Let $E$, $F$ be vector spaces with basis $\{e_1,\dots,e_m\}$, $\{f_1,\dots,f_n\}$. Let $T:E\to F$ be a linear transformation. We say that the matrix $A\in\mathbb{R}^{m\times n}$ represents $T$ with respect to the bases above if the following holds

  • $ \forall v\in E\,\,\,\,\, T(v)=Av$
  • $\forall j \,\,\,\,\,Ae_j=\sum_{j=1}^n a_{ij}f_i$

Is this definition correct? (Note that I am identifying finite dimensional vector spaces with $\mathbb{R}^n$.)


2 Answers 2


Well, sort of.

Notice first that for $v \in E$, $Av$ may not even be defined since $E$ is not necessarily $\mathbb{R}^n$ and then how do you multiply a matrix with $v$?

I believe that you want to use $\mathbb{R}^n$ and $\mathbb{R}^m$ as a coordinate system for $E$ and $F$ with the bases are $B_1 := \{e_1, ... e_m\}$ and $B_2 := \{f_1, ... , f_n\}$.

Then $A \in \mathbb{R}^{n \times m}$ is said to represent $T$ $\iff$ for every $i, [Te_i]_{B_2} = A[e_i]_{B_1} \iff $ for every $v \in E, [Tv]_{B_2} = A[v]_{B_1}$.

So the problem in your definition is mainly the lack of coordinates.

  • $\begingroup$ How to infer the second property in my question from your definition? $\endgroup$
    – avati91
    Jul 18, 2015 at 16:47
  • $\begingroup$ This is the second property: for every $i, [Te_i]_{B_2} = A[e_i]_{B_1} $ $\endgroup$
    – amirbd89
    Jul 18, 2015 at 17:41

Notice, the dimension of your $A$ is swapped. It should be $n\times m$, if you multiply $A$ from the left and identify $\mathbb R^m$ with columns, like most people do.

If you just want to express that $A$ is the matrix representation of $T$ in a concise way: $$ \forall x_1,\dotsc,x_m\in\mathbb R : T\left( \sum_{i=1}^m x_i e_i \right) = \sum_{j=1}^n \left( \sum_{i=1}^m a_{j,i} x_i \right) f_j. $$ Due to linearity of $T$, it is equivalent to $$ \forall 1\le i \le m: T(e_i) = \sum_{j=1}^n a_{j,i} f_j. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .