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The problem statement is,

Find a linear transformation $T: \mathbb R^3 \to \mathbb R^3$ such that the set of all vectors satisfying $4x_1-3x_2+x_3=0$ is the (i) null space of $T$ (ii) range of $T$.

For (i),

I found out the basis of the null space for the system,

$$\begin{pmatrix}4&-3&1\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} =0$$

viz, $\mathbf v_1 = \begin{pmatrix}3\\4\\0\end{pmatrix} , \mathbf v_2 = \begin{pmatrix}-1\\0\\4\end{pmatrix}$

So, basically, I have to find linear transformation such that $T\begin{pmatrix}3\\4\\0\end{pmatrix} =0$ and $T\begin{pmatrix}-1\\0\\4\end{pmatrix}=0$ such that vector $\mathbf v \in span\{\mathbf v_1, \mathbf v_2\}$ satisfies $T\left(\mathbf v\right) =0$

Now, I'm stuck at this point. I'm not able to describe the such linear transformations.

Further, for (ii), I don't know how to approach the problem.

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For (i), consider the matrix

$$ A = \begin{pmatrix} 4 & -3 & 1 \\ 4 & -3 & 1 \\ 4 & -3 & 1 \end{pmatrix}. $$

Denoting by $T_A \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ the corresponding linear map $T_A(v) = Av$, we have

$$ T_A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4x_1 - 3x_2 + x_3 \\ 4x_1 - 3x_2 + x_3 \\ 4x_1 - 3x_2 + x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$

so $\ker(T_A)$ is precisely the solution subspace of the equation $4x_1 - 3x_2 + x_3$.

For (ii), consider the matrix

$$ B = \begin{pmatrix} 3 & -1 & 0 \\ 4 & 0 & 0 \\ 0 & 4 & 0 \end{pmatrix} $$

whose columns are $\bf{v_1},v_2$ and the zero vector. The image of $T_B$ is spanned by the columns of $B$ and thus again is equal to the solution subspace of your equation.

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  • $\begingroup$ So, answer to (i) will be $\alpha A$ where $\alpha \in \mathbb R - \{0\}$? And how can be sure that the only solution to the problem is the above stated? $\endgroup$ – crimson Feb 7 '16 at 22:56
  • $\begingroup$ You did not write that you want to find all the possible $T$'s whose kernels are precisely the solution subspace. That is a different question. You can also multiply each row of $A$ by a different non-zero real number. $\endgroup$ – levap Feb 7 '16 at 22:58
  • $\begingroup$ My bad! But if we have to find the all the possible $T$'s, then is there any other solution? $\endgroup$ – crimson Feb 7 '16 at 23:01
  • $\begingroup$ All the possible $A$'s are given by $\begin{pmatrix} 4a & -3a & a \\ 4b & -3b & b \\ 4c & -3c & c \end{pmatrix}$ for $a,b,c \in \mathbb{R}$ where at least one of $a,b,c$ is non-zero. $\endgroup$ – levap Feb 7 '16 at 23:40
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Any linear transformation is completely determined by its action on a basis.

For the first question extend the set $\{v_1, v_2\}$ to a basis for $\mathbb{R}^3$ by adjoining a vector $v_3$ and specify a linear transformation with the required kernel by setting $T(v_3)$ to be a vector all of whose components are nonzero.

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  • $\begingroup$ I get that I have to extend the set to basis for $\mathbb R^3$ but I still don't get the method of evaluating the transformation. $\endgroup$ – crimson Feb 8 '16 at 6:52

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