If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# im(T): Image of a transformation

Showing that the image of a subspace under a transformation is also a subspace. Definition of the image of a Transformation. Created by Sal Khan.

## Want to join the conversation?

• What is the difference between a "subspace" and a "subset"? • A subspace is a subset that needs to be closed under addition and multiplication. That means if you take two members of the subspace and add them together, you'll still be in the subspace. And if you multiply a member of the subspace by a scalar, you'll still be in the subspace. If these two conditions aren't met, your set is not a subspace.

All subspaces are subsets, but not all subsets are subspaces. Just think of it as a definition. The intuition behind it is in Sals videos about subspaces.
• Please correct my understanding below about what are the matrices A, x, T(X) called? They are mentioned at .
A is the transformation matrix, x is the matrix being transformed (or the image that is being transformed) and T(X) is the transformed matrix? • Agree with the above, but for more clarity: A would be the transformation matrix of the transformation T, x is a generic vector or a vector that lives in R^n. Here it is indeed equal to the vector being transformed, you can sometimes see it as a free variable vector in the transformation. However, that depends on the definition of your actual domain of transformation, it limits the extent of x, while T(x) is the transformed vector, or the image of x, sometimes called, that lives both in the codomain and range of T.
It is essential to understand the relation of column space of A and the image of T though, just as a sequel.
• Why does the zero vector always have to be in a subspace V? • Gavriel,

Remember the three rules that Sal gave for the definition of a subspace? They were:
1. Contains the 0 vector
2. Closed under scalar multiplication

Well, imagine a vector A that is in your subspace, and is NOT equal to zero. If rule #2 holds, then the 0 vector must be in your subspace, because if the subspace is closed under scalar multiplication that means that vector A multiplied by ANY scalar must also be in the subspace. Well suppose we multiply by the scalar 0? We would get the 0 vector. So for rule #2 to hold, the subspace must include the 0 vector. In other words, rule #1 must hold if rule #2 is to hold.

And honestly, rule #1 also must hold if rule #3 is to hold. After all, if A is a vector in our subspace, and so is -1*A (from rule #2) then the subspace must also include a zero vector because if vector addition holds, then the sum of any two vectors in our subspace must ALSO be in our subspace. Well, if A and -A are both in our subspace, then so must A+ (-A)… which is of course, the zero vector.

In fact, I'm not even sure why Sal lists it as a rule of subspaces that they include the 0 vector because with rule #2 or rule #3 they have to include the zero vector anyways.
• Can someone please explain more about how the image of the transformation (im(T)) is equivalent to the column space (C(A)) of the matrix that transformation can be represented as? • We can fully define a linear transformation by deciding where it sends the basis vectors. Once we've done that, we can express the transformation as a matrix by writing the basis vectors as a row of column vectors, then replacing each by the vector we send it to.
e.g. the transformation that sends <1,0> to <3, 5, 2> and <0, 1> to <1, 8, 3> can be written as
3, 1
5, 8
2, 3

So reading off the column vectors lets us know a few vectors that our transformation definitely hits, since the basis vectors get mapped to them. Then linearity lets us map to every linear combination of these column vectors, i.e. the column space.
• Is Im(T) the same as Im(A) if T is the transformation and A the standard matrix? • • At , am I right that zero vector condition is not totaly redundand, and zero vector must be a member of subspace to prevent empty set {} from being a valid subspace? Empty set {} is closed under addition and multiplication, but we have no other vector to multiply it by zero to get zero vector. • It is sort of redundant to explicitly include the zero vector. However, in those textbooks that only list two properties (closure under scalar multiplication and closure under vector addition), you'll notice that they explicitly state that V is a non-empty subset of a certain vector space (R^n, for instance). So yes, you can (if you want) avoid mentioning that the zero vector is a member of V but ONLY if you explicitly mention that V is non-empty (in order to avoid the nullset being a valid subspace, as you indicated in your question).

Personally, I prefer to explicitly state that the zero vector is a member of V because:
1.) I'm more likely to forget to mention that V is a non-empty subset
2.) The absolute EASIEST way to prove that a subset is NOT a subspace is to show that the zero vector is not an element (and explicitly mentioning that the zero vector must be a member of a certain set in order to make it a valid subspace reminds me to check that part first).   