Rank-Nullity Theorem: Struggling to Reconcile ${\rm Rank}(A) +{\rm Nul}(A) = n$ and $\dim(V) = \dim({\rm Im}(f)) +\dim({\rm ker}(f))$.

Question

I am struggling to reconcile the two forms of the rank-nullity theorem:

1. $\operatorname{Rank}(\mathbf{A}) + \operatorname{Null}(\mathbf{A}) = n$,
2. $\dim(\mathbf{V}) = \dim(\operatorname{Im}(f)) + \dim(\operatorname{Ker}(f))$

In attempting to reconcile these, I have collected the following facts:

• If we have $\mathbf{A}\mathbf{x} = \mathbf{b}$, then $\mathbf{A}\mathbf{x} = f(\mathbf{x})$ when thinking about it from the perspective of linear transformations. This means that the image of $f$ is $\mathbf{b}$, so we have $\operatorname{Im}(f) = \mathbf{b}$.
• The rank of a matrix is equal to the dimension of its column space. Therefore, the rank of a matrix is equal to the number of leading ones in the matrix in reduced row echelon form. This is the $\operatorname{Rank}(A)$ part.
• Dimension represents the number of elements in a basis.
• Therefore, we have that $\dim(\operatorname{Im}(f)) = \dim(\mathbf{b})$.

I do not see how $\operatorname{Rank}(\mathbf{A}) = \dim(\operatorname{Im}(f))$. In fact, I not even see how a statement like $\dim(\operatorname{Im}(f))$ even makes any sense? $\operatorname{Rank}(\mathbf{A})$ makes sense, since we can calculate this by getting $\mathbf{A}$ into RREF, but how does $\dim(\operatorname{Im}(f)) = \dim(\mathbf{b})$ make any sense? I do not see the connection here.

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

The image of $f$ isn't $b$. You don't seem to know the definition of the image of a function.

The image is the set of all possible outputs of the function $f$ as $x$ ranges over all vectors in the domain.

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Edit, to address the question in the comments below my answer.

The image of $f$ is the span of the columns of $A$. The rank of $A$ is the number of linearly independent columns of $A$. Hence the rank is the size of a basis of the image, namely the basis consisting of the linearly independent columns of $A$. Since the rank is equal to the cardinality of a basis of the image, the rank is equal to the dimension of the image.

Answer 2

For the part you do not understand: since there is an isomorphism $$\overline{\varphi} : V/\text{Ker}(\varphi) \stackrel{\sim}{=} \varphi(V)$$ we have: \begin{equation} \text{dim}V/\text{Ker}(\varphi) = \text{dim}\varphi(V) \end{equation} By consequence of first isomorphism theorem, we have \begin{equation} \text{dim}V = \text{dim}\text{Ker}(\varphi) + \text{dim}V/\text{Ker}(\varphi) \end{equation} So it follows that, $$\dim V = \dim Ker(\varphi)+\dim \varphi(V)$$

And the above $\varphi$ is a linear transformation and of course $\varphi(V) = \text{Im}(V)$

Well, i am not sure if you have learnt the first isomorphism theorem or not, but this is the proper understanding of how the so called rank nullity is derived. So first things first, remember this theorem.

The next is easy to understand, we have well known mapping which is $$T_A: \mathbb{R}^n \to \mathbb{R}^n$$ which maps $$x \mapsto Ax$$ in which $x$ is a column vector in the $\mathbb{R}^n$ space and $A$ is THE matrix that is in question. Now : i hope you can easily verify the fact that in this linear transformation, we have dimension of kernel $T$ is merely the dimension of the nullspace of $A$ and the dimension of the image of $T$ is merely the dimension of the column space of $A$. And hence we see that indeed $$\text{rank}A = \dim \text{Im}T_A$$

My answer is still slightly vague and does not provide any intuition, but to piece it all together requires you to understand the fundamentals and of course the first isomorphism theorem.