# 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..

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.

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 $$A=\tilde A C$$.

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.