Consider the matrix:
$A= \begin{bmatrix} 1 & 1 & 0&5 \\ 3 & 4 & -1&16 \\ 4 & 8 & 1&14 \\ 4&6&-1&20 \end{bmatrix} $
Let $\mathcal{S}$ be the set of the solutions of the homogeneous system $AX=0$ and $\mathcal{T}$ the set of all the vectors $B \in \mathbb{R^4}$ s.t. $AX=B$ has solution(s).
Give an example of an endomorphism $F: \mathbb{R^4} \to \mathbb{R^4}$ s.t $Ker(F)=\mathcal{S}$ and $Im(F)=\mathcal{T}$. Give an example of another endomorphism $G: \mathbb{R^4} \to \mathbb{R^4}$ s.t. $Im(G)=Im(F)$, but $Ker(G) \neq Ker(F)$.
I would say that $F$ is right represented by $A$ because $Ker(F)=\{X \in\ \mathbb{R^4} : F(X)=0\}=\{X \in \mathbb{R^4}: AX=0\}$ and, in order that $\mathcal{T}$ is the set of alla vectors $B$ s.t. $(A|B)$ has solution, must be $\mathcal{T} \in span\{A^1 A^2A^3\}$, ($A^4$ is linearly dependent), and $Im(F)$ is the span of the linearly independent columns of the matrix which represents the endomorphism.
About $G$, should be $\mathcal{T}$ still in the span of columns, but requiring that $A^4$ is linearly independent.
Am I right? I am sorry if this question is slighty basic level, but I am studying this on my own to be prepared for uni next year.
You have the basic idea right: $A$ is a rank-3 matrix, so you can adjust the kernel without changing the image by changing the last column of $A$. The replacement must be a linear combination of the other columns. If it’s linearly independent of them, you will have increased the rank of the matrix to 4 and grown its image.
It’s not hard to work out that $\ker(A)$ is spanned by $(6,-1,-2,-1)^T$, so a simple choice is to zero out the last column. The kernel of the resulting matrix is obviously the span of $(0,0,0,1)^T$, which is a different subspace of the domain. In fact, almost any linear combination of the first three columns will produce a different kernel: it’s not too difficult to show that if $A^4 = c_1A^1+c_2A^2+c_3A^3$, then the kernel is spanned by $(a,b,c,-1)^T$.