Why is it that when finding the matrix representation of a linear transformation $T: V \rightarrow W$ where $V$ has basis $\{v_1, v_2, ..., v_n\}$ and $W$ has basis $\{w_1,w_2,...,w_m\}$, the resulting matrix sometimes has columns $T(v_1), T(v_2),...,T(v_n)$ and other times it has columns representing the coefficients of the linear transformation required to map the $w_k$ onto each $T(V_k)$?

  • $\begingroup$ Can you provide an example of the latter? If I’m correctly understanding what you mean, the two are in fact the same. $\endgroup$ – amd Jul 14 at 6:10

Let $\textsf{T}: \textsf{V}\to \textsf{W}$ be a linear transformation and let $\beta =\{v_1,v_2,\dots,v_n\}$ and $\gamma =\{w_1,w_2,\dots,w_m\}$ basis for $\textsf{V}$ and $\textsf{W}$, respectively.

To compute the matrix associated with $\textsf{T}$ respect to $\beta$ and $\gamma$, we always, that is to say always, obtain the vector $\textsf{T}(v_j)$ and put them as a linear combination of the elements that conform $\gamma$, like this : $$\textsf{T}(v_j)=A_{1j}w_1+A_{2j}w_2+\cdots+A_{mj}w_m$$ then, we put the coefficients $$\begin{pmatrix} A_{1j} \\ A_{2j} \\ \vdots \\ A_{mj} \end{pmatrix}$$ in the $j$-th column of the matrix $A:= [\textsf{T}]_{\beta}^{\gamma}$ (for $j=1,2,\dots,n$).

Now, for your specific doubt, consider $\textsf{T}: \mathbb{R}^2 \to \mathbb{R}^2$ given by $\textsf{T}(a,b)=(-b,a)$ (doesn't matter) and the basis $\beta=\{(2,1),(3,5)\}$ and $\gamma=\{e_1,e_2\}$ (the standard basis).

Above I said, we must first to compute $\textsf{T}(2,1)$ and the result put it as a linear combination of the elements in the second basis. In this case $$\textsf{T}(2,1)=(-1,2)=(-1)e_1+2e_2$$ As you can see, the coefficients that we will put in the first column of the matrix will look exactly like the vector $(-1,2)$ (maybe this is what you meant). To finish, note also that $$\textsf{T}(3,5)=(-5,3)=(-5)e_1+3e_2$$ So $$[\textsf{T}]_{\beta}^{\gamma} =\begin{pmatrix} -1 & -5 \\ 2 & 3 \end{pmatrix}$$

(an apology for my english, I hope this has clarified your doubt a bit)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.