Answer to Question #162389 in Linear Algebra for Nikhil

Question

Define: R^3→R^3 by

T(x,y,z)=(x-y+z,x+y,y+z)

Let v1= (1,1,1), v2= (0,1,1), v3= (0,0,1). Find a matrix of T with respect to the basis {v1,v2,v3}. Futher check T is invertible or not.

Answer 1

The map $T$ with respect to the standard basis $e_1=(1,0,0)$, $e_2=(0,1,0)$, $e_3=(0,0,1)$ can be presented in the form: $\left(\begin{matrix}x-y+z\\x+y\\y+z\end{matrix}\right)=\left(\begin{matrix}1&-1&1\\1 &1&0\\0&1&1\end{matrix}\right)\left(\begin{matrix}x\\y\\z\end{matrix}\right)$. In order to find a matrix with respect to the basis $v_1$,$v_2$ and $v_3$ , we need to find such matrix $T'$ that $x'v_1+y'v_2+z'v_3=T'(xv_1+yv_2+zv_3)$. We have: $xv_1+yv_2+zv_3=(x,x+y,x+y+z)^{\top}$ In case we act on the vector with the map $T$ we will get: $(x-(x+y)+x+y+z,2x+y,2x+2y+z)=(x+z,2x+y,2x+2y+z)$. It has the following representation in basis $v_1,v_2,v_3:$ $(x+z,2x+y,2x+2y+z)=(x+z)v_1+(2x+y-(x+z))v_2+(2x+2y+z-(x+z)-(x+y-z))v_3=(x+z)v_1+(x+y-z)v_2+(y+z)v_3$

The respective matrix has the form: $T'=\left(\begin{matrix}1&0&1\\1&1&-1\\0&1&1\end{matrix}\right)$