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For each of the following, show that T is a linear transformation and find bases for ker(T) and range(T) then compute the nullity and rank of T, verify the rank equation and determin whether t is one to one or onto

(a) $T: \mathbb Q^3 \to \mathbb Q^2$ defined by $T([x_1, x_2, x_3]) = [x_1 - x_2, 2x_3]$

Solution:

Let $x, y \in \mathbb Q^3$ and $\lambda \in \mathbb Q$. T is a linear transformation because $T(x+y) = T([x_1 + y_1, x_2 + y_2, x_3 + y_3]) = [x_1 + y_1 - (x_2 + y_2), 2(x_3 + y_3)] = [x_1 - x_2, 2x_3] + [y_1 + y_2, 2y_3] = T(x) + T(y)$

$T(\lambda x) = T([\lambda x_1, \lambda x_2, \lambda x_3]) = [\lambda x_1 - \lambda x_2, \lambda(2x_3)] = \lambda[x_1 - x_2, 2x_3] = \lambda T(x)$

$T(x) = 0 \leftrightarrow x_1 - x_2 = 0, 2x_3 = 0 \to x_3 = 0, x_1 = x_2$

$ker(T) = \{[x_1, x_1, 0] x_1 \in \mathbb Q \} = \{x_1[1, 1, 0] | x_1 \in \mathbb Q \} = sp([1,1,0])$ A basis for ker(T) is $\{[1,1,0]\}$ and nullity(T) = 1.

(Below I don't get)

$range(T) = sp(T(e_1), T(e_2), T(e_3)) = sp([1,0], [-1,0], [0,1]) = sp([1,0], [0,1]) = \mathbb Q ^2$. Hench $\{e_2, e_2\}$ is a basis of $range(T)$ and $rank(T) = 2$

Now rank equation: $1 + 2 = 3 = dim \mathbb Q^3$. Since $range(T) = \mathbb Q^2$ T is onto. Since $ker(T) \neq \{0\}$, T is not one-to-one.


I like don't get anything after I put "Below I don't get". Howed they get those basis for range(T) and why is T onto.

Also in a matrix for ker(T) can it be written as below?

\begin{array}{cc} 1 & -1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ \end{array}

if i let $x_2 = s \in \mathbb R = 1$, then

$x_1 = 1, x_2 = 1, x_3 = 0$

therefore $\{[1,1,0]\}$ is a basis for ker(T). Avoiding all those other statements

EDIT: Shouldn't $range(T) = sp([1,0], [-1,0], [0,2])$. Is the solution a bit wrong?

Also should it be $\{e_1, e_2 \}$ is a basis for range(T)?

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If the map $T$ is linear, it has an associated matrix (relative to the canonical bases); the matrix must have as columns $T(e_1)$, $T(e_2)$ and $T(e_3)$, $\{e_1,e_2,e_3\}$ denoting the canonical basis.

Writing all vectors as columns, $$ T(e_1)=T\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \qquad T(e_2)=T\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}-1\\0\end{bmatrix} \qquad T(e_3)=T\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}0\\2\end{bmatrix} $$ Hence, $T$ is linear if and only if $$ T\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} $$ for every vector $\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$. This is readily verified as true.

Gaussian elimination on the matrix gives the reduced row-echelon form $$ \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ which tells you that $\{T(e_1),T(e_3)\}$ is a basis for the range of $T$. But, as the rank of $T$ is $2$, the range is the whole $\mathbb{Q}^2$ and any basis is also a valid choice.

In a more general case, a basis of the range is determined by the vectors corresponding to the pivot columns, in this case first and third column.

A vector belongs to the kernel if and only if $x_1=x_2$ and $x_3=0$, so a basis for the kernel is given by $$ \begin{bmatrix}1\\1\\0\end{bmatrix} $$

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