Mixing
if $f,g$ are linear transformations, find basis of $ operatorname{im}(f) cap ker(f circ g) $ and $...
$begingroup$
I have done this task, however, my solutions are suspiciously strange. Could someone verify if my reasoning is okay? Possibly something to advise ...
If all is ok (I hope), this post would be nice pattern for future visitors
Task
Given are linear transformations:
$f in L(mathbb R^3, mathbb R[t]_2) $
$$ vec{x} = [x_1,x_2,x_3]^T rightarrow f(vec{x})(t)=(x_1+x_3)t^2 + x_2$$
and
$g in L(mathbb R[t]_2,mathbb R^3) $
$$p in mathbb R[t]_2 rightarrow g(p) = [p(-1),p'(0),p(1)]^T $$
Find basis of subspace:
a) $ operatorname{im}(f) cap ker(f circ g) $
b) $ operatorname{im}(g circ f) + ker(f) $
My solution
a)
Take a random polynomial:
$$p(t) = at^2 + bt + c $$
$$p(1) = a + b + c $$
$$p(-1) = a - b +c $$
$$p'(0) = b $$
so $ g(p) = [a-b+c,b,a+b+c]^T $
Ok now we are looking for $ fcirc g $
$$f(g(p)) = 2(a+c)t^2 + b $$
Ok, now I want its kernel:
$$f(g(p)) = 0 leftrightarrow a = - c wedge b = 0$$
so $ker(fcirc g) = operatorname{span}[1,0,-1]^T $
I need also $ operatorname{im}(f)$ so $$ operatorname{im}(f) = operatorname{span}([1,0,0]^T,[0,1,0]^T)$$
but
$$ ([1,0,0]^T,[0,1,0]^T,[1,0,-1]^T $$
are linearly independent so $ operatorname{im}(f) cap ker(f circ g) = 0 $
b)
Now I am looking for $ operatorname{im}(g circ f)$
$$g(f(vec{x})) = [x_1+x_2+x_3,0,x_1+x_2+x_3]^T = operatorname{span}([1,0,1]^T) = operatorname{im}(g circ f)$$
Now $ker(f)$
$$ f(vec{x})(t)=(x_1+x_3)t^2 + x_2 = 0 leftrightarrow x_1 = -x_3 wedge x_2 = 0$$
so
$$ker(f) = operatorname{span}([1,0,-1]^T)$$
Ok now we are looking for:
$$ operatorname{im}(g circ f) + ker(f) = operatorname{span}([1,0,-1]^T,[1,0,1]^T) $$
but $[1,0,-1]^T,[1,0,1]^T $ are linearly independent so
$ operatorname{span}([1,0,-1]^T,[1,0,1]^T) $ is a basis of $ operatorname{im}(g circ f) + ker(f) $
Thanks for your time!
linear-algebra functions proof-verification linear-transformations
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
$endgroup$
add a comment |
$begingroup$
I have done this task, however, my solutions are suspiciously strange. Could someone verify if my reasoning is okay? Possibly something to advise ...
If all is ok (I hope), this post would be nice pattern for future visitors
Task
Given are linear transformations:
$f in L(mathbb R^3, mathbb R[t]_2) $
$$ vec{x} = [x_1,x_2,x_3]^T rightarrow f(vec{x})(t)=(x_1+x_3)t^2 + x_2$$
and
$g in L(mathbb R[t]_2,mathbb R^3) $
$$p in mathbb R[t]_2 rightarrow g(p) = [p(-1),p'(0),p(1)]^T $$
Find basis of subspace:
a) $ operatorname{im}(f) cap ker(f circ g) $
b) $ operatorname{im}(g circ f) + ker(f) $
My solution
a)
Take a random polynomial:
$$p(t) = at^2 + bt + c $$
$$p(1) = a + b + c $$
$$p(-1) = a - b +c $$
$$p'(0) = b $$
so $ g(p) = [a-b+c,b,a+b+c]^T $
Ok now we are looking for $ fcirc g $
$$f(g(p)) = 2(a+c)t^2 + b $$
Ok, now I want its kernel:
$$f(g(p)) = 0 leftrightarrow a = - c wedge b = 0$$
so $ker(fcirc g) = operatorname{span}[1,0,-1]^T $
I need also $ operatorname{im}(f)$ so $$ operatorname{im}(f) = operatorname{span}([1,0,0]^T,[0,1,0]^T)$$
but
$$ ([1,0,0]^T,[0,1,0]^T,[1,0,-1]^T $$
are linearly independent so $ operatorname{im}(f) cap ker(f circ g) = 0 $
b)
Now I am looking for $ operatorname{im}(g circ f)$
$$g(f(vec{x})) = [x_1+x_2+x_3,0,x_1+x_2+x_3]^T = operatorname{span}([1,0,1]^T) = operatorname{im}(g circ f)$$
Now $ker(f)$
$$ f(vec{x})(t)=(x_1+x_3)t^2 + x_2 = 0 leftrightarrow x_1 = -x_3 wedge x_2 = 0$$
so
$$ker(f) = operatorname{span}([1,0,-1]^T)$$
Ok now we are looking for:
$$ operatorname{im}(g circ f) + ker(f) = operatorname{span}([1,0,-1]^T,[1,0,1]^T) $$
but $[1,0,-1]^T,[1,0,1]^T $ are linearly independent so
$ operatorname{span}([1,0,-1]^T,[1,0,1]^T) $ is a basis of $ operatorname{im}(g circ f) + ker(f) $
Thanks for your time!
linear-algebra functions proof-verification linear-transformations
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
$endgroup$
$begingroup$
I think you meant to write that $gin L(mathbb R[t]_2, mathbb R^3)$. Otherwise the compositions (and the way $g$ was defined) don't make sense.
$endgroup$
– Math1000
Jan 5 at 15:27
add a comment |
$begingroup$
I have done this task, however, my solutions are suspiciously strange. Could someone verify if my reasoning is okay? Possibly something to advise ...
If all is ok (I hope), this post would be nice pattern for future visitors
Task
Given are linear transformations:
$f in L(mathbb R^3, mathbb R[t]_2) $
$$ vec{x} = [x_1,x_2,x_3]^T rightarrow f(vec{x})(t)=(x_1+x_3)t^2 + x_2$$
and
$g in L(mathbb R[t]_2,mathbb R^3) $
$$p in mathbb R[t]_2 rightarrow g(p) = [p(-1),p'(0),p(1)]^T $$
Find basis of subspace:
a) $ operatorname{im}(f) cap ker(f circ g) $
b) $ operatorname{im}(g circ f) + ker(f) $
My solution
a)
Take a random polynomial:
$$p(t) = at^2 + bt + c $$
$$p(1) = a + b + c $$
$$p(-1) = a - b +c $$
$$p'(0) = b $$
so $ g(p) = [a-b+c,b,a+b+c]^T $
Ok now we are looking for $ fcirc g $
$$f(g(p)) = 2(a+c)t^2 + b $$
Ok, now I want its kernel:
$$f(g(p)) = 0 leftrightarrow a = - c wedge b = 0$$
so $ker(fcirc g) = operatorname{span}[1,0,-1]^T $
I need also $ operatorname{im}(f)$ so $$ operatorname{im}(f) = operatorname{span}([1,0,0]^T,[0,1,0]^T)$$
but
$$ ([1,0,0]^T,[0,1,0]^T,[1,0,-1]^T $$
are linearly independent so $ operatorname{im}(f) cap ker(f circ g) = 0 $
b)
Now I am looking for $ operatorname{im}(g circ f)$
$$g(f(vec{x})) = [x_1+x_2+x_3,0,x_1+x_2+x_3]^T = operatorname{span}([1,0,1]^T) = operatorname{im}(g circ f)$$
Now $ker(f)$
$$ f(vec{x})(t)=(x_1+x_3)t^2 + x_2 = 0 leftrightarrow x_1 = -x_3 wedge x_2 = 0$$
so
$$ker(f) = operatorname{span}([1,0,-1]^T)$$
Ok now we are looking for:
$$ operatorname{im}(g circ f) + ker(f) = operatorname{span}([1,0,-1]^T,[1,0,1]^T) $$
but $[1,0,-1]^T,[1,0,1]^T $ are linearly independent so
$ operatorname{span}([1,0,-1]^T,[1,0,1]^T) $ is a basis of $ operatorname{im}(g circ f) + ker(f) $
Thanks for your time!
linear-algebra functions proof-verification linear-transformations
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
$endgroup$
I have done this task, however, my solutions are suspiciously strange. Could someone verify if my reasoning is okay? Possibly something to advise ...
If all is ok (I hope), this post would be nice pattern for future visitors
Task
Given are linear transformations:
$f in L(mathbb R^3, mathbb R[t]_2) $
$$ vec{x} = [x_1,x_2,x_3]^T rightarrow f(vec{x})(t)=(x_1+x_3)t^2 + x_2$$
and
$g in L(mathbb R[t]_2,mathbb R^3) $
$$p in mathbb R[t]_2 rightarrow g(p) = [p(-1),p'(0),p(1)]^T $$
Find basis of subspace:
a) $ operatorname{im}(f) cap ker(f circ g) $
b) $ operatorname{im}(g circ f) + ker(f) $
My solution
a)
Take a random polynomial:
$$p(t) = at^2 + bt + c $$
$$p(1) = a + b + c $$
$$p(-1) = a - b +c $$
$$p'(0) = b $$
so $ g(p) = [a-b+c,b,a+b+c]^T $
Ok now we are looking for $ fcirc g $
$$f(g(p)) = 2(a+c)t^2 + b $$
Ok, now I want its kernel:
$$f(g(p)) = 0 leftrightarrow a = - c wedge b = 0$$
so $ker(fcirc g) = operatorname{span}[1,0,-1]^T $
I need also $ operatorname{im}(f)$ so $$ operatorname{im}(f) = operatorname{span}([1,0,0]^T,[0,1,0]^T)$$
but
$$ ([1,0,0]^T,[0,1,0]^T,[1,0,-1]^T $$
are linearly independent so $ operatorname{im}(f) cap ker(f circ g) = 0 $
b)
Now I am looking for $ operatorname{im}(g circ f)$
$$g(f(vec{x})) = [x_1+x_2+x_3,0,x_1+x_2+x_3]^T = operatorname{span}([1,0,1]^T) = operatorname{im}(g circ f)$$
Now $ker(f)$
$$ f(vec{x})(t)=(x_1+x_3)t^2 + x_2 = 0 leftrightarrow x_1 = -x_3 wedge x_2 = 0$$
so
$$ker(f) = operatorname{span}([1,0,-1]^T)$$
Ok now we are looking for:
$$ operatorname{im}(g circ f) + ker(f) = operatorname{span}([1,0,-1]^T,[1,0,1]^T) $$
but $[1,0,-1]^T,[1,0,1]^T $ are linearly independent so
$ operatorname{span}([1,0,-1]^T,[1,0,1]^T) $ is a basis of $ operatorname{im}(g circ f) + ker(f) $
Thanks for your time!
linear-algebra functions proof-verification linear-transformations
linear-algebra functions proof-verification linear-transformations
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
edited Jan 5 at 20:07
VirtualUser
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
asked Jan 5 at 14:43
VirtualUserVirtualUser
67912
67912
$begingroup$
I think you meant to write that $gin L(mathbb R[t]_2, mathbb R^3)$. Otherwise the compositions (and the way $g$ was defined) don't make sense.
$endgroup$
– Math1000
Jan 5 at 15:27
add a comment |
$begingroup$
I think you meant to write that $gin L(mathbb R[t]_2, mathbb R^3)$. Otherwise the compositions (and the way $g$ was defined) don't make sense.
$endgroup$
– Math1000
Jan 5 at 15:27
$begingroup$
I think you meant to write that $gin L(mathbb R[t]_2, mathbb R^3)$. Otherwise the compositions (and the way $g$ was defined) don't make sense.
$endgroup$
– Math1000
Jan 5 at 15:27
$begingroup$
I think you meant to write that $gin L(mathbb R[t]_2, mathbb R^3)$. Otherwise the compositions (and the way $g$ was defined) don't make sense.
$endgroup$
– Math1000
Jan 5 at 15:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ So $text{Im}(f)$ is spanned by ${1,x^2}={(1,0,0),(0,0,1)}$. Clearly, $ker(fcirc g)=text{span}{(1,0,-1)}subsettext{span}{(1,0,0),(0,0,1)}=text{Im}(f)$, so the answer to part $(a)$ is $ker(fcirc g)$.
Answer to part $(b)$ is correct.
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
$endgroup$
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
add a comment |
$begingroup$
One thing I learned early: the usual terminology here is an 'arbitary' polynomial rather than a 'random' one. 'Random' implies you want to bring in probability/measure theory, which you don't.
Now onto more serious matters. It is important that when talking about $ker(f circ g)$ and $im(f)$ you recognise the space they actually live in, which is $mathbb{R}[t]_2$. Your answers aren't in $mathbb{R}[t]_2$ but $mathbb{R}^3$, so they fail to answer the question and also cause confusion. It is vital that we correct your results to be subsets of the space they should be. For instance...
$im(f)= {ax^2+b|a,b in mathbb{R}}$, for instance. Can you do the same for $ker (f circ g)$?
For the second question, can you identify $im(g circ f)$ and $ker(f)$ as subspaces of $mathbb{R}^3$?
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
$endgroup$
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ So $text{Im}(f)$ is spanned by ${1,x^2}={(1,0,0),(0,0,1)}$. Clearly, $ker(fcirc g)=text{span}{(1,0,-1)}subsettext{span}{(1,0,0),(0,0,1)}=text{Im}(f)$, so the answer to part $(a)$ is $ker(fcirc g)$.
Answer to part $(b)$ is correct.
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
$endgroup$
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
add a comment |
$begingroup$
$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ So $text{Im}(f)$ is spanned by ${1,x^2}={(1,0,0),(0,0,1)}$. Clearly, $ker(fcirc g)=text{span}{(1,0,-1)}subsettext{span}{(1,0,0),(0,0,1)}=text{Im}(f)$, so the answer to part $(a)$ is $ker(fcirc g)$.
Answer to part $(b)$ is correct.
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
$endgroup$
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
add a comment |
$begingroup$
$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ So $text{Im}(f)$ is spanned by ${1,x^2}={(1,0,0),(0,0,1)}$. Clearly, $ker(fcirc g)=text{span}{(1,0,-1)}subsettext{span}{(1,0,0),(0,0,1)}=text{Im}(f)$, so the answer to part $(a)$ is $ker(fcirc g)$.
Answer to part $(b)$ is correct.
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
$endgroup$
$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ So $text{Im}(f)$ is spanned by ${1,x^2}={(1,0,0),(0,0,1)}$. Clearly, $ker(fcirc g)=text{span}{(1,0,-1)}subsettext{span}{(1,0,0),(0,0,1)}=text{Im}(f)$, so the answer to part $(a)$ is $ker(fcirc g)$.
Answer to part $(b)$ is correct.
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
answered Jan 5 at 15:26
Shubham JohriShubham Johri
4,920717
4,920717
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
add a comment |
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
$begingroup$
Ah, thanks for checking! ;)
$endgroup$
– VirtualUser
Jan 5 at 17:33
add a comment |
$begingroup$
One thing I learned early: the usual terminology here is an 'arbitary' polynomial rather than a 'random' one. 'Random' implies you want to bring in probability/measure theory, which you don't.
Now onto more serious matters. It is important that when talking about $ker(f circ g)$ and $im(f)$ you recognise the space they actually live in, which is $mathbb{R}[t]_2$. Your answers aren't in $mathbb{R}[t]_2$ but $mathbb{R}^3$, so they fail to answer the question and also cause confusion. It is vital that we correct your results to be subsets of the space they should be. For instance...
$im(f)= {ax^2+b|a,b in mathbb{R}}$, for instance. Can you do the same for $ker (f circ g)$?
For the second question, can you identify $im(g circ f)$ and $ker(f)$ as subspaces of $mathbb{R}^3$?
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
$endgroup$
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
add a comment |
$begingroup$
One thing I learned early: the usual terminology here is an 'arbitary' polynomial rather than a 'random' one. 'Random' implies you want to bring in probability/measure theory, which you don't.
Now onto more serious matters. It is important that when talking about $ker(f circ g)$ and $im(f)$ you recognise the space they actually live in, which is $mathbb{R}[t]_2$. Your answers aren't in $mathbb{R}[t]_2$ but $mathbb{R}^3$, so they fail to answer the question and also cause confusion. It is vital that we correct your results to be subsets of the space they should be. For instance...
$im(f)= {ax^2+b|a,b in mathbb{R}}$, for instance. Can you do the same for $ker (f circ g)$?
For the second question, can you identify $im(g circ f)$ and $ker(f)$ as subspaces of $mathbb{R}^3$?
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
$endgroup$
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
add a comment |
$begingroup$
One thing I learned early: the usual terminology here is an 'arbitary' polynomial rather than a 'random' one. 'Random' implies you want to bring in probability/measure theory, which you don't.
Now onto more serious matters. It is important that when talking about $ker(f circ g)$ and $im(f)$ you recognise the space they actually live in, which is $mathbb{R}[t]_2$. Your answers aren't in $mathbb{R}[t]_2$ but $mathbb{R}^3$, so they fail to answer the question and also cause confusion. It is vital that we correct your results to be subsets of the space they should be. For instance...
$im(f)= {ax^2+b|a,b in mathbb{R}}$, for instance. Can you do the same for $ker (f circ g)$?
For the second question, can you identify $im(g circ f)$ and $ker(f)$ as subspaces of $mathbb{R}^3$?
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
$endgroup$
One thing I learned early: the usual terminology here is an 'arbitary' polynomial rather than a 'random' one. 'Random' implies you want to bring in probability/measure theory, which you don't.
Now onto more serious matters. It is important that when talking about $ker(f circ g)$ and $im(f)$ you recognise the space they actually live in, which is $mathbb{R}[t]_2$. Your answers aren't in $mathbb{R}[t]_2$ but $mathbb{R}^3$, so they fail to answer the question and also cause confusion. It is vital that we correct your results to be subsets of the space they should be. For instance...
$im(f)= {ax^2+b|a,b in mathbb{R}}$, for instance. Can you do the same for $ker (f circ g)$?
For the second question, can you identify $im(g circ f)$ and $ker(f)$ as subspaces of $mathbb{R}^3$?
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
answered Jan 5 at 15:36
ChessanatorChessanator
2,0091411
2,0091411
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
add a comment |
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
Yes, I've just done that in my solution
$endgroup$
– VirtualUser
Jan 5 at 16:55
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
$begingroup$
@Chessanator For $(a)$, the basis vectors are $1=(1,0,0),x=(0,1,0),x^2=(0,0,1)$. The notation might be confusing but it is not wrong.
$endgroup$
– Shubham Johri
Jan 5 at 17:49
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$begingroup$
I think you meant to write that $gin L(mathbb R[t]_2, mathbb R^3)$. Otherwise the compositions (and the way $g$ was defined) don't make sense.
$endgroup$
– Math1000
Jan 5 at 15:27