So from my understanding of Rank, that its the row space of of a Matrix, count how many rows there are, that are not the same if you multiply them by a constant.

My understanding that a Kernel means that the function is equal to $~0~$, For example $~T(x,y) = (x+2,y)~$, the Kernel would be $~(-2,0)~$, since that gets the $~0~$ vector.

Range is all the columns that when the Matrix is reduced to row echelon form, they got a leading $~1~$.

Nullity is when I multiply a vector or matrix and get $~0~$ as an answer

So if I'm looking for the Rank of the Kernel of $~T~$ that is in $~\mathbb R^4~$, that makes no sense since the Kernel of $~T~$ is a vector, not a matrix, same with the range, unless its only the first element of the vector. Then the nullity of rank$(T) * 0 = 0~$, so that would be the nullity?

I'm really confused. I tried googling this and all I see is weird Greek letters that say this is an element of that, and its all written in a way that for me to be able to read it, I would have to understand the concepts in retrospect.

  • $\begingroup$ rank is not necessarily how many rows there are; it is the dimension of the row (or column) space; kernel is the subspace mapped to the $0$ vector; If $T(x,y)=(x+2,y)$ were a linear transformation then we'd have $T(2x,2y)=2T(x,y)$ for all $(x,y)$ $\endgroup$ – J. W. Tanner Jul 22 '19 at 22:07
  • $\begingroup$ @Aron Mlodkowski The "nullity" and the "rank" are nonnegative integers. "Kernel", "row space" and "column space" are vector spaces. $\endgroup$ – Dzoooks Jul 22 '19 at 22:12

Let's go back to definitions. Range is a set-theoretic concept. Given a set map $f:X \to Y$, we denote the image or range as the set of values in $Y$, i.e. $\{y\in Y|y=f(x) \text{ for some }x\in X\}$.

Your other concepts live in the world of linear algebra. Let me define things abstractly, rather than by matrices. Given a linear transformation $T:V\to W$, we define the nullspace to the set of vectors $v\in V$ such that $T(v)=0$, i.e. $T^-(0)$. It turns out both the range and nullspace of a transformation $T$ happen to be subspaces in their appropriate domains. Since dimension is a well defined concept (over fields), we can ask for the dimensions of these two subspaces. The dimension of the range is defined to be the rank, while the dimension of the nullspace is defined to be the nullity.

How does all this relate to matrices? Recall the data of a matrix of a transformation $T$ is that the columns of the matrix denote where a basis of $V$ lands after applying $T$. Extending the transformation linearly (i.e. You know what $T(e_1),...T(e_n)$ is, so $T(a_1e_1+...+a_ne_n):=a_1T(e_1)+...+a_nT(e_n)$), uniquely(linear independence of e_i!) extends (e_i spans V!) $T$ to the entirety of $V$.

Knowing this, we can start translating all our definitions from the abstract into statements about matrices. For example, we see the range of a matrix is the Span of the columns. The rank of a matrix would then the the number of linearly independent columns. For example, the 2x2 matrix consisting of all 1's will have rank 1 despite being a transformation two 2-dimensional spaces.

How would we define nullspace of a matrix? Denoting the matrix by $M$, it would be the subspace of vectors $v\in V$ such that $Mv=0$. Concretely, this means for you to solve a system of equations, and is what linear algebra was built to do. Now to define nullity of a matrix, we can use the rank-nullity theorem which tells us $\dim(V)=rk(T)+nul(T)$, so we can define nullity of the matrix as $\dim(V)-rk(T)$.

Some conceptual mistakes I saw in your post: you're confusing nullity with nullspace. The former is a natural number, while the latter is a subspace of V, NOT in general a vector (unless the kernel is just 0). The phrase the rank of the kernel makes no sense; it only makes sense to talk about the rank of a transformation. Similar, the nullity of the rank also makes no sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.