16 Matrix Multiplication
Matrix multiplication (symbol: \(\times\)) is perhaps the more complex of the matrix algebra operations we will cover. It is a bit involved, but relatively easy once you get the hang of it. We will begin with a simple example before doing more complicated stuff.
16.1 Matrix Multiplication Rules
First, we will let out the basic rules of matrix multiplication:
- You can always multiply two matrices as long as the number of columns of the first matrix equal the rows of the second matrix. To check whether this is the case, all you have to do is put the two matrices side by side and list their dimensions.
\[ \mathbf{A}_{3 \times \mathbf{5}} \times \mathbf{B}_{\mathbf{5} \times 6} \] - The two little “fives” in bold are called the inner dimensions of the two matrices. The little “three” on the left and the little “six” on the right are called the outer dimensions. So another way of stating the first rule of matrix multiplication is that the product of two matrices is defined as long as their inner dimensions equal to one another when you line them up from left to right.
When the number of columns of a matrix equal the number of rows of another matrix so that their inner dimensions match we say that the the two matrices are conformable. When this is not the case, we say the matrices are non-conformable.
Thus, another way of stating the first rule is that only the product of conformable matrices is defined. If the matrices are not conformable then their product is not defined (e.g., there is no answer to the question of what we get if we multiply them!).
This means that unlike numbers or the matrix dot product, where the order of the two things you are multiplying doesn’t matter (\(4 \times 3 = 3 \times 4\) or \(\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}\)), in matrix multiplication it does matter. Alas, for any two matrices \(\mathbf{A}\) and \(\mathbf{B}\),
\[ \mathbf{A} \times \mathbf{B} \neq \mathbf{B} \times \mathbf{A} \]
- When you multiply a matrix times another matrix, the resulting matrix will have number of rows equal to the number of rows of the first matrix and number of columns equal to the number of columns of the second matrix. Thus:
\[ \mathbf{A}_{3 \times \mathbf{5}} \times \mathbf{B}_{\mathbf{5} \times 6} = \mathbf{C}_{3 \times 6} \tag{16.1}\]
- Equation eq-matmult0 says that the product of a three by five matrix \(\mathbf{A}\) (three rows and five columns) times a five by six matrix \(\mathbf{B}\) (five rows and six columns) is a third matrix \(\mathbf{C}\) with three rows and six columns. Another way of saying this last rule is that the product of two conformable matrices will have dimensions equal to their outer dimensions.
16.2 Multiplying a Matrix Times its Transpose
- By definition, as discussed in sec-trans, the rows of a matrix are equal to the columns of its transpose, and vice versa. The product of a matrix times its transpose and the transpose times the original matrix is always defined, no matter what the dimensions of the original matrix are. Thus,
\[ \mathbf{A} \times \mathbf{A}^T = defined! \]
\[ \mathbf{A}^T \times \mathbf{A} = defined! \]
- When you multiply a matrix times its transpose, the resulting matrix will be a square matrix with number of rows and columns equal to the number of rows of the original matrix. For instance, say matrix \(\mathbf{A}_{5 \times 3}\) is of dimensions \(5 \times 3\) (like the matrix shown in Table tbl-trans-1). Then its transpose \(A^T_{3 \times 5}\) will be of dimensions \(3 \times 5\) (like the matrix shown in Table tbl-trans-2). That means the product of the matrix times its transpose will be:
\[ \mathbf{A}_{5 \times 3} \times \mathbf{A}_{3 \times 5}^T = \mathbf{B}_{5 \times 5} \tag{16.2}\]
- Equation eq-matmult1 says that a five by three matrix multiplied by its transposed yields a square matrix \(\mathbf{B}\) of dimensions five by five (a square matrix with five rows and five columns). In the same way,
\[ \mathbf{A}_{3 \times 5}^T \times \mathbf{A}_{5 \times 3} = \mathbf{B}_{3 \times 3} \tag{16.3}\]
- Equation eq-matmult2 says that the transpose of a five by three matrix multiplied by the original yields a product matrix \(\mathbf{B}\) of dimensions three by three (a square matrix with three rows and three columns).
16.3 Matrix Powers
- You can multiply a matrix times itself to get matrix powers but only if matrix is a square matrix (has the same number of rows and columns). Thus,
\[ \mathbf{A}^2 = \mathbf{A} \times \mathbf{A} \] \[ \mathbf{A}^3 = \mathbf{A} \times \mathbf{A} \times \mathbf{A} \] \[ \mathbf{A}^4 = \mathbf{A} \times \mathbf{A} \times \mathbf{A} \times \mathbf{A} \] \[ \mathbf{A}^n = \mathbf{A} \times \mathbf{A} \times \mathbf{A} \times \mathbf{A} \ldots \]
For all square matrices \(\mathbf{A}\) of any dimension. Since matrices used to represent social networks, like the adjacency matrix are square matrices, that means that you can always find the powers of an adjacency matrix.
When you multiply a square matrix times another square matrix of the same dimensions, the resulting matrix is of the same dimensions as the original two matrices. Thus,
\[ \mathbf{A}_{5 \times 5} \times \mathbf{A}_{5 \times 5} = \mathbf{A}^2_{5 \times 5} \]
Table 16.1: A matrix and its transpose
| 3 | 4 | 5 |
| 7 | 9 | 3 |
| 4 | 6 | 2 |
| 5 | 3 | 4 |
| 2 | 5 | 4 |
| 3 | 7 | 4 | 5 | 2 |
| 4 | 9 | 6 | 3 | 5 |
| 5 | 3 | 2 | 4 | 4 |
16.4 Matrix Multiplication Examples
Now let’s see some examples of how matrix multiplication works. Table tbl-prod shows the result of multiplying the matrix shown in Table tbl-trans-1 times its transpose, shown in Table tbl-trans-2.
| 50 | 72 | 46 | 47 | 46 |
| 72 | 139 | 88 | 74 | 71 |
| 46 | 88 | 56 | 46 | 46 |
| 47 | 74 | 46 | 50 | 41 |
| 46 | 71 | 46 | 41 | 45 |
Now where the heck did these numbers come from? Don’t panic. We’ll break it down. First, let’s begin with the number \(50\) in cell corresponding to the first row and first column of Table tbl-prod. To find out where this number came from, let’s look at the first row of Table tbl-trans-1, composed of the vector \(\{3, 4, 5\}\), and the first-column of Table tbl-trans-2, composed of the same vector \(\{3, 4, 5\}\). Now, the number \(50\) comes from the fact that we multiply each of the corresponding entries of the two vectors, and then add them up, as follows:
\[ (3 \times 3) + (4 \times 4) + (5 \times 5) = 9 + 16 + 25 = 50 \]
Neat! Now let’s see where the number \(74\) in the fourth row and second column of Table tbl-prod came from. For that we look at the entries in the fourth row of Table tbl-trans-1, composed of the vector \(\{5, 3, 4\}\) and the second column of Table tbl-trans-2 composed of the vector \(\{7, 9, 3\}\). Like before, we take the first number of the first vector and multiply it by the first number of the second vector, the second number of the first vector and multiply it by the second number of the second vector, and the third number of the first vector and multiply it by the third number of the second vector and add up the results:
\[ (5 \times 7) + (3 \times 9) + (4 \times 3) = 35 + 27 + 12 = 74 \] And we keep on going like this to get each of the twenty five numbers in Table tbl-prod (there are twenty five numbers because Table tbl-prod has five rows and five columns and five times five equal twenty five). In general terms, the number in the \(i^{th}\) row and \(j^{th}\) column of Table tbl-prod is equal to the sum of the products of the numbers in the \(i^{th}\) row of the Table tbl-trans-1 and the \(j^{th}\) column of Table tbl-trans-2.
Note that the resulting product matrix shown in Table tbl-prod is symmetric. The same numbers that appear in the upper-triangle also appear in the lower triangle, such that \(b_{ij} = b_{ji}\). So once you know the numbers in one of the triangles, you can fill up the numbers in the other one without having to do all the multiplying and adding up!
Now, let’s multiply the matrix in Table tbl-trans-2 times the matrix in Table tbl-trans-1. As the rules of matrix multiplication show, this will result in a matrix of dimensions \(3 \times 3\) because Table tbl-trans-2 has three rows and Table tbl-trans-1 has three columns. This is shown in Table tbl-prod2.
| 103 | 124 | 72 |
| 124 | 167 | 91 |
| 72 | 91 | 70 |
Like before, if we want to figure out where the number \(72\) in the third row and first column of Table tbl-prod2 came from, we go to the first row of Table tbl-trans-2 composed of the vector \(\{5, 3, 2, 4, 4\}\) and the first column of Table tbl-trans-1, composed of the vector \(\{3, 7, 4, 5, 2\}\) match up each number in terms of order, multiplying them and add up the result:
\[ (5 \times 3) + (3 \times 7) + (2 \times 4) + (4 \times 5) + (4 \times 2) = \]
\[ 15 + 21 + 8 + 20 + 8 = 72 \]
Table 16.4: Powers of a matrix.
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 2 | 0 |
| 1 | 1 | 2 | 1 |
| 2 | 1 | 2 | 2 |
| 1 | 1 | 1 | 2 |
| 2 | 2 | 3 | 3 |
| 3 | 2 | 4 | 3 |
| 4 | 3 | 5 | 4 |
| 3 | 2 | 4 | 2 |
Matrix powers work the same as regular matrix multiplication, except that we are working on just one matrix not two. So for instance, the number \(2\) in the first row and third column of Table tbl-pow-2 comes from the numbers in the first row of Table tbl-pow-1 (\(\{0, 1, 0, 1\}\)) and the numbers in the third column of Table tbl-pow-1 (\(\{0, 1, 1, 1\}\)). We line them up, multiplying them, and add them:
\[ (0 \times 1) + (1 \times 1) + (0 \times 1) + (1 \times 1) = 0 + 1 + 0 + 1 = 2 \] Since we are working with a binary matrix, the product of each of the cell entries will be either a zero (when at least one of the entries is zero) or a one (when both entries are one).
To get the cubed entries in Table tbl-pow-3, we just take Table tbl-pow-2 as the first matrix and Table tbl-pow-1 as the second matrix, and do matrix multiplication magic. Thus, to get the number \(4\) in the third row and fourth column of Table tbl-pow-3, we take the numbers in the third row of Table tbl-pow-2 \(\{2, 1, 2, 2\}\) and the numbers in the fourth column of Table tbl-pow-1 \(\{1, 0, 1, 0\}\), line them up, multiply them, and add them:
\[ (2 \times 1) + (1 \times 0) + (2 \times 1) + (1 \times 0) = 2 + 0 + 2 + 0 = 4 \]
Pretty easy!
16.5 Matrix Multiplication of Vectors
Recall from sec-degset that a vector is a sequence of numbers of a given length. So for instance, the vector \(\mathbf{a} = \{2, 4, 7, 2, 4\}\) is a vector of length five.
Well, and here comes the big reveal, it turns out that another way to think of a vector, is as a special case of matrix. That is, a matrix with one row, and as many columns as the length of the vector! This is a called a row vector. So the row vector \(\mathbf{a}\) vector can be thought of as a matrix of dimensions \(1 \times 5\) (one row and five columns) or \(\mathbf{a}_{1 \times 5}\).
In matrix form:
| 2 | 4 | 7 | 2 | 4 |
Since vectors are matrices, we can perform the same type of matrix operations on them as we did with matrices. For instance, we can compute the transpose of a vector. In the case of \(\mathbf{a}\), the transpose \(\mathbf{a}^T\) is:
| 2 |
| 4 |
| 7 |
| 2 |
| 4 |
The transpose of a row vector is called (you may have guessed) a column vector. The column vector in Table tbl-transvec is a matrix with five rows and one column.
This also means that the same rules of matrix multiplication apply. For instance, we can always multiply a row vector times a column vector, because it is the equivalent of multiplying a matrix times its transpose, and we have already seen in sec-mattrans, that this can always be done:
\[ \mathbf{a}_{1 \times 5} \times \mathbf{a}^T_{5 \times 1} = b_{1 \times 1} \tag{16.4}\]
Equation eq-vecmult1 says that the product of the \(1 \times 5\) row vector \(\mathbf{a}\) times a \(5 \times 1\) column vector \(\mathbf{a}^T\) is a \(1 \times 1\) “matrix” otherwise known as a scalar (that is, a regular old number). We’ve already seen examples of this, because in regular matrix multiplication, each cell of the product matrix is a scalar obtained from multiplying the corresponding terms taken from a row of the first matrix (which is a row vector) times those of the column of the second matrix (which is a column vector).
So in this case this would be:
\[ (2 \times 2) + (4 \times 4) + (7 \times 7) + (2 \times 2) + (4 \times 4) = \]
\[ 4 + 16 + 49 + 4 + 16 = 89 \]
The first rule of vector matrix multiplication is that you can always multiply a row vector times a column vector (even when their entries are not the same) as long as they are the same length (e.g., the number of columns of the row vector equal the number of rows of the column vector).
The second rule of vector matrix multiplication is that when you multiply a row vector times another a column vector the result is always scalar (a single number).
Now notice that if we change the order, and multiply the transpose of a vector times the original? This should be allowed because it conforms to the rules that we have already discussed:
\[ \mathbf{a}^T_{5 \times 1} \times \mathbf{a}_{1 \times 5} = B_{5 \times 5} \tag{16.5}\]
This matrix multiplication is defined because the inner dimensions of the two matrices (the column and row vectors) are the same (one). But note that, according to the rules of matrix multiplication, when you multiply the transpose of a vector times the original, the result is a square matrix, with dimensions \(n \times n\) where \(n\) is the length of the original row vector (the number of columns). In our example if the original vector is \(\mathbf{a} = \{2, 4, 7, 2, 4\}\), then \(\mathbf{a}^T \times \mathbf{a}\) is equal to the matrix shown in Table tbl-transvec2.
| 4 | 8 | 14 | 4 | 8 |
| 8 | 16 | 28 | 8 | 16 |
| 14 | 28 | 49 | 14 | 28 |
| 4 | 8 | 14 | 4 | 8 |
| 8 | 16 | 28 | 8 | 16 |
- So, the third and final rule of vector matrix multiplication is that when you multiply a column vector times a row vector of the same length, the result is a square matrix of row and column dimensions equal to the length of the original vectors.
16.6 Multiplying a Vector Times A Matrix (and Vice Versa)
Since vectors are just matrices, it means that we can always multiply a vector times a matrix (and a matrix times a vector), as long as we follow the matrix multiplication rules laid out in sec-multrules.
16.6.1 Row Vector Times Matrix
For instance, take the row vector \(\mathbf{b} = \{4, 9, 3, 5\}\) and the binary matrix \(\mathbf{A}\) shown in Table tbl-pow-1. Because the row vector \(\mathbf{b}\) is of dimensions \(1 \times 4\) and matrix \(\mathbf{A}\) is of dimensions \(4 \times 4\), it is possible to multiply the vector times the matrix as follows:
\[ \mathbf{b}_{1 \times 4} \times \mathbf{A}_{4 \times 4} = \mathbf{c}_{1 \times 4} \tag{16.6}\]
Equation eq-matvec says that the product of a \(1 \times 4\) row vector times a \(4 \times 4\) square matrix is another row vector of dimensions equal to the original row vector. The result for this example is shown in Table tbl-vecmat1.
Table 16.8: Row vector resulting from multiplying a row vector times a square matrix
| 4 | 9 | 3 | 5 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 |
| 8 | 13 | 17 | 7 |
Of course, it is also possible to multiply a row vector times a rectangular matrix (where the number of rows is not necessarily equal to the number of columns), as long as the number of rows of the rectangular matrix equals the length of the original row vector. For instance, take a row vector \(\mathbf{a}_{1 \times 5}\) shown in Table tbl-vecmat2-1 and a matrix \(B_{5 \times 3}\). Its product would be given by:
\[ \mathbf{a}_{1 \times 5} \times \mathbf{B}_{5 \times 3} = \mathbf{c}_{1 \times 3} \tag{16.7}\]
Equation eq-matvec2 says that the product of a row vector of dimensions \(1 \times 5\) and a matrix of dimensions \(5 \times 3\) is another row vector \(\mathbf{c}\) of dimensions (\(1 \times 3\)). A numerical example corresponding to this situation is shown in Table tbl-vecmat2.
Table 16.9: Row vector resulting from multiplying row vector times a rectangular matrix
| 2 | 4 | 7 | 2 | 4 |
| 8 | 13 | 14 |
| 10 | 12 | 3 |
| 5 | 19 | 15 |
| 15 | 6 | 11 |
| 18 | 10 | 16 |
| 193 | 259 | 231 |
To get the “179” entry in row one and column one of Table tbl-vecmat2, we take the entries of the row vector shown in Table tbl-vecmat2-1 and multiply them by the corresponding entries in the first column of the matrix shown in Table tbl-vecmat2-2 and add up the results:
\[ (2 \times 2) + (4 \times 5) + (7 \times 11) + (2 \times 15) + (4 \times 12) = \]
\[ 4 + 20 + 77 + 30 + 48 = 179 \]
And so on for the other two entries in Table tbl-vecmat2-3. So the main rule of multiplying a row vector times a matrix with number of rows equal to the length of the row vector is that the result will always be another row vector of length equal to the number of columns of the matrix.
16.6.2 Matrix Times Column Vector
In the same way, we can always multiply a matrix times a column vector, as long as the the number of columns of the matrix is equal to the length of the column vector. For instance, take the binary square matrix \(A_{5 \times 5}\) shown in Table tbl-vecmat3-1 and the column vector \(\mathbf{b}_{5 \times 1}\) shown in Table tbl-vecmat2-2. Their product \(\mathbf{c}\) would be given by:
\[ \mathbf{A}_{5 \times 5} \times \mathbf{b}_{5 \times 1} = \mathbf{c}_{5 \times 1} \tag{16.8}\]
Equation eq-matvec3 says that the product of a matrix of dimensions \(5 \times 5\) and a column vector of dimensions \(5 \times 1\) is another column vector \(\mathbf{c}\) of dimensions equal to the original column vector (\(5 \times 1\)). A numerical example of this situation is shown in Table tbl-vecmat3.
Table 16.10: Column vector resulting from multiplying a square matrix times a column vector
| 9 | 19 | 19 | 17 | 16 |
| 15 | 3 | 9 | 1 | 7 |
| 8 | 16 | 11 | 20 | 1 |
| 2 | 10 | 1 | 19 | 13 |
| 4 | 6 | 17 | 7 | 6 |
| 2 |
| 4 |
| 7 |
| 2 |
| 4 |
| 325 |
| 135 |
| 201 |
| 141 |
| 189 |
16.6.3 Matrix Times the All Ones Vector
In matrix multiplication, there is a special row and column vector called the all ones vector. As you may have guessed this is a vector of all ones, of some length \(n\). For instance and all ones row vector of length five is \(\mathbf{1}_{1 \times 5} = \{1, 1, 1, 1, 1\}\) (the symbol for the all ones vector is a boldface “1”). We can also get the transpose of this all ones row vector to get the all ones column vector \(\mathbf{1}^T\).
Why do we care about vectors full of ones? Well, it turns out that the all one row and column vectors have a neat property when we multiplied by matrices. We already know, from the rules of vector matrix multiplication reviewed earlier, that the product of a row vector times a square matrix is always a row vector of the same length as the original, and the product of a square matrix times a column vector is always a column vector of the same length as the original.
Let’s say we a matrix \(\mathbf{A}\) of dimensions \(5 \times 5\), and we multiplied the all ones row vector of length five times this matrix, which would result in the row vector \(\mathbf{b}\). This would be given by the formula:
\[ \mathbf{1}_{1 \times 5} \times \mathbf{A}_{5 \times 5} = b_{1 \times 5} \tag{16.9}\]
A numerical example of the situation depicted in Equation eq-allones1 is shown in Table tbl-allones1.
Table 16.11: Row vector resulting from multiplying the all ones row vector times a square binary matrix
| 1 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 3 | 5 | 2 | 3 | 1 |
If you look at the resulting row vector in Table tbl-allones1-3, we can see that the result of multiplying the all ones row vector times a matrix is a vector that contains the column sums of the matrix entries! So the “2” in position 1 of Table tbl-allones1-3 comes from adding up the numbers in the first column of the matrix, the “1” in position 2 of Table tbl-allones1-3 comes from adding the numbers in the second column and so forth.
Table 16.12: Row vector resulting from multiplying the all ones row vector times a square binary matrix
| 0 | 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 |
| 1 |
| 1 |
| 1 |
| 1 |
| 2 |
| 3 |
| 4 |
| 2 |
| 3 |
In the same way, if we multiply the same matrix times the all ones column vector, we get the results shown in Table tbl-allones2. We can see that the result of multiplying a matrix times the all ones column vector, is another column vector contains the row sums of the original matrix! So, the “2” in the first position of the column vector comes from adding the numbers in the first row of the matrix, the “3” comes from adding the numbers in the second row, and so forth.
16.7 The Identity Matrix
The last “interesting” matrix we will cover is called the identity matrix. This is a square matrix, usually written using the symbol \(\mathbf{I}\) of dimensions \(n \times n\). This matrix will have “1” in every diagonal cell, and “0” in every off-diagonal cell. For instance, an identity matrix of dimensions \(5 \times 5\) is shown Table tbl-idmat.
| 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 0 | 1 |
The interesting thing about this matrix is that when you multiply it times another square matrix of the same dimensions, the result is always the original matrix! So it plays the role that the number “1” plays in regular number multiplication, in matrix algebra. This means that, for any square matrix \(\mathbf{A}\):
\[ \mathbf{A} \times \mathbf{I} = \mathbf{A} \tag{16.10}\]
And also,
\[ \mathbf{I} \times \mathbf{A} = \mathbf{A} \tag{16.11}\]
Neat!