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Section 1.2 Matrices
Matrices can be viewed in two complementary ways: as arrays of data and as functions acting on vectors. We begin with basic matrix arithmetic, then use matrix-vector multiplication to describe linear maps.
Subsection Matrices as arrays of numbers
Definition 1.2.1 . Matrices.
An \(m \times n\) matrix \(A\) is a rectangular array of \(m \cdot n\) real numbers arranged in \(m\) (horizontal) rows and \(n\) (vertical) columns :
\begin{equation*}
A = \begin{bmatrix}
a_{11} \amp a_{12} \amp \cdots \amp a_{1n} \\
a_{21} \amp a_{22} \amp \cdots \amp a_{2n} \\
\vdots \amp \vdots \amp \ddots \amp \vdots \\
a_{m1} \amp a_{m2} \amp \cdots \amp a_{mn} \\
\end{bmatrix}
\end{equation*}
The \(i\) -th row\(A\) is
\begin{equation*}
\begin{bmatrix}
a_{i1} \amp a_{i2} \amp \cdots \amp a_{in}
\end{bmatrix},
\end{equation*}
and the \(j\) -th column\(A\) is
\begin{equation*}
\begin{bmatrix}
a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj}
\end{bmatrix}.
\end{equation*}
The number
\(a_{ij}\text{,}\) which is in the
\(i\) -th row and
\(j\) -th column of
\(A\text{,}\) is the
\((i,j)\) -entry of
\(A\text{,}\) and we often write
\(A = [a_{ij}]\text{.}\) We say
\(A\) is an "
\(m\) by \(n\) " matrix.
Activity 1.2.2 . Reading matrix entries (U1-LO3).
Consider the matrix
\begin{equation*}
A = \begin{bmatrix}
1 \amp 2 \amp 3 \\
5 \amp -8 \amp -13
\end{bmatrix}
\end{equation*}
\(A\) is a \(2 \times 3\) matrix. Compute the following:
(a) (b) (c) (d) (e) An important application of matrices (but far from the only one!) is to record data. The following table gives a few examples.
Table 1.2.3. Some data matrices
Monochrome image
\(X_{ij}\) is the pixel value in row \(i\) and column \(j\text{.}\)
Rainfall data
\(A_{ij}\) is the rainfall at location \(j\) on day \(i\text{.}\)
Asset returns
\(R_{ij}\) is the return of asset \(i\) in period \(j\text{.}\)
Feature matrix
\(X_{ij}\) is the value of feature \(j\) for entity \(i\text{.}\)
Subsection Operations on matrices
Definition 1.2.4 . Equality of matrices.
Two matrices
\(A\) and
\(B\) are
equal if they have the same size and all the corresponding entries are equal.
Activity 1.2.5 . Unknown entries in equal matrices (U1-LO3).
Suppose
\begin{equation*}
A = \begin{bmatrix} 3 \amp y \\ 4 \amp -7 \end{bmatrix}\quad\text{and}\quad B = \begin{bmatrix} 6+x \amp -2 \\ 4 \amp -7 \end{bmatrix}
\end{equation*}
and \(A = B\text{.}\) Find \(x\) and \(y\text{.}\)
Solution .
Since \(A = B\text{,}\) all entries of \(A\) must equal the corresponding entries in \(B\text{.}\) So it must be true, comparing corresponding entries, that
\begin{align*}
3 \amp = 6 + x\\
y \amp = -2
\end{align*}
Therefore, \(x = -3\) and \(y = -2\text{.}\)
Definition 1.2.6 . Sums of matrices.
If
\(A = [a_{ij}]\) and
\(B = [b_{ij}]\) are both
\(m \times n\) matrices, then their
sum \(A+B\) is the matrix
\(C = [c_{ij}]\) where
\(c_{ij} = a_{ij} + b_{ij}\text{.}\)
Activity 1.2.7 . Adding matrices (U1-LO3).
For the matrices
\begin{equation*}
A = \begin{bmatrix}
1 \amp 2 \amp 3 \\
5 \amp -8 \amp -13
\end{bmatrix}
\quad\text{and}\quad
B = \begin{bmatrix}
0 \amp 2 \amp 1 \\
1 \amp 3 \amp -4
\end{bmatrix}
\end{equation*}
calculate \(A+B\text{.}\)
Solution .
\begin{equation*}
\begin{aligned}
A+B
\amp=
\begin{bmatrix}
1+0 \amp 2+2 \amp 3+1\\
5+1 \amp -8+3 \amp -13+(-4)
\end{bmatrix}\\
\amp=
\begin{bmatrix}
1 \amp 4 \amp 4\\
6 \amp -5 \amp -17
\end{bmatrix}.
\end{aligned}
\end{equation*}
Definition 1.2.9 . Scalar multiples of matrices.
If
\(A = [a_{ij}]\) is an
\(m \times n\) matrix and
\(r\) is a real number, then the
scalar multiple of
\(A\) by
\(r\text{,}\) written
\(rA\text{,}\) is the
\(m \times n\) matrix
\(C = [c_{ij}]\text{,}\) where
\(c_{ij} = r a_{ij}\text{,}\) that is,
\(C\) is the matrix obtained by multiplying every entry of
\(A\) by
\(r\text{.}\)
Activity 1.2.10 . Scaling a matrix (U1-LO3).
Solution .
\begin{equation*}
-2A = \begin{bmatrix} -2 \amp -4 \amp -6 \\ -10 \amp 16 \amp 26 \end{bmatrix}
\end{equation*}
Theorem 1.2.12 .
Let \(A\text{,}\) \(B\text{,}\) and \(C\) be \(m \times n\) matrices.
\(A + B = B + A\text{,}\) i.e., matrix addition is
commutative .
\(A + (B + C) = (A + B) + C\text{,}\) i.e., matrix addition is
associative .
Let
\(r\) and
\(s\) be real numbers. Then:
\(r(sA) = (rs)A\text{.}\)
\((r+s)A = rA + sA\text{.}\)
\(r(A+B) = rA + rB\text{.}\)
Why is this true?.
We prove Property 1 only, i.e., the commutativity of addition. Let \(A = [a_{ij}]\) and \(B = [b_{ij}]\text{.}\) Then:
\begin{align*}
A + B \amp = [a_{ij} + b_{ij}]\\
\amp = [b_{ij} + a_{ij}] \amp \text{(since real numbers are commutative)}\\
\amp= B + A
\end{align*}
Definition 1.2.13 .
If \(A = [a_{ij}]\) is an \(m \times n\) matrix, then the transpose of \(A\text{,}\) denoted \(A^T = [a_{ij}^T]\text{,}\) is the \(n \times m\) matrix defined by
\begin{equation*}
a_{ij}^T = a_{ji}
\end{equation*}
In other words, the transpose of \(A\) is obtained by interchanging the rows and the columns of \(A\text{.}\)
Activity 1.2.14 . Transposing matrices (U1-LO3).
Compute the transpose for each of the given matrices:
(a)
\begin{equation*}
A = \begin{bmatrix}
1 \amp 2 \amp 3 \\
5 \amp -8 \amp -13
\end{bmatrix}.
\end{equation*}
Solution .
\begin{equation*}
A^T = \begin{bmatrix} 1 \amp 5 \\ 2 \amp -8 \\ 3 \amp -13 \end{bmatrix}
\end{equation*}
(b)
\begin{equation*}
B = \begin{bmatrix}
5 \amp 2 \amp 3 \\
6 \amp 2 \amp 3 \\
-1 \amp -2 \amp 3
\end{bmatrix}
\end{equation*}
Solution .
\begin{equation*}
B^T = \begin{bmatrix} 5 \amp 6 \amp -1 \\ 2 \amp 2 \amp -2 \\ 3 \amp 3 \amp 3 \end{bmatrix}
\end{equation*}
(c)
\begin{equation*}
C = \begin{bmatrix}
10 \\ 20 \\ 30
\end{bmatrix}.
\end{equation*}
Solution .
\begin{equation*}
C^T = \begin{bmatrix} 10 \amp 20 \amp 30 \end{bmatrix}
\end{equation*}
Definition 1.2.16 . Main diagonal.
If
\(A=[a_{ij}]\) is an
\(m\times n\) matrix, the elements
\(a_{11}, a_{22}, a_{33},\ldots\) are called the
main diagonal of
\(A\text{.}\) A matrix
\(A\) is called
diagonal if its only nonzero entries occur on its main diagonal.
Below are four matrices of various dimensions, with the main diagonal written in bold font.
\begin{equation*}
\begin{bmatrix} \mathbf{a_{11}} \amp a_{12} \amp a_{13} \\ a_{21} \amp \mathbf{a_{22}} \amp a_{23} \\ a_{31} \amp a_{32} \amp \mathbf{a_{33}} \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} \mathbf{a_{11}} \amp a_{12} \\ a_{21} \amp \mathbf{a_{22}} \\ a_{31} \amp a_{32} \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} \mathbf{a_{11}} \amp a_{12} \amp a_{13} \\ a_{21} \amp \mathbf{a_{22}} \amp a_{23} \end{bmatrix}
\end{equation*}
\begin{equation*}
\begin{bmatrix} \mathbf{a_{11}} \\ a_{21} \end{bmatrix}
\end{equation*}
Definition 1.2.17 . Identity matrix.
The
identity matrix \(I_n\) is the
\(n\times n\) matrix with
\(1\) s on the main diagonal and zeros elsewhere.
Forming the transpose of a matrix
\(A\) can be viewed as
flipping \(A\) about its main diagonal.
Theorem 1.2.18 .
If \(r\) is a scalar and \(A\) and \(B\) are matrices of the appropriate sizes, then:
\((A + B)^T = A^T + B^T\text{.}\)
\((rA)^T = rA^T\text{.}\)
Why is this true?.
Let
\(A = [a_{ij}]\) and
\(B = [b_{ij}]\text{.}\) Then
\(A + B = [c_{ij}]\) where
\(c_{ij} = a_{ij} + b_{ij}\text{.}\)
Then
\begin{align*}
(A + B)^T \amp = [c_{ij}^T]\\
\amp = [c_{ji}] \amp \text{By definition of transpose}\\
\amp = [a_{ji} + b_{ji}] \amp \text{Since $c_{ij} = a_{ij} + b_{ij}$}\\
\amp = [a_{ji}] + [b_{ji}] \amp \text{By definition of matrix addition}\\
\amp = [a_{ij}^T] + [b_{ij}^T] \amp \text{By definition of transpose}\\
\amp = A^T + B^T
\end{align*}
Therefore, \((A + B)^T = A^T + B^T\text{.}\)
Definition 1.2.19 . Symmetric matrices.
A matrix
\(A\) with real entries is called
symmetric if
\(A^T = A\text{.}\)
Activity 1.2.21 . Checking matrix symmetry (U1-LO3).
Determine whether each of the following matrices is symmetric or not symmetric:
(a)
\(A = \begin{bmatrix} 0 \amp 2 \amp -3 \\ -2 \amp 0 \amp 5 \\ 3 \amp -5 \amp 0 \end{bmatrix}\)
Solution .
Interchanging rows and columns gives
\begin{equation*}
A^T = \begin{bmatrix} 0 \amp -2 \amp 3 \\ 2 \amp 0 \amp -5 \\ -3 \amp 5 \amp 0 \end{bmatrix}.
\end{equation*}
Since \(A^T \ne A\text{,}\) \(A\) is not symmetric.
(b)
\(B = \begin{bmatrix} 3 \amp 5 \amp 2 \\ 5 \amp 1 \amp 4 \\ 2 \amp 4 \amp -1 \end{bmatrix}\)
Solution .
Interchanging rows and columns gives
\begin{equation*}
B^T = \begin{bmatrix} 3 \amp 5 \amp 2 \\ 5 \amp 1 \amp 4 \\ 2 \amp 4 \amp -1 \end{bmatrix}.
\end{equation*}
\(B\) is symmetric since \(B^T = B\text{.}\)
(c)
\(C = \begin{bmatrix} 1 \amp 2 \amp -3 \\ -2 \amp 0 \amp 5 \\ 3 \amp 5 \amp 0 \end{bmatrix}\)
Solution .
Interchanging rows and columns gives
\begin{equation*}
C^T = \begin{bmatrix} 1 \amp -2 \amp 3 \\ 2 \amp 0 \amp 5 \\ -3 \amp 5 \amp 0 \end{bmatrix}.
\end{equation*}
Since \(C^T \ne C\text{,}\) \(C\) is not symmetric.
(d)
\(D = \begin{bmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}\)
Solution .
Interchanging rows and columns gives
\begin{equation*}
D^T = \begin{bmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}.
\end{equation*}
Since \(D^T = D\text{,}\) \(D\) is symmetric.
