Let $Y = \{y_1, y_2, y_3\}$ be a basis for $R^3$ where $y_1 = (1, 1, 1)$, $y_2 = (4, 1, 1)$ and $y_3 = (1, 1, 2)$.

Let $W = \{w_1,w_2\}$, $w_1 = (1, 1)$ and $w_2 = (2, 4)$ be a basis in $R^2$.

Need to find the matrix of the linear transformation $T$ in these bases that maps

$Ty_1 = (4, 5)$, $Ty_2 = (1, 3)$ and $Ty_3 = (7, 1)$.

Note that all coordinates are w.r.t. the standard basis.

Is the answer

$\left( \begin{array}{ccc} -1 & 2 & 3 \\ -\frac23 & \frac{29}{3} & -4 \end{array} \right)$ ?

The Problem

  • 1
    $\begingroup$ It is unclear what you are doing. Notably the values for $w_1$ and $w_2$ do not appear to be used at all. $\endgroup$ – Marc van Leeuwen Aug 25 '15 at 14:00
  • $\begingroup$ @MarcvanLeeuwen, added the image of actual picture, I'm not able to understand it myself. $\endgroup$ – user261816 Aug 25 '15 at 14:25
  • $\begingroup$ The quoted text is fairly clear. The linear transformation is given by its images of the vectors $y_i$ (not of the standard bases), and you need to find its matrix w.r.t. the bases $Y$ and $W$. Since the initial information already give the matrix $\pmatrix{4&1&7\\5&3&1}$ of $T$ w.r.t. the bases $Y$ (not the standard basis!) and the standard basis of $\Bbb R^2$, you only need to perform change of basis at arrival. $\endgroup$ – Marc van Leeuwen Aug 25 '15 at 14:57

We have the linear map $T$ as $u\mapsto v$. If we fix two bases we can define the matrix $A$ that describes the mapping as $Ax=z$ where $x$ and $z$ are the coordinate vectors of $u$ and $v$ respectively in those bases, i.e. $$ u=[y_1\ y_2\ y_3]x\ \ \Bigl(=\sum_{k=1}^3x_ky_k\Bigr),\qquad v=[w_1\ w_2]z\ \ (=z_1w_1+z_2w_2). $$ It gives the following solution for $Tu=v$ $$ T\underbrace{[y_1\ y_2\ y_3]x}_{u}=\underbrace{[w_1\ w_2]z}_{v}=[w_1\ w_2]\underbrace{Ax}_{z} \quad\Leftrightarrow\quad T[y_1\ y_2\ y_3]=[w_1\ w_2]A\quad\Leftrightarrow $$ $$ \Leftrightarrow\quad T[y_1\ y_2\ y_3]=\left[\matrix{4 & 1 & 7\\5 & 3 & 1}\right]=[w_1\ w_2]A= \left[\matrix{1 & 2\\1 & 4}\right]A\quad\Rightarrow\quad $$

$$ \quad\Rightarrow\quad A=\left[\matrix{1 & 2\\1 & 4}\right]^{-1}\left[\matrix{4 & 1 & 7\\5 & 3 & 1}\right]. $$

UPDATE: In general, if we had $Tu_k=v_k$, i.e. $$ TU=T[u_1\ u_2\ u_3]=[v_1\ v_2]=V $$ then we would need to change $U=YS$ and calculate $$ TYS=V\quad\Rightarrow\quad TY=VS^{-1}=WA\quad\Rightarrow\quad A=W^{-1}VS^{-1}=W^{-1}VU^{-1}Y. $$ P.S. Note that the matrix $VU^{-1}$ here is actually the old matrix of the map, i.e. in the original (standard) basis. This is how all matrices for the same linear map look like in different bases, that is if we have changed the old bases to the new ones with transformation matrices $S$ as $$ Y_\text{new}=Y_\text{old}S_y\qquad \text{and}\qquad W_\text{new}=W_\text{old}S_w $$ then $$ A_\text{new}=S_w^{-1}A_\text{old}S_y. $$

  • $\begingroup$ But then, basis of Y doesn't matter. I've uploaded the image of original problem. $\endgroup$ – user261816 Aug 25 '15 at 14:30
  • $\begingroup$ @user261816 Oh, yes, it does. Otherwise we would need to post-multiply the result with one more matrix. $\endgroup$ – A.Γ. Aug 25 '15 at 14:33
  • $\begingroup$ The concrete values of the $Y$ basis vectors indeed do not influence the answer, due to the way the problem is formulated. In fact you don't even need to know the the space at departure is $\Bbb R^3$, just that $Y=[y_1,y_2,y_3]$ is some basis of it. $\endgroup$ – Marc van Leeuwen Aug 25 '15 at 15:01
  • $\begingroup$ @user261816 I don't understand what $T_y$ means. The problem is standard "find the matrix of the linear map in a given basis" where the map is given by a matrix in another basis. Since you already know $Ty_k$ it makes your life a bit easier. $\endgroup$ – A.Γ. Aug 25 '15 at 15:04

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