Great questions—these get to the heart of why elementary matrices are such a powerful tool in linear algebra. Let’s break each one down clearly, no jargon overload.

1. Why does applying an elementary row/column transformation to the identity matrix $I$ give the corresponding elementary matrix? #

First off, this is by definition! We literally define an elementary matrix as the matrix you get when you perform exactly one elementary row (or column) transformation on the identity matrix $I$. It’s not some mysterious coincidence—we made up this term to label the matrices that bridge "manual row/column operations" and "matrix multiplication." The real magic isn’t the definition itself, but what these matrices do when you multiply them with other matrices (which we’ll get to next).

2. Why is this conclusion valid theoretically, and how do we know the elementary matrix works for all other matrices? #

To get here, let’s recall what matrix multiplication actually does. When you left-multiply a matrix $A$ by another matrix $E$, each row of the resulting matrix $EA$ is a linear combination of the rows of $A$, using the entries of the corresponding row in $E$ as coefficients.

Now, since $E$ comes from the identity matrix $I$, its rows are just the standard basis vectors (the rows of $I$) modified by one elementary transformation. Let’s use examples to make this concrete:

• Suppose $E$ is the matrix you get by swapping rows $i$ and $j$ of $I$. Then $E$’s $i$-th row is the $j$-th standard basis vector (all zeros except a 1 in the $j$-th position), and its $j$-th row is the $i$-th standard basis vector. Left-multiplying $A$ by $E$ will swap the $i$-th and $j$-th rows of $A$—exactly the same elementary transformation we applied to $I$.
• If $E$ is $I$ with the $i$-th row multiplied by a non-zero constant $k$, then left-multiplying $A$ by $E$ will multiply the $i$-th row of $A$ by $k$.
• If $E$ is $I$ with $k$ times the $j$-th row added to the $i$-th row, left-multiplying $A$ by $E$ will add $k$ times $A$’s $j$-th row to its $i$-th row.

This works for any matrix $A$ because matrix multiplication follows the same linear combination rule no matter what $A$ is. The structure of $E$ encodes exactly the elementary transformation, and the multiplication applies that transformation universally—no exceptions.

For column transformations, it’s the same idea but with right-multiplication: $AE$ applies the elementary column transformation encoded in $E$ to $A$, since matrix multiplication on the right uses columns of $E$ as coefficients for linear combinations of $A$’s columns.

3. Can we prove the elementary matrix works for all matrices without verifying each case? #

Absolutely! We can use the properties of the identity matrix and the linearity of matrix multiplication to prove this generally, no need to test every possible matrix.

First, remember the identity matrix’s defining property: for any matrix $A$, $IA = AI = A$. It’s the multiplicative identity, just like the number 1 is for regular multiplication.

Let’s formalize this for row transformations (column transformations follow the same logic with right-multiplication):
Let $E_o$ be an elementary row transformation, and let $E_m$ be the matrix obtained by applying $E_o$ to $I$. We need to show that for any matrix $A$, $E_m A$ is equal to the matrix you get by applying $E_o$ directly to $A$.

We can break this down using the row perspective of matrix multiplication:

• For any matrix $M$, let $row_i(M)$ denote the $i$-th row of $M$.
• By definition of matrix multiplication, $row_i(E_m A) = row_i(E_m) \cdot A$ (this is the dot product of $E_m$’s $i$-th row with $A$’s columns, which is a linear combination of $A$’s rows).
• Since $E_m$ is $I$ with $E_o$ applied, $row_i(E_m) = E_o(row_i(I))$. And $row_i(I)$ is the standard basis vector $e_i$ (all zeros except a 1 in the $i$-th position), so $row_i(I) \cdot A = row_i(A)$.
• Now, consider what $E_o$ does to $A$: applying $E_o$ to $A$ gives a matrix $A'$ where $row_i(A') = E_o(row_i(A))$.

Since every elementary row transformation is a linear operation (swapping rows, scaling a row, adding a multiple of one row to another all preserve linearity), we can connect these dots:
$$row_i(E_m A) = E_o(row_i(I)) \cdot A = E_o(row_i(I) \cdot A) = E_o(row_i(A)) = row_i(A')$$

This holds for every row $i$, so $E_m A = A'$—exactly what we wanted to prove. Since this logic applies to all three types of elementary row transformations, and matrix multiplication works the same way for any $A$, we don’t need to verify individual cases.

The key takeaway here is that the identity matrix’s rows are the building blocks of linear combinations, so modifying those blocks (via elementary transformations) and then multiplying encodes exactly that transformation for any matrix.

内容的提问来源于stack exchange，提问作者John