Use $\cal L$ to denote a linear transformation on some vector space. We know any matrix $\bf{A}$ can be viewed as a linear transformation by defining $\cal L:= \cal L(\bf{v})= Av$ where $\bf{v}$ is a vector. I am curious is any linear transformation can be represented by a matrix? If so, why? If not, can anyone help provide a counterexample?

In particular, projection of vectors onto a vector $\bf{b}$ is a linear transformation defined by ${\cal L(\bf{v})}=({\mathbf{v}} \cdot \frac{{\mathbf{b}}}{{|{\mathbf{b}}|}})\frac{{\mathbf{b}}}{{|{\mathbf{b}}|}}$. Can this linear transformation be represented by a matrix?

Thank you!

  • 4
    $\begingroup$ Any linear transformation between two finite-dimensional vector spaces can be represented by a matrix. Fixing bases of the vector spaces determined the matrix uniquely. $\endgroup$ May 14, 2016 at 13:51
  • $\begingroup$ For your particular case, the matrix is given by the outer matrix product $A = \frac{b}{|b|} \cdot (\frac{b}{|b|})^T$. Note that $\cal{L}(v) = (v \cdot \frac{b}{|b|})\frac{b}{|b|} = \frac{v \cdot b}{b \cdot b}b$. $\endgroup$ May 29, 2019 at 12:44

1 Answer 1


Here is a detailed version of what Travis said:

Let $T:V\rightarrow W$ be a linear transformation between two finite-dimensional vector spaces. Say $\dim(V)=n$ and $\dim(W)=m$. Let $\left\{v_1,\dots , v_n\right\}$ and $\left\{w_1,\dots ,w_m\right\}$ be bases of $V$ and $W$ respectively. Then $T(v_j)=\sum_{i=1}^m a_{ij}w_i$ for unique coefficients $a_{ij}$. Define the matrix $A=(a_{ij})_{1\leq i\leq m, 1\leq j\leq n}$.

We know that $V\cong \mathbb{R}^n$ via the coordinate map $\phi:V\rightarrow \mathbb{R}^n:\sum_{i=1}^n\lambda_i v_i\mapsto (\lambda_1,\dots , \lambda_n)$ and similarly $W\cong \mathbb{R}^m$. We now define a linear map $S:\mathbb{R}^n\rightarrow \mathbb{R}^m:v\rightarrow Av$.

Under the coordinate map, we have that $v_i=e_i$ where $e_i$ is the column with zeroes everywhere except that it is one at the $i$-th component. We then find that $$S(v_j)=Av_j=Ae_j= \begin{pmatrix} a_{1j}\\a_{2j}\\\vdots\\ a_{mj} \end{pmatrix}=\sum_{i=1}^ma_{ij}e_i=\sum_{i=1}^ma_{ij}w_i=T(v_j).$$ In this calculation the proper identificiations are understood. It follows that $S=T$ as desired.

This is perhaps one of the most important ideas in linear algebra. Notice that the matrix $A$ is unique after fixing bases in $V$ and $W$. Given other bases, the linear map $T$ can be represented by some other matrix that is similar to $A$. Using this one can define various concepts for linear maps that are preserved under similarity, for example, one can define the trace and determinant of a linear map using its matrix representation.


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