ads Laws of Matrix Algebra

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.

Table 5.3.1. Laws of Matrix Algebra

(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

Example 5.3.2. More Precise Statement of one Law.

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:

\(\displaystyle A B\)
\(\displaystyle A + B\)
\(\displaystyle A^2 + A B + B A + B ^2\)
\(\displaystyle B^{-1}A^{-1}\)
\(\displaystyle A^2 + B A\)

5.

Let \(A\) and \(B\) be \(n\times n\) matrices of real numbers. Is \(A^2-B^2= (A-B)(A+B)\text{?}\) Explain.

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