In linear algebra, to convert the linear transformation or operator into matrix form, it's not hard, get the standard basis of domain and substitute in the transformation and write the image of the standard basis as a linear combination of standard basis of codomain and finally get the constants (c i's) and write that in column wise in the matrix format, we will get the matrix format of the linear transformation with respect to the standard basis of the domain and codomain. My question here is "what about the reverse part?, can we find the linear transformation if the matrix is given?" I have tried multiplying the matrix with the standard basis of Mn(R), but I am getting the columns as the image.. What should I do? To get the transformation.

  • $\begingroup$ Bear in mind that the matrix represents a linear transformation relative to the choice of (ordered) basis. Similarity of matrices is essentially saying two different matrices can represent the same linear transformation, but with respect to a different choice of ordered basis. $\endgroup$ – hardmath Oct 22 '15 at 12:44
  • $\begingroup$ if you gave a matrix, assign to a basis vector in the domain the linear combination in the codomain's basis using the column's data $\endgroup$ – janmarqz Oct 22 '15 at 12:47
  • $\begingroup$ Sorry.. I can't get you.. Will you please explain $\endgroup$ – Sam Christopher Oct 22 '15 at 12:49

In order to convert between a linear transformation and a matrix, one needs to specify the ordered basis for the domain and also for the codomain.

The Question assumes a "standard basis" for the domain (and also for the codomain), but this is not in general applicable.

That said, if $\mathscr{B} = \{u_1,\ldots,u_n\}$ is chosen as an ordered basis for vector space $U$, and $\mathscr{B}' = \{v_1,\ldots,v_m\}$ is chosen as an ordered basis for vector space $V$, then with respect to that choice a linear transformation $T:U\to V$ is uniquely represented by an $m\times n$ matrix $M$. The same facts which determine $M$ from the actions of $T$ on the bases also allow us to reconstruct the actions of $T$ from the matrix $M$.

Begin by expressing $u$ as a linear combination of the basis elements for $U$:

$$ u = \sum_{i=1}^n c_i u_i $$

Then multiplying column $(c_1,\ldots,c_n)^T$ by the matrix representation gives us the coefficients that represent $T(u)$ in terms of the basis elements for $V$:

$$ T(u) = \sum_{j=1}^m b_j v_j $$


$$ \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} = M \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} $$

In this way the linear tranformation $T$ is completely determined by the choice of bases $\mathscr{B},\mathscr{B}'$ and the matrix representation $M$.


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