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I just want to make sure that what I'm doing is correct. Here's the question:

Determine a linear transformation $T$: $\mathbb{R^3} \rightarrow \mathbb{R^2}$ with kernel $W$:

$W$ = {$(x,y,z)$ | $x-2y+z=0$}

So what I did to define a transformation considering that

$T (1,0,0) = (0,0)$

$T (0,1,0) = (0,1)$

$T (0,0,1) = (1,0)$

I concluded that

$T(x(1,0,0) + y(0,1,0) + z(0,0,1))=(z,y)$

My doubt is in my next step...

So the kernel of the transformation is $z=0 \wedge y=0$

But because W should be the kernel and by W $z=-x+2y$ we have that our kernel is $(x,0,-x)$ so it's the vector (1,0,-1)...

Can someone please help me verify if this is correct?

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wrong part of your answer is defining $ T (1,0,0) = (0,0), T (0,1,0) = (0,1), T (0,0,1) = (1,0)$. using these definitions, the constraint of null space wont be satisfied. a simple way i know for solving this category is as below.

assume $(x\; y\; z)^T$ exists in null space of our transformation. then it can be written as: $$\begin{pmatrix} x\\y\\z \end{pmatrix} = t \begin{pmatrix} 1\\-2\\1 \end{pmatrix}$$ this gives us two equations $x=z$ and $y=-2x$. these two equations are spanning row space of our transformation. writing them in following form and we get: $$\begin{matrix} (1)x & (0)y & (-1)z & = & 0 \\ (2)x & (1)y & (0)z & = & 0\\ \end{matrix}$$ which gives us the transformation matrix as: $$T = \begin{pmatrix} 1 & 0 & -1\\ 2 & 1 & 0\\ \end{pmatrix}$$

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  • $\begingroup$ What? How to you take those equation for the matrix? Anyway I'm trying to find an answer not involving matrices... $\endgroup$ Apr 25 '16 at 14:20
  • $\begingroup$ But thank you ! $\endgroup$ Apr 25 '16 at 14:20
  • $\begingroup$ range of row space of $T$ is the answer and that's why i took those equations as rows of $T$. $\endgroup$ Apr 25 '16 at 14:29
  • $\begingroup$ Ok I think I got it... What I did was to express $W$ as vectors: (2,1,0) and (-1,0,1)... I transform it into equations with zero as the equality and I got to y=-2x and z=x. So my conclusion is that the transformation might be T(x,y,z) = (x,-2x,x)... Is that correct? $\endgroup$ Apr 25 '16 at 15:19
  • $\begingroup$ I concluded it can't be correct because the transformation as IR^2 as image... $\endgroup$ Apr 25 '16 at 16:49
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Firstly, we can find a basis for the nullspace. We have that: $$W \begin{array}[t]{l}= \{ (x,y,z)^T: x-2y+z = 0, x,y,z \in \mathbb R \}\\ = \{ (x,y,z)^T: z = -x+2y, x, y, z\in \mathbb R\}\\ = \{ (x,y,-x+2y)^T: x,y \in \mathbb R\}\\ = \{x\cdot (1,0,-1)^T + y\cdot(0,1,2)^T:x,y \in \mathbb R\}. \end{array}$$

Thus, $$W = \langle \underbrace{(1,0,-1)^T}_{\vec e_1},\, \underbrace{(0,1,2)^T}_{\vec e_2}\rangle.$$

From rank-nullity theorem we have that: $$\dim (\text{Im} T) + \dim (\ker T) = 3 \implies \dim (\text{Im} T)=1.$$ This means that our $2\times 3$ transformation matrix will have rank $1.$ Thus, $$A = \begin{bmatrix} a & b & c \\ \lambda a & \lambda b &\lambda c \end{bmatrix}.$$ So we need to define $a,b,c$.

From $\begin{bmatrix} a & b & c \\ \lambda a & \lambda b &\lambda c \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies a = c$

From $\begin{bmatrix} a & b & c \\ \lambda a & \lambda b &\lambda c \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies b = -2c.$

Thus, our matrix $A$ is of the form: $$A = \begin{bmatrix} a & -2a & a\\ \lambda a & - 2\lambda a& \lambda a\end{bmatrix}.$$

Choosing $a = 1$ and $\lambda = 3$ , we have that:

$$T\begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 1 & - 2 & 1\\3 & -6 & 3\end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} x-2y+z \\ 3x-6y+ 3z \end{bmatrix}$$

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