Monthly Archives: August 2010

ZZ[x] and QQ[x] are not isomorphic

Prove that the rings \mathbb{Z}[x] and \mathbb{Q}[x] are not isomorphic.


We begin with some lemmas.

Lemma 1: Let R and S be rings with 1. If \varphi : R \rightarrow S is a surjective ring homomorphism, then \varphi(1_R) = 1_S. Proof: Let s \in S. Now s = \varphi(r) for some r \in R. Note that \varphi(1_R) \cdot s = \varphi(1_R \cdot r) = \varphi(r) = s, and likewise s \varphi(1_R) = s. Since the 1 in S is unique, we have \varphi(1_R) = 1_S. \square

Lemma 2: Let R and S be rings with 1. If \varphi : R \rightarrow S is a ring homomorphism such that \varphi(1) = 1, then \varphi[R^\times] \subseteq S^\times. Proof: If u \in R^\times is a unit, we have uv = 1 for some v \in R. Then \varphi(u) \varphi(v) = 1, so that \varphi(u) is a unit. \square

Lemma 3: Let R and S be rings with 1. If \varphi : R \rightarrow S is a surjective ring homomorphism, then \varphi[R^\times] = S^\times. Proof: The (\subseteq) direction follows from the previous lemma. Now let w \in S^\times. Since \varphi is surjective, w = \varphi(u) for some u \in R. Note that 1_S = wz for some z \in S^\times. Then 1_R = u \varphi^{-1}(z), and thus u \in R^\times. \square

Lemma 4: Let R and S be rings with 1. If \varphi : R \rightarrow S is a ring isomorphism, then \varphi|_{R^\times} : R^\times \rightarrow S^\times is a multiplicative group isomorphism. Proof: By the previous lemma, \varphi|_{R^\times} indeed maps R^\times surjectively to S^\times. Moreover, as the restriction of an injective function, \varphi|_{R^\times} is injective. Finally, \varphi|_{R^\times} is a multiplicative homomorphism since \varphi is. \square

Note that \mathbb{Z}[x] and \mathbb{Q}[x] are rings with 1. If \varphi : \mathbb{Z}[x] \rightarrow \mathbb{Q}[x] is a ring isomorphism, then the restriction \psi of \varphi to the units is a group isomorphism \psi : \mathbb{Z}[x]^\times \rightarrow \mathbb{Q}[x]^\times. However, note that by Proposition 4, \mathbb{Z}[x] has 2 units while \mathbb{Q}[x] has infinitely many units; thus we have a contradiction.

2ZZ and 3ZZ are not isomorphic as rings

Prove that the rings 2\mathbb{Z} and 3\mathbb{Z} are not isomorphic.


Suppose \varphi : 2\mathbb{Z} \rightarrow 3\mathbb{Z} is an isomorphism. Then \varphi(2) = 3a for some integer a. Now \varphi(4) = \varphi(2 \cdot 2) = \varphi(2) \cdot \varphi(2) = 9a^2, and also \varphi(4) = \varphi(2 + 2) = \varphi(2) + \varphi(2) = 6a, so that 6a = 9a^2. Since \mathbb{Z} is an integral domain, we have 2 = 3a. However, this equation has no solutions in the integers, and we have a contradiction. Thus no such isomorphism exists.

Characterize the center of a group ring

Let \mathcal{K} = \{k_1, \ldots, k_m \} be a conjugacy class in the finite group G. Let R be a ring with 1.

  1. Prove that the element K = \sum_{i=1}^m k_i is in the center of the group ring R[G]. [Hint: Check that g^{-1}Kg = K for all g \in G.]
  2. Let \mathcal{K}_1, \ldots, \mathcal{K}_n be the conjugacy classes of G, and for each i, let K_i be the sum of the elements in \mathcal{K}_i (as described in part 1). Prove that an element \alpha \in R[G] is in the center if and only if \alpha = \sum_{i=1}^n a_iK_i for some elements a_i \in Z(R).

  1. Let g \in G. Note that conjugation by g permutes the elements of \mathcal{K}, so that (as an element of R[G]) we have g^{-1}Kg = g^{-1} \left( \sum_{i=1}^m k_i \right) g = \sum_{i=1}^m g^{-1}k_ig = \sum_{i=1}^m k_i = K. Thus gK = Kg for all g \in G. Then for all M = \sum_{i=1}^t r_i g_i \in R[G], we see the following.
    KM  =  \left( \displaystyle\sum_{i=1}^m k_i \right) \left( \displaystyle\sum_{j=1}^t r_j g_j \right)
     =  \displaystyle\sum_{i=1}^m \displaystyle\sum_{j=1}^t r_j k_i g_j
     =  \displaystyle\sum_{j=1}^t \displaystyle\sum_{i=1}^m r_j k_i g_j
     =  \displaystyle\sum_{j=1}^t r_j \left( \displaystyle\sum_{i=1}^m k_i \right) g_j
     =  \displaystyle\sum_{j=1}^t r_j K g_j
     =  \displaystyle\sum_{j=1}^t r_j g_j K
     =  \left( \displaystyle\sum_{j=1}^t r_j g_j \right) K
     =  MK

    Thus K \in Z(R[G]).

  2. First we show that N = \sum_{i=1}^n a_i K_i \in Z(R[G]). Let M = \sum_{j=1}^t r_j g_j \in R[G]. Then we have the following.
    NM  =  \left( \displaystyle\sum_{i=1}^n a_i K_i \right) M
     =  \displaystyle\sum_{i=1}^n a_i K_i M
     =  \displaystyle\sum_{i=1}^n a_i M K_i
     =  \displaystyle\sum_{i=1}^n a_i \left( \displaystyle\sum_{j=1}^t r_j g_j \right) K_i
     =  \displaystyle\sum_{i=1}^n \left( \displaystyle\sum_{j=1}^t a_i r_j g_j \right) K_i
     =  \displaystyle\sum_{i=1}^n \left( \displaystyle\sum_{j=1}^t r_j a_i g_j \right) K_i
     =  \displaystyle\sum_{i=1}^n \left( \displaystyle\sum_{j=1}^t r_j g_j \right) a_i K_i
     =  \left(\displaystyle\sum_{j=1}^t r_j g_j \right) \left( \displaystyle\sum_{i=1}^n a_i K_i \right)
     =  MN

    Thus N \in Z(R[G]).

    Now let M = \sum_{i=1}^t r_i g_i \in Z(R[G]). First, let s \in R be arbitrary. By examining each coefficient of Ms = sM, we see that r_i \in Z(R) for all i. Now recall that G acts transitively (by conjugation) on each of its conjugacy classes. If K is a conjugacy class of G and g_1,g_2 \in K, then we have h^{-1}g_1h = g_2 for some h \in G. The coefficient of g_2 in M = g^{-1}Mg is r_2 on one hand and r_1 on the other, so that in fact r_1 = r_2. In fact the coefficient of each g_i \in K is r_1, and we have M = \sum_{i=1}^m r_i K_i.

Exhibit an element in the center of a group ring

Let R be a ring with 1 \neq 0, and let G = \{g_1, \ldots, g_n \} be a finite group. Prove that the element N = \sum_{i=1}^n g_i is in the center of the group ring R[G].


Let M = \sum_{i=1}^n r_i g_i be an element of R[G]. Note that for each g_i \in G, the action of g_i on G by conjugation permutes the subscripts. Then we have the following.

NM  =  \left( \displaystyle\sum_{i=1}^n g_i \right) \left( \displaystyle\sum_{j=1}^n r_jg_j \right)
 =  \displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_jg_ig_j
 =  \displaystyle\sum_{j=1}^n \displaystyle\sum_{i=1}^n r_j g_j g_j^{-1} g_ig_j
 =  \displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_j^{-1}g_ig_j \right)
 =  \displaystyle\sum_{j=1}^n r_j g_j \left( \displaystyle\sum_{i=1}^n g_i \right)
 =  \left( \displaystyle\sum_{j=1}^n r_jg_j \right) \left( \displaystyle\sum_{i=1}^n g_i \right)
 =  MN.

Thus N \in Z(R[G]).

Compute in a group ring

Consider the following elements of the group ring \mathbb{Z}/(3)[S_3]: \alpha = 1(2\ 3) + 2(1\ 2\ 3) and \beta = 2(2\ 3) + 2(1\ 3\ 2). Compute \alpha + \beta, 2\alpha - 3\beta, \alpha\beta, \beta\alpha, and \alpha^2.


Evidently,

  1. \alpha + \beta = 2(1\ 2\ 3) + 2(1\ 3\ 2)
  2. 2\alpha - 3\beta = 2\alpha = 2(2\ 3) + (1\ 2\ 3)
  3. \alpha\beta = 2(1\ 2) + 1(1\ 2\ 3)
  4. \beta\alpha = 1(1\ 3) + 2(1\ 3\ 2)
  5. \alpha^2 = 1(1) + 2(1\ 2) + 2(1\ 3) + 1(1\ 3\ 2)

Compute in a group ring

Consider the following elements of the integral group ring \mathbb{Z}[S_3]: \alpha = 3(1\ 2) - 5(2\ 3) + 14(1\ 2\ 3) and \beta = 6(1) + 2(2\ 3) - 7(1\ 3\ 2). Compute the following elements: \alpha + \beta, 2\alpha - 3\beta, \alpha\beta, \beta\alpha, and \alpha^2.


Evidently,

  1. \alpha + \beta = 6(1) + 3(1\ 2) - 3(2\ 3) + 14(1\ 2\ 3) - 7(1\ 3\ 2)
  2. 2\alpha - 3\beta = -18(1) + 6(1\ 2) - 16(2\ 3) + 28(1\ 2\ 3) + 21(1\ 3\ 2)
  3. \alpha\beta = -108(1) + 81(1\ 2) - 21(1\ 3) - 30(2\ 3) + 90(1\ 2\ 3)
  4. \beta\alpha = -108(1) + 18(1\ 2) + 63(1\ 3) - 51(2\ 3) + 84(1\ 2\ 3) + 6(1\ 3\ 2)
  5. \alpha^2 = 34(1) - 70(1\ 2) - 28(1\ 3) + 42(2\ 3) - 15(1\ 2\ 3) + 181(1\ 3\ 2)

Compute in a group ring over Dih(8)

Let \alpha = r + r^2 - 2s and \beta = -3r^2 + rs be elements of the group ring \mathbb{Z}[D_8]. Compute the following: \beta\alpha, \alpha^2, \alpha\beta - \beta\alpha, and \beta\alpha\beta.


Evidently,

  1. \beta\alpha = -3 - 2r - 3r^3 + s + 6r^2s + r^3s
  2. \alpha^2 = 5 + r^2 + 2r^3 - 4r^2s - 4r^3s
  3. \alpha\beta - \beta\alpha = 2r - 2r^3 - s + r^2s
  4. \beta\alpha\beta = 15r + 10r^2 + 7r^3 - 21s - 6rs - 5r^2s

Strictly upper (lower) triangular matrices are zero divisors

Let S be any ring and let n \geq 2 be an integer. Prove that if A is any strictly upper triangular matrix in M_n(S) then A^n = 0. (A strictly upper triangular matrix is a matrix whose entries on and below the main diagonal are zero.)


Before approaching this problem, we will introduce some “structural” operations on matrices and prove some basic properties.

Definition: Let X be a set.

  1. Suppose A = [a_{i,j}] \in \mathsf{Mat}_{k,n}(X) and B = [b_{i,j}] \in \mathsf{Mat}_{k,m}(X). We define a matrix [A | B] \in \mathsf{Mat}_{k,n+m}(X) as follows: ([A|B])_{i,j} = a_{i,j} if 1 \leq j \leq n and b_{i,j-n} otherwise.
  2. Suppose A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(X) and B = [b_{i,j}] \in \mathsf{Mat}_{m,k}(X). We define a matrix \left[ \frac{A}{B} \right] \in \mathsf{Mat}_{n+m,k}(X) as follows: \left(\left[ \frac{A}{B} \right]\right)_{i,j} = a_{i,j} if 1 \leq i \leq n and b_{i-n,j} otherwise.

Lemma: Let X be a set, A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(X), B = [b_{i,j}] \in \mathsf{Mat}_{n,\ell}(X), C = [c_{i,j}] \in \mathsf{Mat}_{m,k}(X), and D = [d_{i,j}] \in \mathsf{Mat}_{m,\ell}(X). Then

\left[ \dfrac{[A | B]}{[C | D]} \right] = \left[ \left[ \dfrac{A}{C} \right] \bigg| \left[ \dfrac{B}{D} \right] \right].

Proof: Let \mathcal{S} denote the matrix on the left hand side of the equals sign and \mathcal{T} the matrix on the right. We consider four possibilities for (i,j).

  1. Suppose 0 < i \leq n and 0 < j \leq k. Then (\mathcal{S})_{i,j} = a_{i,j} = (\mathcal{T})_{i,j}.
  2. Suppose 0 < i \leq n and k < j \leq j+\ell. Then (\mathcal{S})_{i,j} = b_{i,j} = (\mathcal{T})_{i,j}.
  3. Suppose n < i \leq n+m and 0 < j \leq k. Then (\mathcal{S})_{i,j} = c_{i,j} = (\mathcal{T})_{i,j}.
  4. Suppose n < i \leq n+m and k < j \leq k+\ell. Then (\mathcal{S})_{i,j} = d_{i,j} = (\mathcal{T})_{i,j}.

Thus \mathcal{S} = \mathcal{T}. \square

Since these two operators “abide”, we will drop the inner brackets and write (for example) \left[ \dfrac{A}{C} \bigg| \dfrac{B}{D} \right] for brevity.

Lemma: Let R be a ring.

  1. If A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(R), B = [b_{i,j}] \in \mathsf{Mat}_{k,m}(R), and C = [c_{i,j}] \in \mathsf{Mat}_{k,\ell}(R), then A \cdot [B|C] = [A \cdot B | A \cdot C].
  2. If A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(R), B = [b_{i,j}] \in \mathsf{Mat}_{m,k}(R), and C = [c_{i,j}] \in \mathsf{Mat}_{k,\ell}(R), then \left[ \frac{A}{B} \right] \cdot C = \left[ \frac{A \cdot C}{B \cdot C} \right].

Proof: The (i,j) entry of A \cdot [B|C] is \sum_{p=1}^k a_{i,p}([B|C])_{p,j}. If 0 < j \leq m, then this sum is \sum_{p=1}^k a_{i,p}b_{p,j} = (A \cdot B)_{i,j}. If m < j \leq m+\ell, then this sum is \sum_{p=1}^k a_{i,p}c_{p,j-m} = (A \cdot C)_{i,j-m}. Thus (A \cdot [B|C])_{i,j} = ([A \cdot B|A \cdot C])_{i,j} for all (i,j). The proof of the second statement is analogous. \square

Lemma: Let R be a ring. If A = [a_{i,j}] \in \mathsf{Mat}_{n,k}(R), B = [b_{i,j}] \in \mathsf{Mat}_{n,\ell}(R), C = [c_{i,j}] \in \mathsf{Mat}_{k,m}(R), and D = [d_{i,j}] \in \mathsf{Mat}_{\ell,m}(R), then [A|B] \cdot \left[ \dfrac{C}{D} \right] = AC + BD. Proof: For each (i,j), note the following.

\left( [A|B] \cdot \left[ \dfrac{C}{D} \right] \right)_{i,j}  =  \displaystyle\sum_{p=1}^{k+\ell} \left( [A|B] \right)_{i,p} \left( \left[ \dfrac{C}{D} \right] \right)_{p,j}
 =  \left( \displaystyle\sum_{p=1}^{k} \left( [A|B] \right)_{i,p} \left( \left[ \dfrac{C}{D} \right] \right)_{p,j} \right) + \left( \displaystyle\sum_{p=k+1}^{k+\ell} \left( [A|B] \right)_{i,p} \left( \left[ \dfrac{C}{D} \right] \right)_{p,j} \right)
 =  \left( \displaystyle\sum_{p=1}^k a_{i,p}c_{p,j} \right) + \left( \displaystyle\sum_{p=k+1}^{k+\ell} b_{i,p-k}d_{p-k,j} \right)
 =  (A \cdot C)_{i,j} + (B \cdot D)_{i,j}
 =  (A \cdot C + B \cdot D)_{i,j}

Thus the two matrices are equal. \square

Lemma: Let R be a ring. Let A_1 \in \mathsf{Mat}_{n \times k}(R), B_1 \in \mathsf{Mat}_{n \times \ell}(R), C_1 \in \mathsf{Mat}_{t \times k}(R), D_1 \in \mathsf{Mat}_{t \times \ell}(R), A_2 \in \mathsf{Mat}_{k \times m}(R), B_2 \in \mathsf{Mat}_{k \times p}(R), C_2 \in \mathsf{Mat}_{\ell \times m}(R), and D_2 \in \mathsf{Mat}_{\ell \times p}(R). Then

\left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{A_2 | B_2}{C_2 | D_2} \right] = \left[ \dfrac{A_1A_2 + B_1C_2 | A_1B_2 + B_1D_2}{C_1A_2 + D_1C_2 | C_1B_2 + D_1D_2} \right].

Proof: Using the previous lemmas, we have the following.

\left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{A_2 | B_2}{C_2 | D_2} \right]  =  \left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \left[ \dfrac{A_2}{C_2} \right] \bigg| \left[ \dfrac{B_2}{D_2} \right] \right]
 =  \left[ \left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{A_2}{C_2} \right] \bigg| \left[ \dfrac{A_1 | B_1}{C_1 | D_1} \right] \cdot \left[ \dfrac{B_2}{D_2} \right] \right]
 =  \left[ \left[ \dfrac{[A_1 | B_1]}{[C_1 | D_1]} \right] \cdot \left[ \dfrac{A_2}{C_2} \right] \bigg| \left[ \dfrac{[A_1 | B_1]}{[C_1 | D_1]} \right] \cdot \left[ \dfrac{B_2}{D_2} \right] \right]
 =  \left[ \left[ \dfrac{[A_1|B_1] \cdot \left[ \dfrac{A_2}{C_2} \right]}{[C_1|D_1] \cdot \left[ \dfrac{A_2}{C_2} \right]} \right] \Bigg| \left[ \dfrac{[A_1|B_1] \cdot \left[ \dfrac{B_2}{D_2} \right]}{[C_1|D_1] \cdot \left[  \dfrac{B_2}{D_2} \right]} \right] \right]
 =  \left[ \dfrac{A_1A_2 + B_1C_2 | A_1B_2 + B_1D_2}{C_1A_2 + D_1C_2 | C_1B_2 + D_1D_2} \right]. \square

We now introduce another definition.

Definition: Let R be a ring, n \geq 2, and 1 \leq k \leq n. A matrix M \in \mathsf{Mat}_n(R) is called k-strictly upper triangular if M = \left[ \dfrac{0 | M^\prime}{0_k | 0} \right] where 0_k is the k \times k zero matrix, M^\prime has dimensions (n-k) \times (n-k), and M^\prime is upper triangular.

For example, 1-strictly upper triangular matrices and strictly upper triangular matrices are the same, and an n \times n matrix is zero if and only if it is n-strictly upper triangular.

Lemma: Let R be a ring, n \geq 2, and N a square matrix over R of dimension n. If N is strictly upper triangular and N = \left[ \dfrac{N_1 | N_2}{N_3 | N_4} \right], where N_4 is square, then N_4 is strictly upper triangular. Proof: The elements on or below the main diagonal of N_4 are on or below the main diagonal of N, hence are zero. \square

Lemma: Let R be a ring, n \geq 2, and 1 \leq k \leq n. If A = \left[ \begin{array}{c|c} 0 & A^\prime \\ \hline 0_k & 0 \end{array} \right] is k-strictly upper triangular and A^\prime is strictly upper triangular, then A is k+1-strictly upper triangular. Proof: We have A^\prime = \left[ \begin{array}{c|c} 0 & A^{\prime\prime} \\ \hline 0_1 & 0 \end{array} \right], where A^{\prime\prime} is upper triangular and 0_1 has dimension 1 \times 1. Thus we have the following.

A = \left[ \begin{array}{c|c|c} 0 & 0 & A^{\prime\prime} \\ \hline 0 & 0_1 & 0 \\ \hline 0_k & 0 & 0 \end{array} \right] = \left[ \begin{array}{c|c} 0 & A^{\prime\prime} \\ \hline 0_{k+1} & 0 \end{array} \right],

So that A is k+1-strictly upper triangular. \square

Lemma: Let R be a ring, let n \geq 2, and let M,N \in \mathsf{Mat}_n(R). If M is upper triangular and N is strictly upper triangular, then MN is strictly upper triangular. Proof: Recall that (MN)_{i,j} = \sum_{k=1}^n m_{i,k}n_{k,j}. Suppose i \geq j. Then if k \geq j, n_{k,j} = 0. If k < i, m_{i,k} = 0. Thus (MN)_{i,j} = 0, so that MN is strictly upper triangular. \square

Lemma: Let R be a ring, let n \geq 2, and let 1 \leq k < n. If M,N \in \mathsf{Mat}_n(R) such that M is k-strictly upper triangular and N is strictly upper triangular, then MN is k+1-strictly upper triangular. Proof: Write M = \left[ \begin{array}{c|c} 0 & M^\prime \\ \hline 0_k & 0 \end{array} \right] and N = \left[ \begin{array}{c|c} N_1 & N_2 \\ \hline N_3 & N_4 \end{array} \right], where M and N_4 have dimension n-k \times n-k. Evidently, MN = \left[ \begin{array}{c|c} 0 & M^\prime N_4 \\ \hline 0_k & 0 \end{array} \right]. By the previous lemma, since M^\prime is upper triangular and N_4 is strictly upper triangular, M^\prime N_4 is strictly upper triangular. Thus MN is k+1-strictly upper triangular. \square

Now to the main result.

If A is an n \times n matrix over a ring R, and A is strictly upper triangular, then by an easy induction argument A^k is k-strictly upper triangular. Thus A^n = 0.

The center of a matrix ring over a commutative ring is precisely the scalar matrices

Let R be a commutative ring with 1. Prove that the center of the ring M_n(R) is the set of scalar matrices.


Recall the definition and properties of E_{i,j} from a previous exercise.

We begin with a lemma.

Lemma: E_{p,q}E_{s,t} = E_{p,t} if q = s and 0 otherwise. Proof: If q = s, then the pth row of E_{p,q}E_{s,t} is the sth row of E_{s,t} and all other entries are 0. Thus E_{p,q}E_{s,t} = E_{p,t}. If q \neq s, then the pth row of E_{p,q}E_{s,t} is the qth row of E_{s,t}, which is all zeroes, and all other entries are 0. Thus E_{p,q}E_{s,t} = 0. \square

Now suppose B = [b_{i,j}] \in Z(M_n(R)).

By the previous exercise, note that the (p,t) entry of E_{p,q}BE_{s,t} = E_{p,q}E_{s,t}B is b_{q,s}. By the lemma, if q \neq s, then b_{q,s} = 0. Thus B is a diagonal matrix. Now if q = s, then the (p,t) entry of E_{p,t}B is b_{q,q} on one hand, and b_{t,t} on the other, since the pth row of E_{p,t}B is the tth row of B. Thus b_{q,q} = b_{t,t} for all choices of q and t. Hence B = bI for some b \in R, and we have Z(M_n(R)) \subseteq \{ rI \ |\ r \in R \}.

Conversely, (rI)A = rA = Ar = A(rI). Thus Z(N_n(R)) = \{ rI \ |\ r \in R \}.

Definition and properties of matrices with a single nonzero entry

Let S be a ring with identity 1 \neq 0. Let n be a positive integer and let A = [a_{i,j}] be an n \times n matrix over S. Let E_{i,j} be the element of M_n(S) whose (i,j) entry is 1 and whose other entries are all 0.

  1. Prove that E_{i,j}A is the matrix whose ith row is the jth row of A and all other rows are 0.
  2. Prove that AE_{i,j} is the matrix whose jth column is the ith column of A and all other columns are 0.
  3. Deduce that E_{p,q}AE_{r,s} is the matrix whose (p,s) entry is a_{q,r} and all other entries are 0.

  1. By definition, E_{i,j}A = [c_{p,q}], where c_{p,q} = \sum_{k=1}^n e_{p,k}a_{k,q}. Note that if p \neq i, then e_{p,k} = 0, so that c_{p,q} = 0. If p = i, then c_{p,q} = a_{j,q}; thus the ith row of E_{i,j}A is the jth row of A, and all other entries are 0.
  2. The proof for AE_{i,j} is very similar.
  3. By the above arguments, E_{p,q}A is the matrix whose pth row is the qth row of A, and all other entries are 0. Then E_{p,q}AE_{r,s} is the matrix whose sth column is the rth column of E_{p,q}A, which is all zeroes except for the pth row, whose entry is the (q,r) entry of A, and all other entries are zero.