Section 5.3 Laws of Matrix Algebra
Subsection 5.3.1 The Laws
The following is a summary of the basic laws of matrix operations. Assume that the indicated operations are defined; that is, that the orders of the matrices \(A\text{,}\) \(B\) and \(C\) are such that the operations make sense.
| (1) Commutative Law of Addition | \(A + B = B + A\) |
| (2) Associative Law of Addition | \(A + (B + C) = (A + B) + C\) |
| (3) Distributive Law of a Scalar over Matrices | \(c(A + B) = c A + c B\text{,}\) where \(c \in \mathbb{R}\text{.}\) |
| (4) Distributive Law of Scalars over a Matrix | \(\left(c_1 + c_2 \right)A = c_1A +c_2 A\text{,}\) where \(c_1, c_2 \in \mathbb{R}\text{.}\) |
| (5) Associative Law of Scalar Multiplication | \(c_1 \left(c_2 A\right) =\left(c_1 \cdot c_2 \right)A\text{,}\) where \(c_1, c_2 \in \mathbb{R}\text{.}\) |
| (6) Zero Matrix Annihilates all Products | \(\pmb{0}A = \pmb{0}\text{,}\) where \(\pmb{0}\) is the zero matrix. |
| (7) Zero Scalar Annihilates all Products | \(0 A =\pmb{0}\text{,}\) where 0 on the left is the scalar zero. |
| (8) Zero Matrix is an identity for Addition | \(A + \pmb{0} = A\text{.}\) |
| (9) Negation produces additive inverses | \(A + (-1)A = \pmb{0}\text{.}\) |
| (10) Right Distributive Law of Matrix Multiplication | \((B + C)A = B A + C A\text{.}\) |
| (11) Left Distributive Law of Matrix Multiplication | \(A(B + C) = A B + A C\text{.}\) |
| (12) Associative Law of Multiplication | \(A(B C) = (A B)C\text{.}\) |
| (13) Identity Matrix is a Multiplicative Identity | \(I A = A\) and \(A I = A\text{.}\) |
| (14) Involution Property of Inverses | If \(A^{-1}\) exists,\(\left(A^{-1} \right)^{-1} = A\text{.}\) |
| (15) Inverse of Product Rule | If \(A^{-1}\) and \(B^{-1}\) exist, \((A B)^{-1}= B^{-1}A^{-1}\) |
Subsection 5.3.2 Commentary
If we wished to write out each of the above laws more completely, we would specify the orders of the matrices. For example, Law 10 should read:
Let \(A\text{,}\) \(B\text{,}\) and \(C\) be \(m\times n\text{,}\) \(n\times p\text{,}\) and \(n\times p\) matrices, respectively, then \(A(B + C) = A B + A C\)
Remarks:
- Notice the absence of the “law” \(A B = B A\text{.}\) Why?
- Is it really necessary to have both a right (No. 11) and a left (No. 10) distributive law? Why?
Exercises 5.3.3 Exercises
1.
Rewrite the above laws specifying as in Example 5.3.2 the orders of the matrices.
Answer.
- Let \(A\) and \(B\) be \(m\) by \(n\) matrices. Then \(A+B=B+A\text{,}\)
- Let \(A\text{,}\) \(B\text{,}\) and \(C\) be \(m\) by \(n\) matrices. Then \(A+(B+C)=(A+B)+C\text{.}\)
- Let \(A\) and \(B\) be \(m\) by \(n\) matrices, and let \(c\in \mathbb{R}\text{.}\) Then \(c(A+B)=cA+cB\text{,}\)
- Let \(A\) be an \(m\) by \(n\) matrix, and let \(c_1,c_2\in \mathbb{R}\text{.}\) Then \(\left(c_1+c_2\right)A=c_1A+c_2A\text{.}\)
- Let \(A\) be an \(m\) by \(n\) matrix, and let \(c_1,c_2\in \mathbb{R}\text{.}\) Then \(c_1\left(c_2A\right)=\left(c_1c_2\right)A\)
- Let \(\pmb{0}\) be the zero matrix, of size \(m \textrm{ by } n\text{,}\) and let \(A\) be a matrix of size \(n \textrm{ by } r\text{.}\) Then \(\pmb{0}A=\pmb{0}=\textrm{ the } m \textrm{ by } r \textrm{ zero matrix}\text{.}\)
- Let \(A\) be an \(m \textrm{ by } n\) matrix, and \(0 = \textrm{ the number zero}\text{.}\) Then \(0A=0=\textrm{ the } m \textrm{ by } n \textrm{ zero matrix}\text{.}\)
- Let \(A\) be an \(m \textrm{ by } n\) matrix, and let \(\pmb{0}\) be the \(m \textrm{ by } n\) zero matrix. Then \(A+\pmb{0}=A\text{.}\)
- Let \(A\) be an \(m \textrm{ by } n\) matrix. Then \(A+(- 1)A=\pmb{0}\text{,}\) where \(\pmb{0}\) is the \(m \textrm{ by } n\) zero matrix.
- Let \(A\text{,}\) \(B\text{,}\) and \(C\) be \(m \textrm{ by } n\text{,}\) \(n \textrm{ by } r\text{,}\) and \(n \textrm{ by } r\) matrices respectively. Then \(A(B+C)=AB+AC\text{.}\)
- Let \(A\text{,}\) \(B\text{,}\) and \(C\) be \(m \textrm{ by } n\text{,}\) \(r \textrm{ by } m\text{,}\) and \(r \textrm{ by } m\) matrices respectively. Then \((B+C)A=BA+CA\text{.}\)
- Let \(A\text{,}\) \(B\text{,}\) and \(C\) be \(m \textrm{ by } n\text{,}\) \(n \textrm{ by } r\text{,}\) and \(r \textrm{ by } p\) matrices respectively. Then \(A(BC)=(AB)C\text{.}\)
- Let \(A\) be an \(m \textrm{ by } n\) matrix, \(I_m\) the \(m \textrm{ by } m\) identity matrix, and \(I_n\) the \(n \textrm{ by } n\) identity matrix. Then \(I_mA=AI_n=A\)
- Let \(A\) be an \(n \textrm{ by } n\) matrix. Then if \(A^{-1}\) exists, \(\left(A^{-1}\right)^{-1}=A\text{.}\)
- Let \(A\) and \(B\) be \(n \textrm{ by } n\) matrices. Then if \(A^{-1}\) and \(B^{-1}\) exist, \((AB)^{-1}=B^{-1}A^{-1}\text{.}\)
2.
Verify each of the Laws of Matrix Algebra using examples.
3.
Let \(A = \left(
\begin{array}{cc}
1 & 2 \\
0 & -1 \\
\end{array}
\right)\text{,}\) \(B= \left(
\begin{array}{ccc}
3 & 7 & 6 \\
2 & -1 & 5 \\
\end{array}
\right)\text{,}\) and \(C= \left(
\begin{array}{ccc}
0 & -2 & 4 \\
7 & 1 & 1 \\
\end{array}
\right)\text{.}\) Compute the following as efficiently as possible by using any of the Laws of Matrix Algebra:
- \(\displaystyle A B + A C\)
- \(\displaystyle A^{-1}\)
- \(\displaystyle A(B + C)\)
- \(\displaystyle \left(A^2\right)^{-1}\)
- \(\displaystyle (C + B)^{-1}A^{-1}\)
Answer.
- \(\displaystyle AB+AC=\left( \begin{array}{ccc} 21 & 5 & 22 \\ -9 & 0 & -6 \\ \end{array} \right)\)
- \(\displaystyle A^{-1}=\left( \begin{array}{cc} 1 & 2 \\ 0 & -1 \\ \end{array} \right)=A\)
- \(A(B+C)=A B+ B C\text{,}\) which is given in part (a).
- \(\left(A^2\right)^{-1}=(AA)^{-1}=(A^{-1}A)=I^{-1}=I \quad \) by part c
4.
Let \(A =\left(
\begin{array}{cc}
7 & 4 \\
2 & 1 \\
\end{array}
\right)\) and \(B =\left(
\begin{array}{cc}
3 & 5 \\
2 & 4 \\
\end{array}
\right)\text{.}\) Compute the following as efficiently as possible by using any of the Laws of Matrix Algebra:
- \(\displaystyle A B\)
- \(\displaystyle A + B\)
- \(\displaystyle A^2 + A B + B A + B ^2\)
- \(\displaystyle B^{-1}A^{-1}\)
- \(\displaystyle A^2 + A B\)
5.
Let \(A\) and \(B\) be \(n\times n\) matrices of real numbers. Is \(A^2-B^2= (A-B)(A+B)\text{?}\) Explain.
