Linear Algebra with Vectors

NumPy is designed to do math (and do it well!). This means that NumPy will treat vectors, matrices and tensors in a way that a mathematician would expect. For example, if you had two vectors:

v1 = np.array([4, 5, 2, 7])
v2 = np.array([2, 1, 3, 3])

And you add them together

v1 + v2

The result will be a ndarray where all the elements have been added together.

array([ 6, 6, 5, 10])

In contrast, if we had two Python Lists

list1 = [4, 5, 2, 7]
list2 = [2, 1, 3, 3]

adding them together would just concatenate the lists.

list1 + list2
# output: [4, 5, 2, 7, 2, 1, 3, 3]

Multiplying the two vectors together also results in an element by element operation:

v1 * v2

Gives us array([ 8, 5, 6, 21]) since 4x2=8, 5x1=5 and so on. And for a Python List, this operation would not work at all.

list1 * list2 # error!

Broadcasting

Now, oftentimes you'll want to do some sort of operation between an array and a single number. In mathematics, this single number is often called a scalar. For example, you might want to multiply every value in your NumPy array by 2:

In order to achieve this result, NumPy will make the shape of the smaller array - our scalar - compatible with the larger array. This is what the documentation refers to when it mentions the term "broadcasting".

The same rules about 'expanding' the smaller ndarray hold true for 2 or more dimensions. We can see this with a 2-Dimensional Array:

array_2d = np.array([[1, 2, 3, 4], 
                     [5, 6, 7, 8]])

The scalar operates on an element by element basis.

The documentation on broadcasting also shows us a few more examples:

Matrix Multiplication

But what if we're not multiplying our ndarray by a single number? What if we multiply it by another vector or a 2-dimensional array? In this case, we follow the rules of linear algebra.

a1 = np.array([[1, 3],
               [0, 1],
               [6, 2],
               [9, 7]])

b1 = np.array([[4, 1, 3],
               [5, 8, 5]])

Challenge: Let's multiply a1 with b1. Looking at the Wikipedia example above, work out the values for c12 and c33 on paper. Then use the .matmul() function or the @ operator to check your work.

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Solution: Matrix multiplication with NumPy

The solution code is pretty straightforward

But how did the calculations arrive at 25 for c12 and 28 for c33? Substituting the number into the formula we get:

c12 = 1*1 + 3*8 = 1 + 24 = 25

c33 = 6*3 + 2*5 = 18 + 10 = 28