Mixing
Conditions of a set of vectors, so that a specific linear map is injective/surjective
$begingroup$
Let $m in Bbb N$ and $v_1,dots,v_m in V$ be distinct vectors. Furthermore let $Acolon ={v_1,dots,v_m} $ and $$T: Bbb K^m to V \ T(x_1,dots,x_m) = x_1v_1+dots+x_mv_m$$ be a linear map.
Under what conditions on $A$ is $T$ injective and surjective?
My Solution:
$A$ must be a basis for $V$, because:
Let $A$ be linearly independent: Then $lambda_1v_1+dotslambda_mv_m = 0$ has only the trivial solution.
$T$ is injective iff $ker T = {0}$. Let $k in ker T$, so $k_1v_1+dots+k_mv_m = 0$. Now using equating the coefficients, we get
$$lambda_1v_1+dots+lambda_mv_m = k_1v_1+dots+k_mv_m Rightarrow lambda_1 = k_1 = dots = lambda_m = k_m = 0$$
because of the linearly independence of $A$.
Therefore $ker T = {0}$.
Let $A$ be a spanning set of $V$:
$T$ is surjective iff $im T = V$. Because $A$ is a spanning set and $T$ can be interpreted as mapping all linear combinations of $A$, we get $im T = V$.
Is this correct?
linear-algebra abstract-algebra linear-transformations
asked Jan 20 at 10:55
strelsolstrelsol
245
$endgroup$
add a comment |
$begingroup$
Let $m in Bbb N$ and $v_1,dots,v_m in V$ be distinct vectors. Furthermore let $Acolon ={v_1,dots,v_m} $ and $$T: Bbb K^m to V \ T(x_1,dots,x_m) = x_1v_1+dots+x_mv_m$$ be a linear map.
Under what conditions on $A$ is $T$ injective and surjective?
My Solution:
$A$ must be a basis for $V$, because:
Let $A$ be linearly independent: Then $lambda_1v_1+dotslambda_mv_m = 0$ has only the trivial solution.
$T$ is injective iff $ker T = {0}$. Let $k in ker T$, so $k_1v_1+dots+k_mv_m = 0$. Now using equating the coefficients, we get
$$lambda_1v_1+dots+lambda_mv_m = k_1v_1+dots+k_mv_m Rightarrow lambda_1 = k_1 = dots = lambda_m = k_m = 0$$
because of the linearly independence of $A$.
Therefore $ker T = {0}$.
Let $A$ be a spanning set of $V$:
$T$ is surjective iff $im T = V$. Because $A$ is a spanning set and $T$ can be interpreted as mapping all linear combinations of $A$, we get $im T = V$.
Is this correct?
linear-algebra abstract-algebra linear-transformations
asked Jan 20 at 10:55
strelsolstrelsol
245
$endgroup$
add a comment |
$begingroup$
Let $m in Bbb N$ and $v_1,dots,v_m in V$ be distinct vectors. Furthermore let $Acolon ={v_1,dots,v_m} $ and $$T: Bbb K^m to V \ T(x_1,dots,x_m) = x_1v_1+dots+x_mv_m$$ be a linear map.
Under what conditions on $A$ is $T$ injective and surjective?
My Solution:
$A$ must be a basis for $V$, because:
Let $A$ be linearly independent: Then $lambda_1v_1+dotslambda_mv_m = 0$ has only the trivial solution.
$T$ is injective iff $ker T = {0}$. Let $k in ker T$, so $k_1v_1+dots+k_mv_m = 0$. Now using equating the coefficients, we get
$$lambda_1v_1+dots+lambda_mv_m = k_1v_1+dots+k_mv_m Rightarrow lambda_1 = k_1 = dots = lambda_m = k_m = 0$$
because of the linearly independence of $A$.
Therefore $ker T = {0}$.
Let $A$ be a spanning set of $V$:
$T$ is surjective iff $im T = V$. Because $A$ is a spanning set and $T$ can be interpreted as mapping all linear combinations of $A$, we get $im T = V$.
Is this correct?
linear-algebra abstract-algebra linear-transformations
asked Jan 20 at 10:55
strelsolstrelsol
245
$endgroup$
Let $m in Bbb N$ and $v_1,dots,v_m in V$ be distinct vectors. Furthermore let $Acolon ={v_1,dots,v_m} $ and $$T: Bbb K^m to V \ T(x_1,dots,x_m) = x_1v_1+dots+x_mv_m$$ be a linear map.
Under what conditions on $A$ is $T$ injective and surjective?
My Solution:
$A$ must be a basis for $V$, because:
Let $A$ be linearly independent: Then $lambda_1v_1+dotslambda_mv_m = 0$ has only the trivial solution.
$T$ is injective iff $ker T = {0}$. Let $k in ker T$, so $k_1v_1+dots+k_mv_m = 0$. Now using equating the coefficients, we get
$$lambda_1v_1+dots+lambda_mv_m = k_1v_1+dots+k_mv_m Rightarrow lambda_1 = k_1 = dots = lambda_m = k_m = 0$$
because of the linearly independence of $A$.
Therefore $ker T = {0}$.
Let $A$ be a spanning set of $V$:
$T$ is surjective iff $im T = V$. Because $A$ is a spanning set and $T$ can be interpreted as mapping all linear combinations of $A$, we get $im T = V$.
Is this correct?
linear-algebra abstract-algebra linear-transformations
linear-algebra abstract-algebra linear-transformations
asked Jan 20 at 10:55
strelsolstrelsol
245
asked Jan 20 at 10:55
strelsolstrelsol
245
asked Jan 20 at 10:55
strelsolstrelsol
245
asked Jan 20 at 10:55
strelsolstrelsol
245
asked Jan 20 at 10:55
strelsolstrelsol
245
245
add a comment |
add a comment |
1 Answer
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$begingroup$
Well, your intuition has not led you astray. These are indeed the correct characterisations of injectivity and surjectivity of $T$.
In terms of your write-up, I do have some concerns. The question could be interpreted a few ways, but I seem to think it wants a characterisation (meaning an "if and only if" condition) for $T$ being bijective. In the first part, you only showed that $A$ being linearly independent implied $T$ is injective. You should probably show the other direction too.
As for the second part, I would probably want to see more detail if I were grading this as part of an assignment. Leave no room for doubt in my mind that you know what you're talking about!
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Well, your intuition has not led you astray. These are indeed the correct characterisations of injectivity and surjectivity of $T$.
In terms of your write-up, I do have some concerns. The question could be interpreted a few ways, but I seem to think it wants a characterisation (meaning an "if and only if" condition) for $T$ being bijective. In the first part, you only showed that $A$ being linearly independent implied $T$ is injective. You should probably show the other direction too.
As for the second part, I would probably want to see more detail if I were grading this as part of an assignment. Leave no room for doubt in my mind that you know what you're talking about!
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
$endgroup$
add a comment |
$begingroup$
Well, your intuition has not led you astray. These are indeed the correct characterisations of injectivity and surjectivity of $T$.
In terms of your write-up, I do have some concerns. The question could be interpreted a few ways, but I seem to think it wants a characterisation (meaning an "if and only if" condition) for $T$ being bijective. In the first part, you only showed that $A$ being linearly independent implied $T$ is injective. You should probably show the other direction too.
As for the second part, I would probably want to see more detail if I were grading this as part of an assignment. Leave no room for doubt in my mind that you know what you're talking about!
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
$endgroup$
add a comment |
$begingroup$
Well, your intuition has not led you astray. These are indeed the correct characterisations of injectivity and surjectivity of $T$.
In terms of your write-up, I do have some concerns. The question could be interpreted a few ways, but I seem to think it wants a characterisation (meaning an "if and only if" condition) for $T$ being bijective. In the first part, you only showed that $A$ being linearly independent implied $T$ is injective. You should probably show the other direction too.
As for the second part, I would probably want to see more detail if I were grading this as part of an assignment. Leave no room for doubt in my mind that you know what you're talking about!
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
$endgroup$
Well, your intuition has not led you astray. These are indeed the correct characterisations of injectivity and surjectivity of $T$.
In terms of your write-up, I do have some concerns. The question could be interpreted a few ways, but I seem to think it wants a characterisation (meaning an "if and only if" condition) for $T$ being bijective. In the first part, you only showed that $A$ being linearly independent implied $T$ is injective. You should probably show the other direction too.
As for the second part, I would probably want to see more detail if I were grading this as part of an assignment. Leave no room for doubt in my mind that you know what you're talking about!
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
answered Jan 20 at 11:17
Theo BenditTheo Bendit
19.2k12353
19.2k12353
add a comment |
add a comment |
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