Matrix representation induced by quotient space

Question

someone can help me with this question, I know how to solve ker(A) but I don't know how to develop matrix representation. Thanks!!!!!

Answer 1

The quotient space of $\mathbb{R}^3 / \text{Ker}(A)$ would be elements of the form $\text{Ker}(A)+x$. Thus, you need to find a basis for such elements. A short calculation reveals $\text{rref}(A) = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]$. It follows that $x \in \text{Ker}(A)$ has $x_1=-x_3$ and $x_2=x_3$. Thus, $\text{Ker}(A) = \text{span}\{ (-1,1,1) \}$. Clearly $(1,0,0)$ and $(0,1,0)$ are not in the kernel hence an arbitrary element in the quotient space is parametrized by the first and second coordinates of $\mathbb{R}^3$. Notice, $$ Ae_1 = (1,2,0) = 1e_1+2e_2, \ \ Ae_2 = (-1,1,-3) = -1e_1+e_2-3e_3 $$ we need to convert $e_3$ to a term from the kernel and $e_1,e_2$. Notice, $$ (-1,1,1) \in \text{Ker}(A) \ \ \Rightarrow \ \ (-3,3,3) = (-3,3,0)+(0,0,3) \in \text{Ker}(A) $$ thus, in the quotient space where things in the kernel are set to zero, $(-3,3,0) = (0,0,-3)$ which gives (in the quotient): $$ Ae_2 = (-1,1,-3) = -1e_1+1e_2-3e_3 = -1e_1+1e_2+ (-3,3,0) = -4e_1+4e_2 $$ Thus, in total, if I use $\{ \text{Ker}(A)+e_1,\text{Ker}(A)+e_2 \}$ as a basis for the quotient space then the mapping induced from $A$ in the natural fashion is $2 \times 2$ and, given the calculations above, the matrix of the map represented by $A$ reduced to the kernel is: $$ \left[ \begin{array}{cc} 1 & -4 \\ 2 & 4 \end{array}\right].$$ Now, at this point some soul searching is needed. Why on earth did I choose $e_1,e_2$ as the basis representatives for the quotient? Answer, no particular reason. In principle, you could choose the basis so the matrix derived above turns out to be the identity. In fact, this is always possible provided $A \neq 0$.