Mixing
Linear Maps and Inverses
$begingroup$
$S$ and $T$ are linear transformations of $mathbb{R^3}$ which have inverses. Show that $ST$ has and inverse and that $(ST)^{-1}=T^{-1}S^{-1}$.
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
linear-algebra
asked Jan 9 at 0:11
KathySongKathySong
1,270923
$endgroup$
add a comment |
$begingroup$
$S$ and $T$ are linear transformations of $mathbb{R^3}$ which have inverses. Show that $ST$ has and inverse and that $(ST)^{-1}=T^{-1}S^{-1}$.
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
linear-algebra
asked Jan 9 at 0:11
KathySongKathySong
1,270923
$endgroup$
$begingroup$
Can you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
$S$ and $T$ are linear transformations of $mathbb{R^3}$ which have inverses. Show that $ST$ has and inverse and that $(ST)^{-1}=T^{-1}S^{-1}$.
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
linear-algebra
asked Jan 9 at 0:11
KathySongKathySong
1,270923
$endgroup$
$S$ and $T$ are linear transformations of $mathbb{R^3}$ which have inverses. Show that $ST$ has and inverse and that $(ST)^{-1}=T^{-1}S^{-1}$.
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
linear-algebra
linear-algebra
asked Jan 9 at 0:11
KathySongKathySong
1,270923
asked Jan 9 at 0:11
KathySongKathySong
1,270923
edited Jan 9 at 10:32
KathySong
asked Jan 9 at 0:11
KathySongKathySong
1,270923
asked Jan 9 at 0:11
KathySongKathySong
1,270923
asked Jan 9 at 0:11
KathySongKathySong
1,270923
1,270923
$begingroup$
Can you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
Can you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
$begingroup$
Can you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
$begingroup$
Can you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You've rather overcomplicated the issue: you know that $T^{-1}S^{-1}$ is a linear map that exists (since $S$ and $T$ have inverses), so all you need to is check that it's an inverse for ST. As a hint: this result is true in all kinds of more general spaces (all the way up to an arbitrary associative algebra), so there's no need to be messing around with evaluating your functions.
Also, $S(X)T(Y)$ is not equal to $ST(XY)$, and $T(X) = 0$ implying $X = 0$ is not sufficient for $T$ to be invertible (it also needs to be surjective).
answered Jan 9 at 0:16
user3482749user3482749
4,216919
$endgroup$
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
answered Jan 9 at 0:44
KathySongKathySong
1,270923
$endgroup$
1
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You've rather overcomplicated the issue: you know that $T^{-1}S^{-1}$ is a linear map that exists (since $S$ and $T$ have inverses), so all you need to is check that it's an inverse for ST. As a hint: this result is true in all kinds of more general spaces (all the way up to an arbitrary associative algebra), so there's no need to be messing around with evaluating your functions.
Also, $S(X)T(Y)$ is not equal to $ST(XY)$, and $T(X) = 0$ implying $X = 0$ is not sufficient for $T$ to be invertible (it also needs to be surjective).
answered Jan 9 at 0:16
user3482749user3482749
4,216919
$endgroup$
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
You've rather overcomplicated the issue: you know that $T^{-1}S^{-1}$ is a linear map that exists (since $S$ and $T$ have inverses), so all you need to is check that it's an inverse for ST. As a hint: this result is true in all kinds of more general spaces (all the way up to an arbitrary associative algebra), so there's no need to be messing around with evaluating your functions.
Also, $S(X)T(Y)$ is not equal to $ST(XY)$, and $T(X) = 0$ implying $X = 0$ is not sufficient for $T$ to be invertible (it also needs to be surjective).
answered Jan 9 at 0:16
user3482749user3482749
4,216919
$endgroup$
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
You've rather overcomplicated the issue: you know that $T^{-1}S^{-1}$ is a linear map that exists (since $S$ and $T$ have inverses), so all you need to is check that it's an inverse for ST. As a hint: this result is true in all kinds of more general spaces (all the way up to an arbitrary associative algebra), so there's no need to be messing around with evaluating your functions.
Also, $S(X)T(Y)$ is not equal to $ST(XY)$, and $T(X) = 0$ implying $X = 0$ is not sufficient for $T$ to be invertible (it also needs to be surjective).
answered Jan 9 at 0:16
user3482749user3482749
4,216919
$endgroup$
You've rather overcomplicated the issue: you know that $T^{-1}S^{-1}$ is a linear map that exists (since $S$ and $T$ have inverses), so all you need to is check that it's an inverse for ST. As a hint: this result is true in all kinds of more general spaces (all the way up to an arbitrary associative algebra), so there's no need to be messing around with evaluating your functions.
Also, $S(X)T(Y)$ is not equal to $ST(XY)$, and $T(X) = 0$ implying $X = 0$ is not sufficient for $T$ to be invertible (it also needs to be surjective).
answered Jan 9 at 0:16
user3482749user3482749
4,216919
answered Jan 9 at 0:16
user3482749user3482749
4,216919
answered Jan 9 at 0:16
user3482749user3482749
4,216919
answered Jan 9 at 0:16
user3482749user3482749
4,216919
4,216919
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
$begingroup$
may you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07
add a comment |
$begingroup$
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
answered Jan 9 at 0:44
KathySongKathySong
1,270923
$endgroup$
1
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
add a comment |
$begingroup$
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
answered Jan 9 at 0:44
KathySongKathySong
1,270923
$endgroup$
1
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
add a comment |
$begingroup$
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
answered Jan 9 at 0:44
KathySongKathySong
1,270923
$endgroup$
Let $ST(X)=X$ for any $X in mathbb{R^n}$. Since $S, T$ are linear and these have inverses, then we have $T^{-1}S^{-1}$ is linear an exist so
$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$,
hence $T^{-1}S^{-1}$ is an inverse of $ST$.
Can you check my answer?
answered Jan 9 at 0:44
KathySongKathySong
1,270923
answered Jan 9 at 0:44
KathySongKathySong
1,270923
answered Jan 9 at 0:44
KathySongKathySong
1,270923
answered Jan 9 at 0:44
KathySongKathySong
1,270923
1,270923
1
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
add a comment |
1
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
1
1
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
$begingroup$
I'd like some more detail. Thus far, you've basically just written down "it is because I say so". Something like $(T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}1T = T^{-1}T = 1$, and $(ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = S1S^{-1} = SS^{-1} = 1$, so $T^{-1}S^{-1}$ is the inverse of $ST$.
$endgroup$
– user3482749
Jan 9 at 12:08
add a comment |
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$begingroup$
Can you check my answer?
$endgroup$
– KathySong
Jan 9 at 10:07