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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 of \(A\) is
\begin{equation*}
\begin{bmatrix}
a_{i1} \amp a_{i2} \amp \cdots \amp a_{in}
\end{bmatrix},
\end{equation*}
and the \(j\) -th column of \(A\) 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 1.2.1 .
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)
Subsection Data matrices
An important application of matrices (but far from the only one!) is to record data. The following table gives a few examples.
Table 1.2.2. 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 Matrix operations
Definition 1.2.3 . Equality of matrices.
Two matrices
\(A\) and
\(B\) are
equal if they have the same size and all the corresponding entries are equal.
Activity 1.2.2 .
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 1.2.4 . 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 1.2.3 .
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 1.2.6 . 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 1.2.4 .
Solution .
\begin{equation*}
-2A = \begin{bmatrix} -2 \amp -4 \amp -6 \\ -10 \amp 16 \amp 26 \end{bmatrix}
\end{equation*}
Definition 1.2.7 . Linear combinations of matrices.
If \(A_1, A_2, \ldots, A_k\) are \(m \times n\) matrices and \(c_1, c_2, \ldots, c_k\) are real numbers, then an expression of the form
\begin{equation*}
c_1 A_1 + c_2 A_2 + \cdots + c_k A_k
\end{equation*}
is called a linear combination of \(A_1, A_2, \ldots, A_k\text{.}\) The scalars \(c_1, \ldots, c_k\) are called the coefficients of the linear combination.
This is the same idea as a linear combination of vectors from
Vectors . The only difference is that the objects being combined are matrices of the same size.
Activity 1.2.5 .
Compute the following linear combination of matrices:
\begin{equation*}
4 \begin{bmatrix} 0 \amp 2 \\ -3 \amp 3 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} 4 \amp 2 \\ 6 \amp 2 \end{bmatrix}
\end{equation*}
Solution .
\begin{align*}
\amp= \begin{bmatrix} 0 \amp 8 \\ -12 \amp 12 \end{bmatrix} - \begin{bmatrix} 2 \amp 1 \\ 3 \amp 1 \end{bmatrix} \amp= \begin{bmatrix} -2 \amp 7 \\ -15 \amp 11 \end{bmatrix}
\end{align*}
Theorem 1.2.8 .
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 .
There is a unique
\(m \times n\) matrix
\(O\) such that
\(A + O = A\) for any
\(m \times n\) matrix
\(A\text{.}\) The matrix
\(O\) is called the
\(m \times n\) zero matrix , and is the matrix with zeros in every entry.
For each
\(m \times n\) matrix
\(A\text{,}\) there is a unique
\(m \times n\) matrix
\(D\) such that
\(A + D = O\text{.}\) The matrix
\(D\) must be the matrix
\(-A = (-1)A\text{.}\) The matrix
\(-A\) is called the
negative of
\(A\text{.}\)
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 1.2.9 .
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 1.2.6 .
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 1.2.11 . 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} \\ 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}} \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}} \\ a_{21} \end{bmatrix}
\end{equation*}
Forming the transpose of a matrix
\(A\) can be viewed as
flipping \(A\) about its main diagonal.
Theorem 1.2.12 .
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^T + B^T \amp \text{By definition of transpose}
\end{align*}
Therefore, \((A + B)^T = A^T + B^T\text{.}\)
Definition 1.2.13 . Symmetry and skew-symmetry.
A matrix \(A\) with real entries is called:
Activity 1.2.7 .
Determine whether the following matrices are symmetric, skew symmetric, or neither:
(a)
\(A = \begin{bmatrix} 0 \amp 2 \amp -3 \\ -2 \amp 0 \amp 5 \\ 3 \amp -5 \amp 0 \end{bmatrix}\)
Solution .
\(A\) is skew symmetric since
\(A^T = -A\text{.}\)
(b)
\(B = \begin{bmatrix} 3 \amp 5 \amp 2 \\ 5 \amp 1 \amp 4 \\ 2 \amp 4 \amp -1 \end{bmatrix}\)
Solution .
\(B\) 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 .
\(C\) is neither symmetric nor skew symmetric.
(d)
\(D = \begin{bmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}\)
Solution .
\(D\) is both symmetric and skew symmetric.