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Find a linear mapping $\phi : \mathbb{R}^4 \rightarrow \mathbb{R}^3$

such that,

$Ker$(T) is generated by $(-1,0,0,1)$ and $(1,3,2,0)$, and

$Img(T)$ Is generated by (1,1,1) and (0,-2,1)

In general, how can I find a linear mapping given his kernel and image? Thanks.

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  • $\begingroup$ Extend the basis of the kernel to a basis of $\mathbb{R}^4$, and map the two basis element (not in the kernel) to the basis of the image. $\endgroup$ – Krish Sep 8 '17 at 4:11
  • $\begingroup$ Can you elaborate more. I studied from Grossman Linear Algebra book and is not mentioned the "extend to a basis" concept. $\endgroup$ – Carlitos_30 Sep 8 '17 at 4:31
  • $\begingroup$ See math.stackexchange.com/q/1854485/265466 for an example of how to do this and math.stackexchange.com/q/159135/265466 (from the related questions at right) for a general explanation. $\endgroup$ – amd Sep 8 '17 at 8:12
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1. Extension of Basis:

Let $V$ be a finite dimensional vector space over a field $k$, and let $S:= \{v_1, \cdots , v_r\}$ be any linearly independent set of vectors in $V$.

Claim: The set $S$ can be extended to a basis of $V$.

If $\dim(V)=r$, we are done. If not, then there is a non-zero element $v_{r+1}\in V$ which is not in the subspace (of $V$) spanned by $S$. One can show that $S \cup \{v_{r+1}\}$ is also linearly independent. If $S \cup \{v_{r+1}\}$ spans the whole space $V$, we are done. Otherwise we continue this process. This process will terminate because $V$ is finite dimensional, and thus we get a basis of $V$.

2. Idea/Method:

Let $V, W$ be finite dimensional vector spaces over a field $k$. Suppose we are given a subspace $V' \subset V$ with a basis $\{e_1, \cdots, e_d\}$ and a subspace$W'\subset W$ with a basis $\{f_1, \cdots ,f_r\}$. We want to find a linear transformation $T: V \to W$ with $\ker T= V'$ and $\operatorname{Im}T=W'$. If such $T$ exists, then by Rank-nullity theorem, $\dim V=d+r$. If this equality doesn't hold, then there is no such $T$. So assume $\dim V=d+r$. Now extend the basis of $\ker T$ to a basis of $V$, say, $\{e_1, \cdots ,e_d, e_{d+1}, \cdots ,e_{d+r}\}$. Define a map $$T: V \to W$$ by $$T(e_i)=0, 1 \leq i \leq d$$ $$T(e_{d+i})=f_i, 1 \leq i \leq r.$$ Then $T$ is a linear transformation (why?), and $\ker T=V'$, $\operatorname{Im}T=W'$.

Note: The linear transformation $T$ is uniques only up to conjugate.

3. Example:

Let $e_1=(1,0,0), e_2=(0,1,0) \in \mathbb{R^3}$ and $f_1=(1,0) \in \mathbb{R}^2$. We want a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ such that $\ker T$ has a basis $\{e_1, e_2\}$ and $\operatorname{Im}T$ is generated by $f_1$. Note that, for $e_3=(0,0,1) \in \mathbb{R}^3,\{e_1, e_2, e_3\}$ is a basis of $\mathbb{R}^3$. Now use the above technique to define a linear transformation. We can define $T$ more concretely as follows:

Any $(x,y,z) \in \mathbb{R}^3$ can be written as $xe_1+ye_2+ze_3$. So $$T(x,y,z)=xTe_1+yTe_2+zTe_3=zf_1=(0,z).$$

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  • $\begingroup$ Thanks very much for your time and invaluable help. $\endgroup$ – Carlitos_30 Sep 8 '17 at 10:53
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Assume that $v_1=(1,1,1)^T$ and $v_2=(0,-2,1)^T$. Because $v_1,v_2$ are the basis of the image space. So we can assume that $$ A = [v_1,v_2,a_1v_1+a_2v_2,b_1v_1+b_2v_2] $$ where $a_1,a_2,b_1$ and $b_2$ are real numbers. And we know that $w_1=(−1,0,0,1)^T$ and $w_2=(1,3,2,0)^T$ are the basis of kernel space. So we must have $$ Aw_1 = 0\quad \text{and}\quad Aw_2 = 0 $$ From the first one, we plug in $w_1$, we get that (because $v_1$ and $v_2$ are linearly independent): $$ (b_1-1)v_1+b_2v_2=0\quad\rightarrow\quad b_1=1\quad\text{and}\quad b_2=0 $$ In the same way, we have $a_1=-1/2$ and $a_2=-3/2$. Then we get the transformation matrix $$ A = [v_1,v_2,-1/2v_1-3/2v_2,v_1] $$.

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