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Example: midterm 4b

$\set{W_i| W_i \leq V, i \in I}$

Collection of infinite many distinct subspaces of V.

$W_i \cap W_j = \set{0}$ for $i \neq j, I$ infinite indexing set.

$\cup_{i \in I} W_i$ is NOT a subspace of $V_i$.

Counter example.

$W_i = {(x,y) | y = ix, i \in R}$

Example: HW 3 - 2.1.14

$V, W$ vector space, $T: V \to W$ linear

(a) T one-to-one $\iff$ T carries independent subsets of $V$ onto linearly independent subsets of $W$. Note: Do not assume $S$ is finite.

$T$ one-to-one if $T_x = T_y \implies x = y$.

$T$ onto if for each $z \in W$ there is $x \in V$ s.t. $T_x = z$.

  1. $T \oto \iff N(T) = \set{0}$
  2. $T \ot \iff R(T) = W$
Example: 2.1.37

$T:V \to W$ T not a linear transformation, additive if T(x+y) = T(x) + T(y).

$V, W$ vector space over P.

Then any additive function T is a linear transformation.

we have to prove $T(x) = cTx \forall c \in R, x \in V$.

Proof [expand]

(*) true for any $c \in N$

$c = 1$ trivial $ T(1x) = 1Tx $

$ c=2, T(2x) = T(x + x) = Tx + Tx = 2Tx$

induction, assume (*) is true for some $ n \in N_1 $

$T(nx) = nTx$

$T((n+1) x) = T(nx + x) = T(nx) + Tx$

$nTx +Tx = (n+1) Tx$

By induction it is true $\forall c \in N$

Proof [expand]

$ T(0) = T (0) = T(0) + T(0)$ $\implies T(0) = 0$

$T(0x) = T(0) = 0 = 0 \cdot T(x)$, true for $c = 0$

$0 = T(o) = T(x+(-x)) = T(x) + T(-x)$

$\implies T(-x) = - T(x)$

if $c \in Z, c < 0$ then $-c \in N$

$T(cx) = -T((-c) x) = -(-c) T(x) = cT(x)$(step 1)

$(*) is true for c \in Z$

Proof [expand]

$c \in \R, c = \frac{p}{q} p \in Z q \in N$

$T(px) = pT(x) $(step 2)

$T(q \frac{p}{q} x) = q \cdot T(\frac{p}{q} x)$(step 2)

$c = \frac{p}{q} p \in Z, q \in N$

$pT(x) = qT(\frac{p}{q}x) \implies T(\frac{p}{q} x) = \frac{p}{q}T(x)$

Example: 2.2.16

$V, W$ vector space, $\dim V = \dim V$, $T: V \to W$ linear, Show that there exists ordered basis $\beta, \gamma$ for $V, W$ s.t. $[T]^\gamma_\beta$ is a diagonal matrix.

Recall

$\beta = \set{v_1, v_2, …, v_{n}} , \gamma = \set{w_1, w_2, …, w_{m}}$

$T(v_i) = \sum_{i=1}^{m} a_{ij} W_i$

$A= (a_{ij}) = [T]^\gamma_\beta$

Proof [expand]

let $k = \dim R(T) \leq n$ find a basis $\set{w_1, w_2, …, w_{k}}$ of $R(T)$

Proof [expand]

For each $i \leq i \leq k$

there is $V_i \in V$ s.t. $TV_i = W_i$ since $W \in R(T)$.

Proof [expand]

check \set{V_1, V_2, …, V_{k}} is linearly independent. (use the fact that $\set{w_1, w_2, …, w_{k}}$ is linear independent)

Proof [expand]

$n = dimV = \dim N(T) + \dim R(T)$

$\implies N(T) = n - k$

choose basis of $N(T) \leq V, \set{V_{k+1}, …, V_n}$

Proof [expand]

Define $\beta = \set{v_1, v_2, …, v_{n}}$

check this is a basis of $V$.

enough to prove that $ V_1, V_2, …, V_{n} is linearly independent. Why? $

Proof [expand]

Use replacement theorem to extend $\set{w_1, w_2, …, w_{k}}$ to a basis $\set{w_1, w_2, …, w_{n}}$ of $W$.

Proof [expand]

define $\gamma = \set{w_1, w_2, …, w_{n}}$

prove that $ [T]^\gamma_\beta $ is diagonal.

Example: 2.2.13 extra

$V, W$ is $T, U: V \to W$ $T, U $ nonzero linear transformation, If $R(T) \cap R(U) = \set{0}$ prove: $\set{T, U}$ is a linearly independent subset of $L(V, W)$

Note $L(V, W)$ is the collection of all linear transformation from $V$ to $W$.

Proof [expand]

Assume $ aT + bU = 0 $ for some $a,b$

This means $aT(x) + bU(x) = 0, \forall x \in V$

To prove independent, we have to prove that $a = b = 0 $ suppose not, without loss of generality, assume $a \neq 0$

If $b = 0$, then $aT=0 \to T =0$ contradiction.

(T, V are non-zero maps).

so $a, b \neq 0$.

T is a non-zero map

so $\exists v \in V s.t. T(v) \neq 0$

$a T(v) + bU(v) = 0$

$\implies U(v) = - \frac{a}{b} T(v)$

if $T(v) = W, w \in range(T)$

$Range(T)$ is subspace, $- \frac{a}{b} w \in Range(T) $

$U(v) = - \frac{a}{b} w \in Range(V)$