What is µ?
I'm having trouble with this proof we need to learn -
Let E = µ(I). Prove that µ(A) = EA for every A ∈ M_{nn}
A hint some guy gave is to begin with the ij entry of EA and prove that it equals the ij entry of µ(A), by using the
deﬁnition of matrix multiplication and the deﬁnitions of E and µ(A).
Any help? Thanks a bunch
Let M_{nn} be the set of all real n × n matrices, let 1 ≤ k, l ≤ n be ﬁxed integers with k != l, and let c (!= 0) be
a real number. Consider the function µ : M_{nn} → M_{nn}, whose image µ(A) for each A ∈ M_{nn} is deﬁned
by
for 1 ≤ i, j ≤ n
Oops sorry, this was some information supplied with the theories.
This is just showing that elementary row operations can be cast as matrix multiplication. A pretty standard part of any linear algebra course.
http://en.wikipedia.org/wiki/Elementary_matrix
To be more precise, $\mu$ should be "tagged" with the subscripts k,l because you get a DIFFERENT function for each pair.
I will suppose that k and l are fixed, and unequal. So, $\mu$ is the function that adds c times the k-th row of a matrix A, to the l-th row, and leaves all other rows unchanged.
We wish to show that $\mu(A) = \mu(I)A$. For all rows but the l-th row, A is left unchanged by $\mu$, so we need to show that likewise for any row but the l-th row, the i-th row of $\mu(I)A$ is the same as the i-th row of A.
By the definition of matrix multiplication, the i-th row of $\mu(I)A$ is the dot product of the i-th row of $\mu(I)$ with each column of A in turn, that is:
$$(\mu(I)A)_{ij} = \sum_{r = 1}^n (\mu(I))_{ir}(A)_{rj}$$
If i is not equal to l, then the i-th row of $\mu(I)$ is just the i-th standard (row) vector: (0,...,0,1,0,...,0) with 1 in the i-th place. That is:
$(\mu(I))_{ir} = 0, i \neq r$, and $(\mu(I))_{ii} = 1$. In this case, the only term that survives in the sum is $(\mu(I))_{ii}(A)_{ij} = (A)_{ij}$, so we see that for any row but the l-th, A is indeed left unchanged.
Now we consider what happens for our "exceptional" l-th row. In this case, the l-th row of $\mu(I)$ is: $e_l + ce_k$. For example, if l < k, this looks like:
(0,...,0,1,0,...,0,c,0,...,0).
So now we get TWO terms that survive in the dot product, one is $(A)_{lj}$, and the other is $c(A)_{kj}$, the sum of which is precisely the l,j-th entry of $\mu(A)$.