Question 12

12

Write the set of real sequences as \mathbb R^{\omega}.

(i) Let (x_n) and (y_n) be real sequences and \alpha, \beta \in \mathbb R. Note that:

\begin{align*}T_1(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \alpha x_3 + \beta y_3, \ldots) &= (0, \alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \alpha x_3 + \beta y_3, \ldots) \\&= \alpha(0,x_1,x_2,x_3,\ldots) + \beta(0,y_1,y_2,y_3,\ldots)\\ & = \alpha T_1(x_1,x_2,x_3,\ldots) + \beta T_1(y_1,y_2,y_3,\ldots)\end{align*}

So T_1 is linear.

Note that for:

(0,x_1,x_2,x_3,\ldots) = (0,0,0,0,\ldots)

we must have x_1 = x_2 = x_3 = \ldots = 0, ie. that (x_n) is the zero sequence. So \mathrm {ker}(T_1) = \{0\}.

We also have:

\mathrm{im}(T_1) = \{(x_n) \in \mathbb R^\omega \mid x_1 = 0\} = \{(x_1, x_2, x_3, \ldots) \mid x_i \in \mathbb R, \, x_1 = 0\}

(b)

We have:

\begin{align*}T_2(\alpha (x_n) + \beta (y_n)) & = (\alpha x_1 + \beta y_1 + \alpha x_2 + \beta y_2, 0, \alpha x_3 + \beta y_3, \alpha x_4 + \beta y_4, 0, \ldots) \\ & = \alpha(x_1 + x_2, 0, x_3, x_4, 0, 0, 0, \ldots) + \beta(y_1 + y_2, 0, y_3, y_4, 0, 0, 0, \ldots) \\ & = T_2(\alpha(x_n)) + T_2(\beta(y_n))\end{align*}

So T_2 is linear.

Note that for:

(x_1 + x_2, 0, x_3, x_4, 0, 0, 0, \ldots)

we must have x_1 + x_2 = 0, (ie. x_1 = -x_2) and x_3 = x_4 = 0. x_i for i \ge 5 can be any real.

So:

\mathrm{ker}(T_2) = \{(x_n) \in \mathbb R^\omega \mid x_1 = -x_2, \, x_3 = x_4 = 0\} = \{(x_1, -x_1, 0, 0, x_5, x_6, \ldots) \mid x_i \in \mathbb R\}

Note that as x_1, x_2 vary over the reals, x_1 + x_2 can take any real value, so:

\mathrm{im}(T_2) = \{(x_n) \in \mathbb R^\omega \mid x_2 = 0, \, x_i = 0 \text { for } i \ge 5\} = \{(x_1, 0, x_3, x_4, 0, 0, \ldots) \mid x_i \in \mathbb R\}

©

Yes. Note that since V is infinite dimensioned, the identity map has infinite dimensional image. But clearly it has one dimensional kernel (in fact, the kernel is the zero subspace) so doesn’t quite work as an example. We can however modify it slightly, if instead mapping:

(x_1, x_2, x_3, \ldots) \mapsto (x_1, x_2, x_3, x_4, \ldots)

we map (still linearly):

(x_1, x_2, x_3, \ldots) \mapsto (x_1, 0, x_3, 0, \ldots)

then the kernel is precisely every sequence with all odd entries zeroes. ie.

\mathrm{ker}(T_3) = \{(x_n) \in \mathbb R^\omega \mid x_{2 i + 1} = 0\}

clearly infinite dimensional, since it is spanned by the linearly independent set:

\{(x_n) \in \mathbb R^\omega \mid x_{2n} = 1, \, x_i = 0, \, i \ne n, \, n \in \mathbb N\} = \{(0,1,\ldots), (0,0,0,1,\ldots), (0,0,0,0,0,1,\ldots) \ldots\}

We similarly have:

\mathrm{im}(T_3) = \{(x_n) \in \mathbb R^\omega \mid x_{2i} = 0\}

which is also infinite dimensional.

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