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Exercise 2.1.26

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1.

Since

V = W_{1} \oplus W_{2}

, every vector

x

have an unique representation

x = x_{1} + x_{2}

with

x_{1} \in W_{1}

and

x_{2} \in W_{2}

. So, now we have

T (x + cy) = T (x_{1} + x_{2} + c y_{1} + c y_{2})

= T ((x_{1} + c y_{1}) + (x_{2} + c y_{2})) = x_{1} + c y_{1} = T (x) + cT (y) .

And hence it’s linear.

On the other hand, we have $x = x + 0$ and hence $T (x) = x$ if $x \in W_{1}$ . And if $x \notin W_{1}$ , this means $x = x_{1} + x_{2}$ with $x_{2} \neq 0$ and hence we have $T (x) = x_{1} \neq x_{1} + x_{2}$ .

2.

x_{1} \in W_{1}

then we have

T (x_{1} + 0) = x_{1} \in R (T)

; and we also have

R (T) \subset W_{1}

. If

x_{2} \in W_{2}

then we have

T (x_{2}) = T (0 + x_{2}) = 0

and hence

x_{2} \in N (T)

; and if

x \in N (T)

, we have

x = T (x) + x = 0 + x

and hence

x \in W_{2}

3.

It would be

T (x) = x

by (a).

4.

It would be

T (x) = 0

jephianlin

2011-06-27 00:00

Exercise 2.1.26

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