# Find the Standard Matrix of a linear transformation

Define a linear transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$ by: $$T\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) = \begin{bmatrix}x_2\\x_1\end{bmatrix}$$ Find the standard matrix of $T$, called $A$, and find the basis for $\mbox{range}(A)$, $\mbox{null}(A)$, and the $\mbox{rank}(A)$.

I know that the $\mbox{range}(A)$ is all of the pivot columns in $A$, and the $\mbox{null}(A)$ is defined by $Ax=0$, and the rank theorem states that the $\mbox{rank}(A)+\mbox{dim}(\mbox{null}(A))$ is the number of columns in $A$, so I feel confident I can solve the second part of the problem if I could find $A$, but I need help finding $A$.

Also sorry for the poor formatting, I still don't know how to format questions efficiently on here.

• math.stackexchange.com/help/notation – user9464 Nov 21 '16 at 14:36
• Do you know what does a "standard matrix of $T$" mean? – user9464 Nov 21 '16 at 14:38

It seem to me that the matrix is of form \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.

To elaborate on Jack... A linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is uniquely defined by where a set of basis for $\mathbb{R}^n$ is mapped to in $\mathbb{R}^m$.

The standard matrix is where we take the "standard basis" ( $\{e_1, \ldots, e_n \}$, with $e_i = \{0, \ldots ,1, \ldots ,0 \}$ with $1$ at the ith position (row)) to form a matrix $A$ with the images of these basis vectors as columns. So we have a matrix $$(T(e_1), \ldots, T(e_n))$$

In your case, we consider the image of $\{e_1, e_2\} = \{(1,0), (0,1)\}$ under $T$, and our matrix $A$ would be as defined above. I believe you could solve the problem now for finding $A$.

To solve the the second part, reduce to row echelon form / find number of linearly independent vectors. Hope this helps.

I'm assuming that by "standard matrix of $T$" you mean the matrix representing $T$ with respect to the standard basis of $\mathbb{R}^2$, which is

$$B = \{e_1, e_2\},\quad \mbox{where}\quad e_1 = \begin{pmatrix}1 \\ 0 \end{pmatrix},\quad e_2 = \begin{pmatrix}0 \\ 1 \end{pmatrix}.$$

Representing $T$ w.r.t. $B$ amounts to expressing the image under $T$ of each vector of $B$ as a linear combination of the elements of $B$. In particular matrix $A$, the "standard matrix of $T$", contains in position $(i,j)$ the coefficient of $e_i$ in the expression for $T(e_j)$.

Try writing down $T(e_1)$ and $T(e_2)$ and you'll see it's a piece of cake.

As already noticed, in order to find A is enough to calculate $T(e_1)$ and $T(e_2)$. Supposing that $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ we calculate: $\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \Rightarrow \begin{bmatrix} b \\ d \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and analogously $\begin{bmatrix} a \\ c \end{bmatrix}=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Finaly: $$A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$