Linear transformation $l:\mathbb{R}^3 \mapsto \mathbb{R}$ is determined as follows: $l(1,0,0)=1$; $l(1,4,0)=-1$; $l(0,0,1)=1$. I need to find $\text{Ker}(l)$.

Answer should be $\text{Span}\{(0,2,1),(1,2,0)\}$.

But my answer is $\text{Span}\{(1,4,1),(2,4,0)\}$.

I tried to solve the problem the following way: $$l(x) = x_1 l(1,0,0) + x_2 l(1,4,0) + x_3 l(0,0,1).$$ Then $l(x) = x_1 - x_2 +x_3$. After than solve the system $x_1 - x_2 + x_3 = 0$. I got $$(x_1,x_2,x_3) = \text{Span}\{(0,1,1) (1,1,0)\}.$$ Express to standart coordinates I got: $x = b(1,0,0) + (a+b)(1,4,0) + a(0,0,1)$. So kernel is $\text{Span}\{(1,4,1),(2,4,0)\}$. Can't find the mistake.

  • $\begingroup$ Hint: What is the difference between the two answers? $\endgroup$ – Jessica B Aug 20 '14 at 12:22

Both answers are correct. That is $\mathcal{A}=((0,2,1)^T,(1,2,0)^T)$ and $\mathcal{B}=((1,4,1)^T,(2,4,0)^T)$ are both basis of the Kernel. Indeed for the basis $\mathcal{A}$ we have

$$\begin{pmatrix}0\\2\\1\end{pmatrix} = -\frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\end{pmatrix}$$

Applying the function $l$ and using the properties of a linear map we get

\begin{align*}l\left( \begin{pmatrix}0\\2\\1\end{pmatrix}\right) &=l\left(-\frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\end{pmatrix}\right)\\ &=-\frac12l\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\right)+\frac12l\left(\begin{pmatrix}1\\4\\0\end{pmatrix}\right)+l\left(\begin{pmatrix}0\\0\\1\end{pmatrix}\right)\\ &=-\frac12\cdot 1 + \frac12\cdot (-1)+ 1 =0\end{align*} And

$$\begin{pmatrix}1\\2\\0\end{pmatrix} = \frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+0\begin{pmatrix}0\\0\\1\end{pmatrix}$$ Applying the function $l$ and using as above the properties for linear maps:

\begin{align*} l\left(\begin{pmatrix}1\\2\\0\end{pmatrix} \right) &= \frac12 l\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\right) + \frac12 l\left(\begin{pmatrix}1\\4\\0\end{pmatrix}\right) \\ &=\frac12 \cdot 1 + \frac12 \cdot (-1) \end{align*}

While for the basis $\mathcal{B}$ you can see that $(0,2,4)^T = 2(0,2,1)^T$ and for $$\begin{pmatrix}1\\4\\1\end{pmatrix}=0\cdot\begin{pmatrix}1\\0\\0\end{pmatrix}+\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\end{pmatrix}$$

And again applying $l$

\begin{align*} l\left(\begin{pmatrix}1\\4\\1\end{pmatrix} \right) &= l\left(\begin{pmatrix}1\\4\\0\end{pmatrix} \right) + l\left( \begin{pmatrix}0\\0\\1\end{pmatrix}\right) \\ &= -1 + 1 = 0 \end{align*}

Hope this helps :-)


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