Calculate basis from kernel and image of $\varphi$ given by:
$\varphi:\mathbb{R}^{2\times2} \rightarrow \mathbb{R}^{2\times2}, X\mapsto X\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}-\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}X^{tr}=\begin{pmatrix}
0 & -x_{11}-x_{21}+x_{12}+x_{22} \\
-x_{22}-x_{12}+x_{21}+x_{11} & 0
\end{pmatrix}=:B \qquad \qquad (1)$
Basis from kernel
My thoughts: Simplify $\varphi$ as far as I can and then solve $\varphi\cdot =0$ My simplification is the second line marked with $(1)$. However if I try to solve $Bz=0$ then for $x_{21}+x_{11}=x_{22}+x_{12}$ I get $z_1=z_2=0$, and for $x_{21}+x_{11}\neq x_{22}+x_{12}$ $z_1=z_1, z_2=z_2$.
The Problem: How does this help me? I have two solutions, so what is the basis?
Basis from image
I found in my notes that I could write $\varphi(e_1),…,\varphi(e_4)$ in a matrix (with $e_i$ trivial basis of $\mathbb{R}^{2\times 2}$), reduce it Gauß and then the rows which are not zero rows, create the basis.
The problem: $\varphi\in \mathbb{R}^{2\times 2}$ and you can't write it in a matrix like this.
Question
- Are the methods appropriate or what else should I do?
- If they are ok, how can I solve the problems mentioned above?
Best Answer
There are two ways I see to attack the problem. First, there's the naive method, involving directly applying the definitions of kernel and image. Second, there's the matrix method, involving turning the linear operator into a matrix, then analysing the matrix using row reduction. Both methods are viable. You are using the naive method in the first part, and referring to the matrix method in the second part, but seem unable to figure out how to do it.
To continue the naive method for part 1, you have a system of two linear equations \begin{align*} -x_{11} - x_{21} + x_{12}+x_{22} &= 0 \\ x_{11} + x_{21} - x_{12} - x_{22} &= 0. \end{align*} Note that both equations are just negatives of each other. We can therefore let one of the variables depend on the other three, which can be free. So, in particular, if we let $t = x_{21}$, $s = x_{12}$, and $r = x_{22}$, then $$x_{11} = s + r - t,$$ and hence \begin{align*} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} &= \begin{pmatrix} s + r - t & s \\ t & r \end{pmatrix} \\ &= s\begin{pmatrix} 1 & 1 \\ 0 & 0\end{pmatrix} + t\begin{pmatrix} -1 & 0 \\ 1 & 0\end{pmatrix} + r\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}. \end{align*} This gives you three vectors whose span contain the kernel of $\varphi$. If you verify that all three vectors are also in the kernel of $\varphi$, and that they are linearly independent, then they form a basis for the kernel.
The naive method can also be used for the second part. By your own calculation, an arbitrary transformed matrix takes the form, \begin{align*} \varphi(X) &= \begin{pmatrix} 0 & -x_{11} - x_{21} + x_{12}+x_{22} \\ x_{11} + x_{21} - x_{12} - x_{22} & 0 \end{pmatrix} \\ &= x_{11}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + x_{21}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} + x_{12} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} + x_{22}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. \end{align*} We now have a list of $4$ vectors that span the image of $\varphi$, but you might notice that they're very dependent. Reduce it down to a linearly independent list, and you'll have a basis.
If you want to do the matrix route properly, you'll need to find the matrix for $\varphi$ with respect to a basis (or two, but why complicate things?). It doesn't matter which basis you choose, so we'll pick an easy one: $$\mathcal{B} = \left(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right).$$ We now compute the matrix of $\phi$ with respect to this basis. Namely, we must transform each basis vector individually in turn, and write the resulting matrices as coordinate column vectors with respect to the given basis. These column vectors will form the columns of our matrix for $\varphi$.
Let $(e_{11}, e_{21}, e_{12}, e_{22}) = \mathcal{B})$. Then \begin{align*} \varphi(e_{11}) &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = 0e_{11} + (-1)e_{21} + 1e_{12} + 0e_{22} \\ \varphi(e_{21}) &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = 0e_{11} + (-1)e_{21} + 1e_{12} + 0e_{22} \\ \varphi(e_{11}) &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = 0e_{11} + 1e_{12} + (-1)e_{12} + 0e_{22} \\ \varphi(e_{21}) &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = 0e_{11} + 1e_{22} + (-1)e_{12} + 0e_{22}. \end{align*} We take the coordinates and write them as column vectors, and use them to form our matrix. Our matrix for $\phi$ becomes $$A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -1 & -1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ The nullspace of this matrix will be the set of coordinate column vectors, with respect to $\mathcal{B}$, corresponding to vectors in the kernel of $\phi$. For example, the coordinate column vector $$\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$ belongs to the nullspace of $A$, because the matrix $$0e_{11} + 1e_{21} + 1e_{12} + 0e_{22} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ belongs to the kernel of $\varphi$.
The columnspace of $A$ corresponds to the image of $\varphi$ in much the same way; it consists of coordinate column vectors with respect to $\mathcal{B}$, corresponding to vectors in the image of $\varphi$. If you reduce the columns down to a basis (there's a method involving row reduction), then they will correspond to a basis for the image of $\varphi$.