Linear transformation from $\mathbf{R}^3$ to $\mathbf{R}^3$ given by the matrix $$L_\beta^\beta=\begin{bmatrix}1&0&5 \\ 0&2&2 \\ 1&-2&0\end{bmatrix}$$ with $\beta=\{(1,0,0),(1,1,0),(1,1,1)\}.$

Give the explicit formula: $L(x,y,z)=(..,..,..)$

I don't know how to handle the basis given.. since it's not the standard basis for $\mathbf{R}^3$. Could someone give a hint?

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    $\begingroup$ if you don't like this base change it by multiplying with the matrix that changes the bases,the linear function L will remain the same $\endgroup$ – giannispapav Oct 7 '17 at 11:48
  • $\begingroup$ We didn't cover change of basis matrices in class yet.. @giannispapav $\endgroup$ – Heinz Doofenschmirtz Oct 7 '17 at 11:52

Hint: Express $(x,y,z)$ as a linear combination of your basis vectors $\beta_1,\beta_2,\beta_3$. The coefficients determine the coordinate vector $b= (b_1,b_2,b_3)$ which you can multiply your matrix with. The result $L_\beta^\beta b$ is the coordinate vector of $L(x,y,z)$, which needs to be transformed back to canonical coordinates.

Of course, you can do this with $(1,0,0),(0,1,0), (0,0,1)$ separately instead of the general $(x,y,z)$.

Alternatively, you can calculuate the matrices $B$ and $B^{-1}$ which transform canonical coordinates to coordinates w.r.t $\beta$ and back. (One of these has the vectors in $\beta$ as column). Then calulate $B^{-1}L_\beta^\beta B$, which describes the transformation $L$ w.r.t. canonical coordinates.

  • $\begingroup$ I transformed $(x,y,z)=(x-y)\beta_1+(y-z)\beta_2+(z)\beta_3$, so $L_\beta^\beta b=(x-y+5z,2y,x-3y+2z)$. When I transform this into canonical coordinates, this gives $(x-y+5z)(1,0,0)+(2y)(1,1,0)+(x-3y+2z)(1,1,1)=$ $(2x-2y+7z,x-y+2z,x-3y+2z)$. Is this approach correct? $\endgroup$ – Heinz Doofenschmirtz Oct 7 '17 at 15:15
  • $\begingroup$ Looks good to me. I did not check thorougly, but the general approach looks good. $\endgroup$ – Roland Oct 7 '17 at 15:28

You want to find out what is the result of applying $L$ to $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Can you express these in terms of given basis? For example $(0,1,0) = (1,1,0)-(1,0,0)$. Now $L(a-b) = L(a)-L(b)$. Is this sufficient for you?


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