Prove that the range is a subspace
$begingroup$
In Linear Algebra Done Right, it said
If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.
Proof:
Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.
If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
$$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
Hence $w_1+w_2 in text{range } T$
Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range
Similarly, I don't understand why $lambda w$ is in the range of $T$
linear-algebra
asked Jan 16 at 23:05
JOHN JOHN
3268
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, it said
If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.
Proof:
Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.
If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
$$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
Hence $w_1+w_2 in text{range } T$
Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range
Similarly, I don't understand why $lambda w$ is in the range of $T$
linear-algebra
asked Jan 16 at 23:05
JOHN JOHN
3268
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, it said
If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.
Proof:
Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.
If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
$$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
Hence $w_1+w_2 in text{range } T$
Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range
Similarly, I don't understand why $lambda w$ is in the range of $T$
linear-algebra
asked Jan 16 at 23:05
JOHN JOHN
3268
$endgroup$
In Linear Algebra Done Right, it said
If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.
Proof:
Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.
If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
$$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
Hence $w_1+w_2 in text{range } T$
Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range
Similarly, I don't understand why $lambda w$ is in the range of $T$
linear-algebra
linear-algebra
asked Jan 16 at 23:05
JOHN JOHN
3268
asked Jan 16 at 23:05
JOHN JOHN
3268
asked Jan 16 at 23:05
JOHN JOHN
3268
asked Jan 16 at 23:05
JOHN JOHN
3268
asked Jan 16 at 23:05
JOHN JOHN
3268
3268
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
add a comment |
$begingroup$
Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:14
user3482749user3482749
4,266919
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
add a comment |
$begingroup$
Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
add a comment |
$begingroup$
Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
$endgroup$
Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
answered Jan 16 at 23:15
DeepSeaDeepSea
71.2k54487
71.2k54487
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
add a comment |
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
$begingroup$
$v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
$endgroup$
– JOHN
Jan 17 at 7:31
add a comment |
$begingroup$
Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 16 at 23:13
Kavi Rama MurthyKavi Rama Murthy
59k42161
59k42161
add a comment |
add a comment |
$begingroup$
Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:14
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:14
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:14
user3482749user3482749
4,266919
$endgroup$
Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.
answered Jan 16 at 23:14
user3482749user3482749
4,266919
answered Jan 16 at 23:14
user3482749user3482749
4,266919
answered Jan 16 at 23:14
user3482749user3482749
4,266919
answered Jan 16 at 23:14
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
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