# Find linear operator for given kernel and image

FInd linear map $$A: \Bbb{R^3} \rightarrow \Bbb{R^3}$$ for given kernel and image. $$Ker(A)=L(\begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix})\space ; \space Im(A)=L(\begin{pmatrix} 1 \\ 0 \\ 1 \\ \end{pmatrix}) \\$$ I've been reading some explonations about this kind of a problem but I didn't understand anything about expanding kernel base to the dimmension of $$\Bbb{R^3}$$. But, according to this solution example , if I form matrix $$A$$ like $$\begin{bmatrix} 1 & a&b \\ 0 &c&d\\ 1 &e&f \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \\$$ and $$\begin{bmatrix} 1 & a&b \\ 0 &c&d\\ 1 &e&f \\ \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \\$$ I don't get anything here for first equation, and for second I get $$\begin{bmatrix} 1+a+d \\ b+e\\ 1+c+f \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\\$$ In this case I can't get a matrix of linear transformation like they did in linked example above. If someone can help me with this, but with theese concrete vectors, or to correct this way of solving..

An more explicit way to look at j3M's strategy is: Let's decide we want to split the map $$A$$ into two parts, $$A = \tilde A \circ C$$ such that

$$\begin{array}{ccccc} &C&&\tilde A& \\ \hline (1,0,0) &\mapsto& \mathbf e_1 &\mapsto& 0 \\ (1,1,1) &\mapsto& \mathbf e_2 &\mapsto& 0 \\ b_3 &\mapsto& \mathbf e_3 & \mapsto& (1,0,1) \end{array}$$

It should be clear that if we choose $$b_3$$ such that the three vectors on the left form a basis for $$\mathbb R^3$$, then the composite map will have the required kernel and image. (This is the case for any $$b_3$$ you can invent, as long as its second and third components are different. The answer by j3M takes $$b_3=(0,0,1)$$; the one by ffffforall takes $$b_3=(1,0,1)$$. Incidentally, these lead to the same final result, whereas $$b_3=(0,1,0)$$ would yield a different solution).

Now, given the way we have specified $$\tilde A$$ it is easy to write down its matrix directly.

On the other hand, $$C$$ is the inverse map of something that is easy to write down directly, so if we write that matrix down and then invert it, we get $$C$$.

Finally, multiply the matrices you've found together, to get $$A=\tilde A C$$.

I suggest a different approach. First, consider the basis $$b_1 = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, \ \ b_2 = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, \ \ b_3 = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} \ \ .$$ Now let $$C$$ be the transition matrix between this basis, and the standard basis of $$\mathbb{R}^3$$ (i.e. $$b_i$$ are the rows of $$C^{-1}$$).

So we can take $$\tilde A = \begin{pmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ which is the representation of some transformation with the required property, in basis $$\{b_i\}$$. It follows that the matrix $${A= C^{-1} \tilde A C}$$ (Edit: this is wrong,see Henning Makholm's answer) has the desired property in the standard basis.

First observe that the vectors $$b_1=(1,0,0)$$, $$b_2=(1,1,1)$$, $$b_3=(1,0,1)$$ form a basis for $$\mathbb{R} ^3$$.

To define a linear map, it suffices to define it in terms of bases for the source and target. In our case we can take the same basis, namely $$B=(b_1,b_2,b_3)$$. Now let $$f$$ be the linear endofunction defined by $$f(b_3)=b_3$$ and $$f(b_1)=f(b_2)=0$$. This function satisfies the requirements.

If you want to find the matrix representing $$f$$ with respect to the standard basis, you can use change-of-basis matrices to compute it from the matrix representation of $$f$$ with respect to $$B$$.

Note that there are many other linear maps that satisfy these requirements, but $$f$$ is probably the easiest to define.