{pmatrix}.\]