Why can a 5x4 matrix never be surjective?
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My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?
linear-algebra matrices
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add a comment |
$begingroup$
My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?
linear-algebra matrices
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$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07
add a comment |
$begingroup$
My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?
linear-algebra matrices
$endgroup$
My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?
linear-algebra matrices
linear-algebra matrices
asked Dec 4 '18 at 15:03
Elliot SilverElliot Silver
304
304
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Recall that rank(T)= dim( Im(T)).
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– gimusi
Dec 4 '18 at 15:07
add a comment |
$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07
$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07
$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07
add a comment |
2 Answers
2
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oldest
votes
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A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
$operatorname{nullity}(T)=-1,$a contradiction.
$endgroup$
add a comment |
$begingroup$
HINT
Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.
What is the maximum rank for $Ain M_{5times 4}$?
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I'm supposed to justify it using the rank nullity theorem
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– Elliot Silver
Dec 4 '18 at 15:05
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@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
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– gimusi
Dec 4 '18 at 15:05
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The max rank would be 5?
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– Elliot Silver
Dec 4 '18 at 15:08
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@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
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– gimusi
Dec 4 '18 at 15:09
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Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
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– Elliot Silver
Dec 4 '18 at 15:12
|
show 2 more comments
Your Answer
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2 Answers
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2 Answers
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active
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$begingroup$
A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
$operatorname{nullity}(T)=-1,$a contradiction.
$endgroup$
add a comment |
$begingroup$
A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
$operatorname{nullity}(T)=-1,$a contradiction.
$endgroup$
add a comment |
$begingroup$
A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
$operatorname{nullity}(T)=-1,$a contradiction.
$endgroup$
A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
$operatorname{nullity}(T)=-1,$a contradiction.
edited Dec 4 '18 at 15:22
answered Dec 4 '18 at 15:13
Thomas ShelbyThomas Shelby
2,408221
2,408221
add a comment |
add a comment |
$begingroup$
HINT
Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.
What is the maximum rank for $Ain M_{5times 4}$?
$endgroup$
$begingroup$
I'm supposed to justify it using the rank nullity theorem
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:05
$begingroup$
@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
$endgroup$
– gimusi
Dec 4 '18 at 15:05
$begingroup$
The max rank would be 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:08
$begingroup$
@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
$endgroup$
– gimusi
Dec 4 '18 at 15:09
$begingroup$
Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:12
|
show 2 more comments
$begingroup$
HINT
Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.
What is the maximum rank for $Ain M_{5times 4}$?
$endgroup$
$begingroup$
I'm supposed to justify it using the rank nullity theorem
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:05
$begingroup$
@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
$endgroup$
– gimusi
Dec 4 '18 at 15:05
$begingroup$
The max rank would be 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:08
$begingroup$
@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
$endgroup$
– gimusi
Dec 4 '18 at 15:09
$begingroup$
Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:12
|
show 2 more comments
$begingroup$
HINT
Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.
What is the maximum rank for $Ain M_{5times 4}$?
$endgroup$
HINT
Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.
What is the maximum rank for $Ain M_{5times 4}$?
edited Dec 4 '18 at 15:05
answered Dec 4 '18 at 15:04
gimusigimusi
92.9k94494
92.9k94494
$begingroup$
I'm supposed to justify it using the rank nullity theorem
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:05
$begingroup$
@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
$endgroup$
– gimusi
Dec 4 '18 at 15:05
$begingroup$
The max rank would be 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:08
$begingroup$
@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
$endgroup$
– gimusi
Dec 4 '18 at 15:09
$begingroup$
Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:12
|
show 2 more comments
$begingroup$
I'm supposed to justify it using the rank nullity theorem
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:05
$begingroup$
@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
$endgroup$
– gimusi
Dec 4 '18 at 15:05
$begingroup$
The max rank would be 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:08
$begingroup$
@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
$endgroup$
– gimusi
Dec 4 '18 at 15:09
$begingroup$
Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:12
$begingroup$
I'm supposed to justify it using the rank nullity theorem
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:05
$begingroup$
I'm supposed to justify it using the rank nullity theorem
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:05
$begingroup$
@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
$endgroup$
– gimusi
Dec 4 '18 at 15:05
$begingroup$
@ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
$endgroup$
– gimusi
Dec 4 '18 at 15:05
$begingroup$
The max rank would be 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:08
$begingroup$
The max rank would be 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:08
$begingroup$
@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
$endgroup$
– gimusi
Dec 4 '18 at 15:09
$begingroup$
@ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
$endgroup$
– gimusi
Dec 4 '18 at 15:09
$begingroup$
Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:12
$begingroup$
Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
$endgroup$
– Elliot Silver
Dec 4 '18 at 15:12
|
show 2 more comments
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$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07