Which are linear transformations?
1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.
For this one I have:
Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.
But I don't know if is that quite simple. The other alternative I have is:
If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then
$$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
$$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
$$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
$$T(v)=v$$
Therefore, $T$ is a linear transformation.
Can be any of them?
2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.
For this one I have:
Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
$$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
$$T(v)=0+0+...+0$$
$$T(v)=0$$
Therefore, $T$ is a linear transformation.
Is this right?
3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.
For this last one I have not clue where to start. Help please!
linear-algebra vector-spaces linear-transformations
asked Nov 23 at 19:15
gi2302
103
add a comment |
1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.
For this one I have:
Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.
But I don't know if is that quite simple. The other alternative I have is:
If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then
$$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
$$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
$$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
$$T(v)=v$$
Therefore, $T$ is a linear transformation.
Can be any of them?
2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.
For this one I have:
Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
$$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
$$T(v)=0+0+...+0$$
$$T(v)=0$$
Therefore, $T$ is a linear transformation.
Is this right?
3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.
For this last one I have not clue where to start. Help please!
linear-algebra vector-spaces linear-transformations
asked Nov 23 at 19:15
gi2302
103
add a comment |
1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.
For this one I have:
Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.
But I don't know if is that quite simple. The other alternative I have is:
If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then
$$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
$$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
$$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
$$T(v)=v$$
Therefore, $T$ is a linear transformation.
Can be any of them?
2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.
For this one I have:
Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
$$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
$$T(v)=0+0+...+0$$
$$T(v)=0$$
Therefore, $T$ is a linear transformation.
Is this right?
3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.
For this last one I have not clue where to start. Help please!
linear-algebra vector-spaces linear-transformations
asked Nov 23 at 19:15
gi2302
103
1) Define $T:Vrightarrow V$ for $T(v)=v$, $forall$ $v in V$.
For this one I have:
Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.
But I don't know if is that quite simple. The other alternative I have is:
If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $vin V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then
$$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$
$$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$
$$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$
$$T(v)=v$$
Therefore, $T$ is a linear transformation.
Can be any of them?
2) Let $V$, $W$ vector spaces. Define $T:Vrightarrow W$ for $T(v)=0$, $forall$ $v in V$.
For this one I have:
Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$
$$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$
$$T(v)=0+0+...+0$$
$$T(v)=0$$
Therefore, $T$ is a linear transformation.
Is this right?
3) Define $T: C[0,1]rightarrow R$ for $T(f)=f(0)+1$.
For this last one I have not clue where to start. Help please!
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
asked Nov 23 at 19:15
gi2302
103
asked Nov 23 at 19:15
gi2302
103
asked Nov 23 at 19:15
gi2302
103
asked Nov 23 at 19:15
gi2302
103
asked Nov 23 at 19:15
gi2302
103
103
add a comment |
add a comment |
2 Answers
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- Your first approach if fine. The second one isn't, since it assumes that the mape is linear.
- Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$
- It is not linear, since $T(0)neq0$.
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
add a comment |
You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
add a comment |
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2 Answers
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2 Answers
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- Your first approach if fine. The second one isn't, since it assumes that the mape is linear.
- Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$
- It is not linear, since $T(0)neq0$.
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
add a comment |
- Your first approach if fine. The second one isn't, since it assumes that the mape is linear.
- Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$
- It is not linear, since $T(0)neq0$.
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
add a comment |
- Your first approach if fine. The second one isn't, since it assumes that the mape is linear.
- Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$
- It is not linear, since $T(0)neq0$.
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
- Your first approach if fine. The second one isn't, since it assumes that the mape is linear.
- Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)text{ and }T(cu)=0=ctimes0=cT(u).$$
- It is not linear, since $T(0)neq0$.
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
answered Nov 23 at 19:18
José Carlos Santos
149k22117219
149k22117219
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
add a comment |
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
José Carlos gracias. In the first one, in the second approach if I don't assume $T$ is linear is ok?
– gi2302
Nov 23 at 19:31
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
That means not assuming that $T(c_1v_1+cdots+c_nv_n)=c_1T(v_1)+cdots+c_nT(v_n)$.
– José Carlos Santos
Nov 23 at 19:33
add a comment |
You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
add a comment |
You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
add a comment |
You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1ne kf(0)+k=kT(f)qquadqquad kne 1$$
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
answered Nov 23 at 19:19
Mostafa Ayaz
13.7k3836
13.7k3836
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
add a comment |
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
Thanks!!!!!!!!!
– gi2302
Nov 23 at 19:32
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
You're welcome!!
– Mostafa Ayaz
Nov 23 at 19:36
add a comment |
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