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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..

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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.

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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 $A=\tilde A C$.

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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.

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