Rank of a linear operator

Question

How to find the Rank of Operator A, acting in $\mathbb R^{2\times 2} \ $ by the rule: $$AX=\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}X$$

I'm told: The rank of a linear operator is equal to the rank of its matrix in its arbitrary pair of basis.

Answer 1

The rank of the operator $T: \mathbb{R}^{2\times 2} \to \mathbb{R}^{2\times 2}$, defined by $T(X) = AX$, agrees with the rank of its matrix representation with respect to any pair of (ordered) bases for the domain and codomain.

For simplicity let's use the same basis for both:

$$ \mathscr{B} = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\} $$

With respect to this (ordered) basis, the representation of the linear operator:

$$ A \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ 0 & 0 \end{pmatrix} $$

is $M = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$. The rank of $M$ is two, as is the rank of linear operator $T$.

Answer 2

I think you have not cleared the concept of Rank of a linear transformation, So just to clarify that Here is the theory.

An $n\times m$ matrix $A$ can be used to define a linear transformation $L_A\colon\mathbb{R}^m\to\mathbb{R}^n$ given by $L_A(\mathbf{v}) = A\mathbf{v}$. If we do this, the kernel of $L_A$ equals the nullspace of $A$, and the image of $L_A$ equals the column-space of $A$. In particular, $\mathrm{rank}(A) = \mathrm{rank}(L_A)$, $\mathrm{nullity}(A)=\mathrm{nullity}(L_A)$.

Going the other way, given a linear transformation $T\colon V\to W$, if we pick a basis $\alpha$ for $V$ and a basis $\beta$ for $W$, then we can define a matrix called the *coordinate matrix of $T$ with respect to $\alpha$ and $\beta$, $[T]_{\alpha}^{\beta}$, which is a $\dim(W)\times\dim(V)$ matrix whose columns are the coordinate vectors, relative to $\beta$, of the images under $T$ of the vectors in $\alpha$. The matrix has the property that for every $\mathbf{v}\in V$, $$[T]_{\alpha}^{\beta}[\mathbf{v}]_{\alpha} = [T(\mathbf{v})]_{\beta}$$ where $[\mathbf{v}]_{\alpha}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $\alpha$, and $[T(\mathbf{v})]_{\beta}$ is the coordinate vector of $T(\mathbf{v})$ with respect to $\beta$. If we do this, then the rank of $T$ equals the rank of $[T]_{\alpha}^{\beta}$, and the nullity of $T$ equals the nullity of $[T]_{\alpha}^{\beta}$.

So the notions of rank and nullity for matrices and for linear transformations correspond to one another under the correspondence between matrices and linear transformations.

I hope this must clear all your doubts.