I am given a linear transformation $p:X_1\oplus X_2\to X_1$, which means that $p(x_1+x_2)=x_1$ where $x_1 \in X_1$ and $x_2 \in X_2$. $X_1$ and $X_2$ are subspaces of a real space $X$. I am also given the bases $(e_1, e_2,..., e_m)$ which is the base of $X_1$ and $(e_{m+1}, e_{m+2},..., e_n)$ which is the base of $X_2$. Here vectors $e_1$, $e_2$, ... $e_n$ are vectors from the standard base of $\mathbb R^n$.

I understand that the vectors from $X_1 \oplus X_2$ can be represented as linear combinations of vectors $e_1,...,e_n$ and when I transform that linear combination I should get a vector that is represented as a linear transformation of vectors $e_1, ..., e_m$.

After I do this, I get stuck and don't know how to continue to find the needed matrix. Any ideas would be helpful, since I am kind of clueless on what the next step is.

  • $\begingroup$ Note that typing $X_1$ produces $X_1$, as opposed to $X1$. $\endgroup$ – Omnomnomnom Apr 24 '17 at 21:17
  • $\begingroup$ @Omnomnomnom thanks, I was trying to figure that out, but couldn't find it $\endgroup$ – ivana14 Apr 24 '17 at 21:19

Hint: Note that $$ p(a_1 e_1 + \cdots + a_m e_m + a_{m+1}e_{m+1} + \cdots + a_n e_n) = a_1 e_1 + \cdots + a_ne_n $$ As such, the matrix of the transformation is the matrix $M$ satisfying $$ M \pmatrix{a_1\\ \vdots \\ a_{m} \\ a_{m+1} \\ \vdots \\ a_n} = \pmatrix{a_1\\ \vdots \\ a_{m}} $$

  • $\begingroup$ Can we solve this like this: Let $\ T_{mxn}$ be the matrix of this operator and $\ I_{nxn} $ be the identity matrix formed from vectors $\ \{e_1,e_2,...,e_n\}$ and $\ Y_{mxn}$ be the matrix of the results and we conclude from $\ T*I=Y $ that $\ T=Y$ $\endgroup$ – CTSnake Apr 24 '17 at 21:48
  • $\begingroup$ Excuse my typo; see my latest edit. To your comment: note that the actual vectors $e_i$ from our basis will have nothing to do with our matrix. $\endgroup$ – Omnomnomnom Apr 24 '17 at 21:50
  • $\begingroup$ Oh! I missed the end of your question where you elaborate on the $e_i$. Yes, you're right. The matrix is $$ T = \pmatrix{I & 0} $$ $\endgroup$ – Omnomnomnom Apr 24 '17 at 21:51
  • $\begingroup$ @CTSnake could you elaborate a bit more? I understand the idea but I am not sure why the dimensions of $T$ are $mxn$ $\endgroup$ – ivana14 Apr 25 '17 at 9:01
  • $\begingroup$ @ivana14 when you form the LT matrix the number of rows "m" is equal to the dimension of codomain and the number of columns "n" is the dimension of domain. $\endgroup$ – CTSnake Apr 25 '17 at 11:29

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