Mixing
Prove that $W$ is the direct sum $text{im}(T) oplus ker (S)$
$begingroup$
The Problem:
Suppose $V,W$ are vector spaces over $F$ and $T:V to W, S:W to V$ are linear transformations. Suppose also that $S circ T$ is an isomorphism. Prove that $W$ is the direct sum $text{im}(T) oplus ker (S)$.
My Approach:
Put $A = text{im}(T)$ and $B = ker (S)$. We need to show that $A cap B = {0}$ and $W = A + B$.
For $A cap B = {0}$ (I think I've got this part): Clearly, ${0} subset A cap B$ (as $A cap B$ is a subspace of $W$). Now, if $x in A cap B subset W$, then $x = T(v)$, for some $v in V$; and so $S(x) = S(T(v)) = 0$, since $x$ is in the kernel of $S$. But since $ST$ is an isomorphism, it must map $0$ to $0$ injectively (right?); and so $T(v)$ must be $0$. Thus, since $T(v) = x$, it follows that $x = 0$; i.e., $A cap B = {0}$.
For $W = A + B$: Let $w in W$, and write $w = w_1 + w_2$ (which is obviously permitted). Then, since $S$ and $T$ are linear, $S(w) = S(w_1 + w_2) = S(w_1) + S(w_2) in V$; and so $T(S(w)) = T(S(w_1)) + T(S(w_2)) in W$. Thus, $S(w_1)$ and $S(w_2)$ are in $A$... I'm not really sure where I'm going with this...
linear-algebra
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
add a comment |
$begingroup$
The Problem:
Suppose $V,W$ are vector spaces over $F$ and $T:V to W, S:W to V$ are linear transformations. Suppose also that $S circ T$ is an isomorphism. Prove that $W$ is the direct sum $text{im}(T) oplus ker (S)$.
My Approach:
Put $A = text{im}(T)$ and $B = ker (S)$. We need to show that $A cap B = {0}$ and $W = A + B$.
For $A cap B = {0}$ (I think I've got this part): Clearly, ${0} subset A cap B$ (as $A cap B$ is a subspace of $W$). Now, if $x in A cap B subset W$, then $x = T(v)$, for some $v in V$; and so $S(x) = S(T(v)) = 0$, since $x$ is in the kernel of $S$. But since $ST$ is an isomorphism, it must map $0$ to $0$ injectively (right?); and so $T(v)$ must be $0$. Thus, since $T(v) = x$, it follows that $x = 0$; i.e., $A cap B = {0}$.
For $W = A + B$: Let $w in W$, and write $w = w_1 + w_2$ (which is obviously permitted). Then, since $S$ and $T$ are linear, $S(w) = S(w_1 + w_2) = S(w_1) + S(w_2) in V$; and so $T(S(w)) = T(S(w_1)) + T(S(w_2)) in W$. Thus, $S(w_1)$ and $S(w_2)$ are in $A$... I'm not really sure where I'm going with this...
linear-algebra
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
add a comment |
$begingroup$
The Problem:
Suppose $V,W$ are vector spaces over $F$ and $T:V to W, S:W to V$ are linear transformations. Suppose also that $S circ T$ is an isomorphism. Prove that $W$ is the direct sum $text{im}(T) oplus ker (S)$.
My Approach:
Put $A = text{im}(T)$ and $B = ker (S)$. We need to show that $A cap B = {0}$ and $W = A + B$.
For $A cap B = {0}$ (I think I've got this part): Clearly, ${0} subset A cap B$ (as $A cap B$ is a subspace of $W$). Now, if $x in A cap B subset W$, then $x = T(v)$, for some $v in V$; and so $S(x) = S(T(v)) = 0$, since $x$ is in the kernel of $S$. But since $ST$ is an isomorphism, it must map $0$ to $0$ injectively (right?); and so $T(v)$ must be $0$. Thus, since $T(v) = x$, it follows that $x = 0$; i.e., $A cap B = {0}$.
For $W = A + B$: Let $w in W$, and write $w = w_1 + w_2$ (which is obviously permitted). Then, since $S$ and $T$ are linear, $S(w) = S(w_1 + w_2) = S(w_1) + S(w_2) in V$; and so $T(S(w)) = T(S(w_1)) + T(S(w_2)) in W$. Thus, $S(w_1)$ and $S(w_2)$ are in $A$... I'm not really sure where I'm going with this...
linear-algebra
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
The Problem:
Suppose $V,W$ are vector spaces over $F$ and $T:V to W, S:W to V$ are linear transformations. Suppose also that $S circ T$ is an isomorphism. Prove that $W$ is the direct sum $text{im}(T) oplus ker (S)$.
My Approach:
Put $A = text{im}(T)$ and $B = ker (S)$. We need to show that $A cap B = {0}$ and $W = A + B$.
For $A cap B = {0}$ (I think I've got this part): Clearly, ${0} subset A cap B$ (as $A cap B$ is a subspace of $W$). Now, if $x in A cap B subset W$, then $x = T(v)$, for some $v in V$; and so $S(x) = S(T(v)) = 0$, since $x$ is in the kernel of $S$. But since $ST$ is an isomorphism, it must map $0$ to $0$ injectively (right?); and so $T(v)$ must be $0$. Thus, since $T(v) = x$, it follows that $x = 0$; i.e., $A cap B = {0}$.
For $W = A + B$: Let $w in W$, and write $w = w_1 + w_2$ (which is obviously permitted). Then, since $S$ and $T$ are linear, $S(w) = S(w_1 + w_2) = S(w_1) + S(w_2) in V$; and so $T(S(w)) = T(S(w_1)) + T(S(w_2)) in W$. Thus, $S(w_1)$ and $S(w_2)$ are in $A$... I'm not really sure where I'm going with this...
linear-algebra
linear-algebra
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 15 at 1:33
thisisourconcerndudethisisourconcerndude
1,1271022
1,1271022
add a comment |
add a comment |
2 Answers
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$begingroup$
You have not specified what $w_1$ and $w_2$ are, so it is not clear to me what are you claiming to be obviously permitted. As for the first part, you reasoning is correct.
Here's a hint: take $w in W$. Now, $S(w)$ is an element of $V$, and so by surjectivity of $ST$ we have that $S(w) = ST(y)$ for some $y$ in $V$. Note that if we were indeed successful in writing $w = T(a) + b$ for $a$ in $V$ and $b$ in $ker S$, we should necessarily have that $S(w) = ST(a)$ and $b = w - T(a)$. Can you take it from here? A solution follows:
Via the last observation, since $w = w - T(y) + T(y)$ and $T(y)$ is in the image of $T$, it suffices to see that $w-T(y)$ is in the kernel of $S$, which holds by construction: $S(w-T(y)) = S(w) - ST(y) = 0$.
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
$endgroup$
add a comment |
$begingroup$
Your proof that $A cap B = {0}$ is correct.
To show $W = A + B$, you'll need to consider the inverse of $ST$, call it $R$. Assuming for the moment that $w = a + b$ where $a in A$ and $b in B$ (only for the moment; keeping this step would make the proof circular!), we have
$$S(w) = S(a) + S(b) = S(a) = ST(x),$$
where $a = T(x)$ for some $x in V$, as $a in operatorname{Im} T$. Solving this, we obtain that
$$x = RS(w) implies a = T(x) = TRS(w).$$
This is the expression we need! So, it leaves us to show that
$$w = TRS(w) + (w - TRS(w))$$
is the decomposition we want. Note, of course, $TRS(w) in operatorname{Im} T$. We also have,
$$S(w - TRS(w)) = S(w) - (ST)RS(w) = S(w) - S(w) = 0$$
so $w - TRS(w) in operatorname{ker}(S)$, as needed.
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
$endgroup$
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2 Answers
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2 Answers
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$begingroup$
You have not specified what $w_1$ and $w_2$ are, so it is not clear to me what are you claiming to be obviously permitted. As for the first part, you reasoning is correct.
Here's a hint: take $w in W$. Now, $S(w)$ is an element of $V$, and so by surjectivity of $ST$ we have that $S(w) = ST(y)$ for some $y$ in $V$. Note that if we were indeed successful in writing $w = T(a) + b$ for $a$ in $V$ and $b$ in $ker S$, we should necessarily have that $S(w) = ST(a)$ and $b = w - T(a)$. Can you take it from here? A solution follows:
Via the last observation, since $w = w - T(y) + T(y)$ and $T(y)$ is in the image of $T$, it suffices to see that $w-T(y)$ is in the kernel of $S$, which holds by construction: $S(w-T(y)) = S(w) - ST(y) = 0$.
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
$endgroup$
add a comment |
$begingroup$
You have not specified what $w_1$ and $w_2$ are, so it is not clear to me what are you claiming to be obviously permitted. As for the first part, you reasoning is correct.
Here's a hint: take $w in W$. Now, $S(w)$ is an element of $V$, and so by surjectivity of $ST$ we have that $S(w) = ST(y)$ for some $y$ in $V$. Note that if we were indeed successful in writing $w = T(a) + b$ for $a$ in $V$ and $b$ in $ker S$, we should necessarily have that $S(w) = ST(a)$ and $b = w - T(a)$. Can you take it from here? A solution follows:
Via the last observation, since $w = w - T(y) + T(y)$ and $T(y)$ is in the image of $T$, it suffices to see that $w-T(y)$ is in the kernel of $S$, which holds by construction: $S(w-T(y)) = S(w) - ST(y) = 0$.
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
$endgroup$
add a comment |
$begingroup$
You have not specified what $w_1$ and $w_2$ are, so it is not clear to me what are you claiming to be obviously permitted. As for the first part, you reasoning is correct.
Here's a hint: take $w in W$. Now, $S(w)$ is an element of $V$, and so by surjectivity of $ST$ we have that $S(w) = ST(y)$ for some $y$ in $V$. Note that if we were indeed successful in writing $w = T(a) + b$ for $a$ in $V$ and $b$ in $ker S$, we should necessarily have that $S(w) = ST(a)$ and $b = w - T(a)$. Can you take it from here? A solution follows:
Via the last observation, since $w = w - T(y) + T(y)$ and $T(y)$ is in the image of $T$, it suffices to see that $w-T(y)$ is in the kernel of $S$, which holds by construction: $S(w-T(y)) = S(w) - ST(y) = 0$.
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
$endgroup$
You have not specified what $w_1$ and $w_2$ are, so it is not clear to me what are you claiming to be obviously permitted. As for the first part, you reasoning is correct.
Here's a hint: take $w in W$. Now, $S(w)$ is an element of $V$, and so by surjectivity of $ST$ we have that $S(w) = ST(y)$ for some $y$ in $V$. Note that if we were indeed successful in writing $w = T(a) + b$ for $a$ in $V$ and $b$ in $ker S$, we should necessarily have that $S(w) = ST(a)$ and $b = w - T(a)$. Can you take it from here? A solution follows:
Via the last observation, since $w = w - T(y) + T(y)$ and $T(y)$ is in the image of $T$, it suffices to see that $w-T(y)$ is in the kernel of $S$, which holds by construction: $S(w-T(y)) = S(w) - ST(y) = 0$.
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
edited Jan 15 at 1:53
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
answered Jan 15 at 1:46
Guido A.Guido A.
7,3821730
7,3821730
add a comment |
add a comment |
$begingroup$
Your proof that $A cap B = {0}$ is correct.
To show $W = A + B$, you'll need to consider the inverse of $ST$, call it $R$. Assuming for the moment that $w = a + b$ where $a in A$ and $b in B$ (only for the moment; keeping this step would make the proof circular!), we have
$$S(w) = S(a) + S(b) = S(a) = ST(x),$$
where $a = T(x)$ for some $x in V$, as $a in operatorname{Im} T$. Solving this, we obtain that
$$x = RS(w) implies a = T(x) = TRS(w).$$
This is the expression we need! So, it leaves us to show that
$$w = TRS(w) + (w - TRS(w))$$
is the decomposition we want. Note, of course, $TRS(w) in operatorname{Im} T$. We also have,
$$S(w - TRS(w)) = S(w) - (ST)RS(w) = S(w) - S(w) = 0$$
so $w - TRS(w) in operatorname{ker}(S)$, as needed.
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
$endgroup$
add a comment |
$begingroup$
Your proof that $A cap B = {0}$ is correct.
To show $W = A + B$, you'll need to consider the inverse of $ST$, call it $R$. Assuming for the moment that $w = a + b$ where $a in A$ and $b in B$ (only for the moment; keeping this step would make the proof circular!), we have
$$S(w) = S(a) + S(b) = S(a) = ST(x),$$
where $a = T(x)$ for some $x in V$, as $a in operatorname{Im} T$. Solving this, we obtain that
$$x = RS(w) implies a = T(x) = TRS(w).$$
This is the expression we need! So, it leaves us to show that
$$w = TRS(w) + (w - TRS(w))$$
is the decomposition we want. Note, of course, $TRS(w) in operatorname{Im} T$. We also have,
$$S(w - TRS(w)) = S(w) - (ST)RS(w) = S(w) - S(w) = 0$$
so $w - TRS(w) in operatorname{ker}(S)$, as needed.
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
$endgroup$
add a comment |
$begingroup$
Your proof that $A cap B = {0}$ is correct.
To show $W = A + B$, you'll need to consider the inverse of $ST$, call it $R$. Assuming for the moment that $w = a + b$ where $a in A$ and $b in B$ (only for the moment; keeping this step would make the proof circular!), we have
$$S(w) = S(a) + S(b) = S(a) = ST(x),$$
where $a = T(x)$ for some $x in V$, as $a in operatorname{Im} T$. Solving this, we obtain that
$$x = RS(w) implies a = T(x) = TRS(w).$$
This is the expression we need! So, it leaves us to show that
$$w = TRS(w) + (w - TRS(w))$$
is the decomposition we want. Note, of course, $TRS(w) in operatorname{Im} T$. We also have,
$$S(w - TRS(w)) = S(w) - (ST)RS(w) = S(w) - S(w) = 0$$
so $w - TRS(w) in operatorname{ker}(S)$, as needed.
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
$endgroup$
Your proof that $A cap B = {0}$ is correct.
To show $W = A + B$, you'll need to consider the inverse of $ST$, call it $R$. Assuming for the moment that $w = a + b$ where $a in A$ and $b in B$ (only for the moment; keeping this step would make the proof circular!), we have
$$S(w) = S(a) + S(b) = S(a) = ST(x),$$
where $a = T(x)$ for some $x in V$, as $a in operatorname{Im} T$. Solving this, we obtain that
$$x = RS(w) implies a = T(x) = TRS(w).$$
This is the expression we need! So, it leaves us to show that
$$w = TRS(w) + (w - TRS(w))$$
is the decomposition we want. Note, of course, $TRS(w) in operatorname{Im} T$. We also have,
$$S(w - TRS(w)) = S(w) - (ST)RS(w) = S(w) - S(w) = 0$$
so $w - TRS(w) in operatorname{ker}(S)$, as needed.
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
answered Jan 15 at 1:49
Theo BenditTheo Bendit
18.5k12152
18.5k12152
add a comment |
add a comment |
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