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Find the matrix (in standard base) of the linear transformation $T$: $R^{4}\rightarrow R^{4}$, given: $$Ker(T)=lin \{ |2,1,1,2|^{T}, |1,2,1,1|^{T}\} \\ Im(T)=lin \{ |1,0,1,0|^{T}, |0,1,1,1|^{T}\} $$ I'm not exactly sure how to approach this problem. Could anyone explain how to solve this?

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First, the cannonical base of $\mathbb{R}^4$ is $\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$. You can define a linear transformation by defining what it do to the base. But it will be hard using directly the standard base, so that we will complete $\{(2,1,1,2),(1,2,1,1)\}$ (this set is l.i. so we can do the completion)(note that this will be useful because we will define $T$ to be zero in $\{(2,1,1,2),(1,2,1,1)\}$ and to take $T$ into the wanted image), we take the $\mathbb{R}^4$ base $\{(1,0,0,0),(2,1,1,2),(1,2,1,1), (0,0,0,1)\}$ (if you have trouble finding this you can ask me). So we define: $$T(1,0,0,0)=(1,0,1,0),$$ $$T(2,1,1,2)=(0,0,0,0),$$ $$T(1,2,1,1)=(0,0,0,0),$$ $$T(0,0,0,1)=(0,1,1,1).$$ Now, we have to compute the value of $T$ at the vectors $(0,1,0,0),(0,0,1,0)$. In fact, we have that: $(0,1,0,0)=(1,2,1,1)-(2,1,1,2)+(1,0,0,0)+(0,0,0,1)$, since $T$ is lineal, we have that: \begin{align} T(0,1,0,0) & = T(1,2,1,1)-T(2,1,1,2)+T(1,0,0,0)+T(0,0,0,1) \\ & = T(1,0,0,0)+T(0,0,0,1) \\ & = (1,0,1,0)+(0,1,1,1) \\ & = (1,1,2,1). \end{align} And $(0,0,1,0)=(1,2,1,1)-(1,0,0,0)-(0,0,0,1)-(0,1,0,0)$, so that: \begin{align} T(0,0,1,0) & = T(1,2,1,1)-T(1,0,0,0)-T(0,0,0,1)-T(0,1,0,0) \\ & = -T(1,0,0,0)-T(0,0,0,1)-T(0,1,0,0)\\ & = -(1,0,1,0)-(0,1,1,1)-(1,1,2,1) \\ & = (-2,-2,-4,-2). \end{align} Summarizing the above, we have obtained: \begin{align} T(1,0,0,0) & = (1,0,1,0) \\ T(0,1,0,0) & = (1,1,2,1) \\ T(0,0,1,0) & = (-2,-2,-4,-2) \\ T(0,0,0,1) & = (0,1,1,1). \end{align} Therefore, we have that $im(T)=\mbox{span}\{T(1,0,0,0),T(0,0,0,1)\}=\mbox{span}\{(1,0,1,0),(0,1,1,1)\}$. And the kernel is $\mbox{span}\{(2,1,1,2),(1,2,1,1)\}$ as required. And, $$\left[T\right]_\beta^\beta=\begin{pmatrix} 1 & 1 & -2 & 0 \\ 0 & 1 & -2 & 1 \\ 1 & 2 & -4 & 1 \\ 0 & 1 & 2 & 1 \end{pmatrix},$$ where $\beta$ is the cannonical (i.e. standard) base of $\mathbb{R}^4$ over the field $\mathbb{R}$. Note: For that matrix we have to fix an orden in the basis that's being used.

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  • $\begingroup$ I have trouble understanding two steps. First of all why are we taking {(1,0,0,0),(2,1,1,2),(1,2,1,1),(0,0,0,1)} as a base and not for example {{0,1,0,0),(2,1,1,2),(1,2,1,1),(0,0,0,1)} (it's l.i. and has 4 elements so it's a base too). Secondly, why T transposes necesarilly (1,0,0,0) into (1,0,1,0) and (0,0,0,1) into (0,1,1,1) and not for example the other way around? $\endgroup$ – oskaritooo Sep 3 '17 at 9:42
  • $\begingroup$ If I understand this correctly there exist infinite number of such transformations (that have ker(T)=... and Im(T)=...). You have defined T in that particular way for the sake of convenience. Choosing {{0,1,0,0),(2,1,1,2),(1,2,1,1),(0,0,0,1)} as the base would also lead us to some solution (altough the final matrix would be different). Am I correct? $\endgroup$ – oskaritooo Sep 3 '17 at 17:49
  • $\begingroup$ Hm, there is not reason to taking mine or yours, it don't care :). The main idea is completing $\{(2,1,1,2),(1,2,1,1)\}$ and then sent those vectors to zero (for having the required kernel), and the others 2 (in the order you want) respective to the generators of he image required. There are different ways of do this, in fact if you take a different order of the basis you'll get a different matrix $\endgroup$ – José Alejandro Aburto Araneda Sep 3 '17 at 18:00
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First solve the simpler version of this: $\ker T = \operatorname{span} \{e_1, e_2\}$ and $\operatorname{im} T = \operatorname{span} \{e_1, e_2\}$. Call the matrix that you get from solving this problem $A$.

Then once you find $A$, find change of basis matrices $P$ and $Q$ with the property that $P$ takes the basis for the kernel to $e_1$ and $e_2$ and $Q$ takes the basis for the image to $e_3$ and $e_4$.

Express the matrix that you want in terms of $A$, $P$, and $Q$.

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