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Let $V$ and $W$ be finite-dimensional vector spaces and let $T:V \rightarrow W$ be a linear transformation between them. I have read that

  1. Performing an elementary row operation on the matrix that represents $T$ is equivalent to performing a corresponding change of basis in the range of $T$, and

  2. Performing an elementary column operation on the matrix that represents $T$ is equivalent to performing a corresponding change of basis in the domain of $T$

Admittedly, this is a rather vague formulation but it's all I have. My question is: Can anyone either explain, or provide a reference to, a precise statement of the relationship between change of basis operations and elementary matrices as described above?

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If you understand what the relationship between a matrix and a linear transformation given bases of $V$ and $W$ is, this should be pretty straightforward to work out yourself. Otherwise perhaps you should brush up on that aspect first. –  Qiaochu Yuan Jun 29 '11 at 20:13
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3 Answers 3

HINT: Let be $\{v_{1}, \dots , v_{n} \}$ a basis of $V$ and $\{w_{1}, \dots , w_{m} \}$ a basis of $W$. If a column of the matrix is $^{T}(a_{1i},a_{2i} \dots, a_{mi})$ it means that $T(v_{i}) = a_{1i}w_{1} + a_{2i}w_{2}\dots + a_{mi}w_{m}$, so, for example, exchanging $\{w_{1},w_{2}, \dots , w_{m} \}$ with $\{w_{2},w_{1}, \dots , w_{m} \}$ will transform the column in $^{T}(a_{2i},a_{1i} \dots, a_{mi})$ and so will be $\forall i$. In this way, you have exchanged the first two rows of the matrix.

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Two more examples, with the same notation as Thomas.

If you multiply the $i$-th row by $\lambda \neq 0$, the resulting matrix can be thought as being the matrix associated to $T$ as well, but now the basis on the range of $T$ is $w_1, \dots , \frac{1}{\lambda} w_i , \dots , w_m$.

If you add the $i$-th column to the $j$-th column, now the basis on the domain of $T$ is $v_1, \dots , v_i , \dots , v_i + v_j , \dots , v_n$.

The trickiest case is what happens when you add $i$-th row to the $j$-th row. :-)

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This question is a year old, but in case anyone is still interested, I gave the "recipes" in an answer to Linear Operators, Representative Matrices and Change of Basis .

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