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Theorem

Let A A be an m×n m \times n matrix, and let B B and C C have sizes for which the indicated sums and products are defined.

  1. A(BC)=(AB)C A(BC) = (AB)C (associative law of multiplication)
  2. A(B+C)=AB+AC A(B + C) = AB + AC (left distributive law)
  3. (B+C)A=BA+CA (B + C)A = BA + CA (right distributive law)
  4. r(AB)=(rA)B=A(rB) r(AB) = (rA)B = A(rB) for any scalar r r
  5. ImA=A=AIn I_m A = A = A I_n (identity for matrix multiplication)

Theorem

Let A A and B B denote matrices whose sizes are appropriate for the following sums and products.

  1. (AT)T=A (A^T)^T = A
  2. (A+B)T=AT+BT (A + B)^T = A^T + B^T
  3. For any scalar r r , (rA)T=rAT (rA)^T = rA^T
  4. (AB)T=BTAT (AB)^T = B^T A^T

Theorem

Let A A , B B , and C C be matrices of the same size, and let r r and s s be scalars.

  1. A+B=B+A A + B = B + A
  2. (A+B)+C=A+(B+C) (A + B) + C = A + (B + C)
  3. A+0=A A + 0 = A
  4. r(A+B)=rA+rB r(A + B) = rA + rB
  5. (r+s)A=rA+sA (r + s)A = rA + sA
  6. r(sA)=(rs)A r(sA) = (rs)A

Proof (a):

Let A=[a⃗1a⃗2…a⃗n] A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} , B=[b⃗1b⃗2…b⃗n] B = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}

A+B=[a⃗1a⃗2…a⃗n]+[b⃗1b⃗2…b⃗n] A + B = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} + \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}

=[a⃗1+b⃗1a⃗2+b⃗2…a⃗n+b⃗n] = \begin{bmatrix} \vec{a}_1 + \vec{b}_1 & \vec{a}_2 + \vec{b}_2 & \dots & \vec{a}_n + \vec{b}_n \end{bmatrix}

=[b⃗1+a⃗1b⃗2+a⃗2…b⃗n+a⃗n] = \begin{bmatrix} \vec{b}_1 + \vec{a}_1 & \vec{b}_2 + \vec{a}_2 & \dots & \vec{b}_n + \vec{a}_n \end{bmatrix}

=B+A = B + A

Proof (b):

Let A=[a⃗1a⃗2…a⃗n] A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} , B=[b⃗1b⃗2…b⃗n] B = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix} , and C=[c⃗1c⃗2…c⃗n] C = \begin{bmatrix} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{bmatrix}

(A+B)+C=([a⃗1+b⃗1a⃗2+b⃗2…a⃗n+b⃗n])+[c⃗1c⃗2…c⃗n] (A + B) + C = \left( \begin{bmatrix} \vec{a}_1 + \vec{b}_1 & \vec{a}_2 + \vec{b}_2 & \dots & \vec{a}_n + \vec{b}_n \end{bmatrix} \right) + \begin{bmatrix} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{bmatrix}

=[(a⃗1+b⃗1)+c⃗1(a⃗2+b⃗2)+c⃗2…(a⃗n+b⃗n)+c⃗n] = \begin{bmatrix} (\vec{a}_1 + \vec{b}_1) + \vec{c}_1 & (\vec{a}_2 + \vec{b}_2) + \vec{c}_2 & \dots & (\vec{a}_n + \vec{b}_n) + \vec{c}_n \end{bmatrix}

=[a⃗1+(b⃗1+c⃗1)a⃗2+(b⃗2+c⃗2)…a⃗n+(b⃗n+c⃗n)] = \begin{bmatrix} \vec{a}_1 + (\vec{b}_1 + \vec{c}_1) & \vec{a}_2 + (\vec{b}_2 + \vec{c}_2) & \dots & \vec{a}_n + (\vec{b}_n + \vec{c}_n) \end{bmatrix}

=[a⃗1a⃗2…a⃗n]+([b⃗1b⃗2…b⃗n]+[c⃗1c⃗2…c⃗n]) = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} + \left( \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix} + \begin{bmatrix} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{bmatrix} \right)

=A+(B+C) = A + (B + C)