Some simple matrix properties

5 Some simple matrix properties

Using the definition of matrix addition described above we can easily verify the following properties of matrix addition:

Key Point 2

Basic Properties of Matrices

Matrix addition is commutative : $A + B = B + A$

Matrix addition is associative : $A + (B + C) = (A + B) + C$

The distributive law holds: $k (A + B) = k A + k B$

These Key Point results follow from the fact that $a_{i j} + b_{i j} = b_{i j} + a_{i j}$ etc.

We can also show that the transpose of a matrix satisfies the following simple properties:

Key Point 3

Properties of Transposed Matrices

\begin{array}{rcl} {(A + B)}^{T} & = & A^{T} + B^{T} \\ {(A - B)}^{T} & = & A^{T} - B^{T} \\ {(A^{T})}^{T} & = & A \end{array}

Example 3

Show that ${(A^{T})}^{T} = A$ for the matrix $A = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}]$

Solution

$A^{T} = [\begin{matrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{matrix}]$ so that ${(A^{T})}^{T} = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}] = A$

Task!

For the matrices $A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}], B = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}]$ verify that

(i) $3 (A + B) = 3 A + 3 B$ (ii) ${(A - B)}^{T} = A^{T} - B^{T}$ .

(i) $A + B = [\begin{matrix} 2 & 1 \\ 2 & 5 \end{matrix}]$ ; $3 (A + B) = [\begin{matrix} 6 & 3 \\ 6 & 15 \end{matrix}]$ ; $3 A = [\begin{matrix} 3 & 6 \\ 9 & 12 \end{matrix}]$ ;

$3 B = [\begin{matrix} 3 & - 3 \\ - 3 & 3 \end{matrix}]$ ; $3 A + 3 B = [\begin{matrix} 6 & 3 \\ 6 & 15 \end{matrix}]$ .

(ii) $A - B = [\begin{matrix} 0 & 3 \\ 4 & 3 \end{matrix}]$ ; ${(A - B)}^{T} = [\begin{matrix} 0 & 4 \\ 3 & 3 \end{matrix}]$ ; $A^{T} = [\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}];$

$B^{T} = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}];$ $A^{T} - B^{T} = [\begin{matrix} 0 & 4 \\ 3 & 3 \end{matrix}]$ .

Exercises

Find the coefficient matrix $A$ of the system: $\begin{array}{rcl} 2 x_{1} + 3 x_{2} - x_{3} & = & 1 \\ 4 x_{1} + 4 x_{2} & = & 0 \\ 2 x_{1} - x_{2} - x_{3} & = & 0 \end{array}$
If $B = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 0 & 1 \end{matrix}]$ determine ${(3 A^{T} - B)}^{T}$ .
If $A = [\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}]$ and $B = [\begin{matrix} - 1 & 4 \\ 0 & 1 \\ 2 & 7 \end{matrix}]$ verify that $3 (A^{T} - B) = {(3 A - 3 B^{T})}^{T}$ .

$A = [\begin{matrix} 2 & 3 & - 1 \\ 4 & 4 & 0 \\ 2 & - 1 & - 1 \end{matrix}], A^{T} = [\begin{matrix} 2 & 4 & 2 \\ 3 & 4 & - 1 \\ - 1 & 0 & - 1 \end{matrix}], 3 A^{T} = [\begin{matrix} 6 & 12 & 6 \\ 9 & 12 & - 3 \\ - 3 & 0 & - 3 \end{matrix}]$

$3 A^{T} - B = [\begin{matrix} 5 & 10 & 3 \\ 5 & 7 & - 9 \\ - 3 & 0 & - 4 \end{matrix}] {(3 A^{T} - B)}^{T} = [\begin{matrix} 5 & 5 & - 3 \\ 10 & 7 & 0 \\ 3 & - 9 & - 4 \end{matrix}]$
$A^{T} = [\begin{matrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{matrix}], A^{T} - B = [\begin{matrix} 2 & 0 \\ 2 & 4 \\ 1 & - 1 \end{matrix}], 3 (A^{T} - B) = [\begin{matrix} 6 & 0 \\ 6 & 12 \\ 3 & - 3 \end{matrix}]$

$B^{T} = [\begin{matrix} - 1 & 0 & 2 \\ 4 & 1 & 7 \end{matrix}], 3 A - 3 B^{T} = [\begin{matrix} 3 & 6 & 9 \\ 12 & 15 & 18 \end{matrix}] - [\begin{matrix} - 3 & 0 & 6 \\ 12 & 3 & 21 \end{matrix}] = [\begin{matrix} 6 & 6 & 3 \\ 0 & 12 & - 3 \end{matrix}]$