# Rank of Matrix and dimension of the image of the function corresponding with that matrix

If asked to show for some matrix $A \in Mat(m \times n, \mathbb{F})$ and for some linear function $f: V \rightarrow W$ such that $B$ and $B'$ are the basis vectors for $V$ and $W,$ respectively, that $$\operatorname{dim}(\operatorname{Im}f_A) = \operatorname{rank}(A) \textsf{ and } \operatorname{dim}(\operatorname{Im}f) = \operatorname{rank}(M^B_{B'}(F)),$$

is the response simply the statement of definitions? Isn't it essentially the definition of matrix rank that the rank of a matrix is equivalent to the rank of the linear function corresponding to that matrix. Moreover, how are these two statements different in any significant way? They appear to me to be virtually the same statement. Is there a slight nuance that I'm missing?

I DO KNOW THAT

(1) The rank of a matrix is equivalent to the dimension of the vector space spanned by the columns of that matrix, or the number of linearly independent columns of that matrix, or the number of non-zero column vectors in the reduced row-echelon form of that matrix.

(2) The column vectors are essentially the basis vectors for the image of $f_A,$ the function corresponding with the matrix $A$ (that is, the basis vectors for the subspace of $W$ generated by $V$ through $f_A$), as the rows are the expression of the basis vectors $B$ of $V$ with respect to the basis vectors $B'$ of $W.$

• Yes, the rank of a matrix or linear operator is the dimension of its image. Whether this is the definition of "rank" or an easy proof from the definition depends on what definition you are using. Note that the column vectors span the image (in matrix terminology, the "column space"), but need not be independent, so the best you can say in general is that they contain a basis for the image. – Paul Sinclair Apr 10 '18 at 23:20

Suppose $A=[\alpha_1 \ldots \alpha_m]^T=[\beta_1 \ldots \beta_n]\in Mat(m,n)$. We have
$$Ax= \begin{bmatrix} \alpha_1^Tx \\ \cdots \\ \alpha_m^Tx \\ \end{bmatrix} = [\beta_1 \ldots \beta_n] \begin{bmatrix} x_1 \\ \cdots \\ x_n \\ \end{bmatrix}$$
So the rows span the orthogonal complement of null space. And the equivalence implies $dim(V)-rank(A)=dimNull(A)$!