Image
Kernal
Linear Independence
Basis
Dimension
Rank-Nullity Theorem
Coordinates How to Examples Exercise
How to:
The image of a function consists of all the values the function takes in its codomain. In the case where our function is a Linear Transformation, A*x, we can find our image by looking at the column vectors in A.
Let's say matrix A is composed of column vectors as such: A = [v1 v2 ... vN]. So, when we multiply this by some vector, x = [x1; x2; ...; xN], the result is: x1*v1 + x2*v2 + ... + xN*vN
So, our image is the "span" of vectors v1, v2, ... , vN. Since x1, x2, ... , xN can take ANY scalar value, the image is the set of all combinations of c1*v1 + c2*v2 + ... + cN*vN, where c1 ... cN are arbitrary scalars.
There is a possibility that our image can be simplified. If any of all columns are themselves combinations of scalar multiples of other columns, then it is repetitive to list them. This will be covered in more detail in the Linear Independence section.
Examples:
Let's find the image of our transformation, A*x, where A = [1 3; 4 12].
Our matrix can be thought of as two column vectors, A(:, 1) = [1; 4] and A(:, 2) = [3; 12]. So, for any vector, x = [x1; x2], our result of A*x will be: x1*[1; 4] + x2*[3; 12]. The set of all combinations of values of x1 and x2 will give us our image.
BUT, if we look closely, we see that [3; 12] = 3*[1; 4], that is, v2 is a scalar multiple of v1. So we can change our result to: A*x = x1*[1; 4] + x2*[3; 12] = x1*[1; 4] + 3*x2*[1; 4] = (x1 + 3*x2)*[1; 4]. Since the sum x1 + 3*x2 can still result in any scalar multiple, our image in this case is the line which is an extension of [1; 4] in R^2.
Exercise: