There are only three row operations and so that's what we'll consider.

  1. Suppose we get $E_1$ by multiplying the $i$th row of identity matrix $I$ by some nonzero $k.$ Then $E_1$ will have a $k$ on its leading diagonal. So, $[E_1^T]_{ii} = [E_1]_{ii} = k$ if $[E_1]_{ii} = k$ for some $i$. Otherwise, $[E_1^T]_{ii} = [E_1]_{ii} = 1.$ Also, $[E_1^T]_{ij} = [E_1]_{ji} = 0$ for $i \neq j.$ Thus $E_1^T = E_1.$

  2. Suppose we get $E_2$ by swapping $i$th, $j$th rows of $I.$ Then $[I]_{ii} = 1$ moves to $[E_2]_{ji}$ so that $[E_2]_{ji} = 1$ and $[I]_{jj} = 1$ moves to $[E_2]_{ij}$ so that $[E_2]_{ij} = 1.$ Thus $[E_2^T]_{ij} =[E_2]_{ji} = 1$ if $= [E_2]_{ji} = 1$ and $[E_2^T]_{ij} =[E_2]_{ji} = 0$ if $= [E_2]_{ji} = 0.$ If $i = j,$ then $[E_2^T]_{ij} =[E_2]_{ji} = 1.$ Thus $E_2^T = E_2.$

  3. Suppose we get $E_3$ by adding $k$ times $i$th row of $I$ to $j$th row. If we look at the picture of this transformation, $E_3^T \neq E_3.$ But $E_3^T$ has a single $k$ and so differs from $I$ by a single element which implies $E_3^T$ is elementary.

Does that make sense?


1 Answer 1


Yes, that definitely makes sense, and is all correct. However, you could perhaps word it a little more concisely, personally I like to avoid elementwise matrix notation unless it's really necessary to convey something. Drawing pictures is nice too.

I like your reasoning that any $E_3$ is the identity with a $k$ off the main diagonal, and transposing that we get the identity with a $k$ off the main diagonal, still an $E_3$.

For the other two, there's nothing wrong with what you said, but sometimes short is sweet.

For $E_1$ I'd just say that transposition clearly fixes the diagonal. $E_1$ is a diagonal matrix, therefore it's symmetric.

For $E_2$, swapping the $i$th and $j$th rows of the identity is equivalent to swapping the $i$th and $j$th columns.


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