Mixing
What is the basis for the kernel if the matrix has full rank?
$begingroup$
I just got a problem set back where I lost a few points on a question and I am not sure what I have done wrong.
I was given a matrix $A in mathbb R^{4 times 4}$ and I found that it has full rank, $text{rank}(A)=4$ (this part was marked correct). I was then asked to give a basis for the image and the kernel. I know from the rank-nullity theorem that the kernel has dimension zero so I wrote down:
$$text{Basis for kernel}:left{ begin{pmatrix}0 \0\0\0 end{pmatrix}right}$$
It was marked wrong and the correct answer is supposed to be "the empty set": $left { right}$. However, how can this make any sense. Isn't the zero vector always in the kernel for linear transformations?
linear-algebra
asked Jan 31 at 21:25
NullspaceNullspace
369110
$endgroup$
add a comment |
$begingroup$
I just got a problem set back where I lost a few points on a question and I am not sure what I have done wrong.
I was given a matrix $A in mathbb R^{4 times 4}$ and I found that it has full rank, $text{rank}(A)=4$ (this part was marked correct). I was then asked to give a basis for the image and the kernel. I know from the rank-nullity theorem that the kernel has dimension zero so I wrote down:
$$text{Basis for kernel}:left{ begin{pmatrix}0 \0\0\0 end{pmatrix}right}$$
It was marked wrong and the correct answer is supposed to be "the empty set": $left { right}$. However, how can this make any sense. Isn't the zero vector always in the kernel for linear transformations?
linear-algebra
asked Jan 31 at 21:25
NullspaceNullspace
369110
$endgroup$
1
$begingroup$
Basis sets must be linearly independent. That isn't.
$endgroup$
– user3482749
Jan 31 at 21:28
$begingroup$
Check this answer: math.stackexchange.com/questions/1909645/…
$endgroup$
– Shirish Kulhari
Jan 31 at 21:29
1
$begingroup$
Yes, the zero vector is in the kernel. In fact, a full-rank matrix has only the zero vector in its kernel. However, a vector space that only is the zero vector has dimension zero. The zero vector can't be part of any basis for any vector space as it is linearly dependent. All that's left is to realize that the span of the empty set is the set containing only the zero vector by virtue of the empty linear combination.
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– Thorgott
Jan 31 at 21:30
1
$begingroup$
@Thorgott Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
$begingroup$
I just got a problem set back where I lost a few points on a question and I am not sure what I have done wrong.
I was given a matrix $A in mathbb R^{4 times 4}$ and I found that it has full rank, $text{rank}(A)=4$ (this part was marked correct). I was then asked to give a basis for the image and the kernel. I know from the rank-nullity theorem that the kernel has dimension zero so I wrote down:
$$text{Basis for kernel}:left{ begin{pmatrix}0 \0\0\0 end{pmatrix}right}$$
It was marked wrong and the correct answer is supposed to be "the empty set": $left { right}$. However, how can this make any sense. Isn't the zero vector always in the kernel for linear transformations?
linear-algebra
asked Jan 31 at 21:25
NullspaceNullspace
369110
$endgroup$
I just got a problem set back where I lost a few points on a question and I am not sure what I have done wrong.
I was given a matrix $A in mathbb R^{4 times 4}$ and I found that it has full rank, $text{rank}(A)=4$ (this part was marked correct). I was then asked to give a basis for the image and the kernel. I know from the rank-nullity theorem that the kernel has dimension zero so I wrote down:
$$text{Basis for kernel}:left{ begin{pmatrix}0 \0\0\0 end{pmatrix}right}$$
It was marked wrong and the correct answer is supposed to be "the empty set": $left { right}$. However, how can this make any sense. Isn't the zero vector always in the kernel for linear transformations?
linear-algebra
linear-algebra
asked Jan 31 at 21:25
NullspaceNullspace
369110
asked Jan 31 at 21:25
NullspaceNullspace
369110
asked Jan 31 at 21:25
NullspaceNullspace
369110
asked Jan 31 at 21:25
NullspaceNullspace
369110
asked Jan 31 at 21:25
NullspaceNullspace
369110
369110
1
$begingroup$
Basis sets must be linearly independent. That isn't.
$endgroup$
– user3482749
Jan 31 at 21:28
$begingroup$
Check this answer: math.stackexchange.com/questions/1909645/…
$endgroup$
– Shirish Kulhari
Jan 31 at 21:29
1
$begingroup$
Yes, the zero vector is in the kernel. In fact, a full-rank matrix has only the zero vector in its kernel. However, a vector space that only is the zero vector has dimension zero. The zero vector can't be part of any basis for any vector space as it is linearly dependent. All that's left is to realize that the span of the empty set is the set containing only the zero vector by virtue of the empty linear combination.
$endgroup$
– Thorgott
Jan 31 at 21:30
1
$begingroup$
@Thorgott Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
1
$begingroup$
Basis sets must be linearly independent. That isn't.
$endgroup$
– user3482749
Jan 31 at 21:28
$begingroup$
Check this answer: math.stackexchange.com/questions/1909645/…
$endgroup$
– Shirish Kulhari
Jan 31 at 21:29
1
$begingroup$
Yes, the zero vector is in the kernel. In fact, a full-rank matrix has only the zero vector in its kernel. However, a vector space that only is the zero vector has dimension zero. The zero vector can't be part of any basis for any vector space as it is linearly dependent. All that's left is to realize that the span of the empty set is the set containing only the zero vector by virtue of the empty linear combination.
$endgroup$
– Thorgott
Jan 31 at 21:30
1
$begingroup$
@Thorgott Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
1
1
$begingroup$
Basis sets must be linearly independent. That isn't.
$endgroup$
– user3482749
Jan 31 at 21:28
$begingroup$
Basis sets must be linearly independent. That isn't.
$endgroup$
– user3482749
Jan 31 at 21:28
$begingroup$
Check this answer: math.stackexchange.com/questions/1909645/…
$endgroup$
– Shirish Kulhari
Jan 31 at 21:29
$begingroup$
Check this answer: math.stackexchange.com/questions/1909645/…
$endgroup$
– Shirish Kulhari
Jan 31 at 21:29
1
1
$begingroup$
Yes, the zero vector is in the kernel. In fact, a full-rank matrix has only the zero vector in its kernel. However, a vector space that only is the zero vector has dimension zero. The zero vector can't be part of any basis for any vector space as it is linearly dependent. All that's left is to realize that the span of the empty set is the set containing only the zero vector by virtue of the empty linear combination.
$endgroup$
– Thorgott
Jan 31 at 21:30
$begingroup$
Yes, the zero vector is in the kernel. In fact, a full-rank matrix has only the zero vector in its kernel. However, a vector space that only is the zero vector has dimension zero. The zero vector can't be part of any basis for any vector space as it is linearly dependent. All that's left is to realize that the span of the empty set is the set containing only the zero vector by virtue of the empty linear combination.
$endgroup$
– Thorgott
Jan 31 at 21:30
1
1
$begingroup$
@Thorgott Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
$begingroup$
@Thorgott Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
2 Answers
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$begingroup$
Indeed the kernel equals to the trivial space ${0}$. However the zero vector is linearly dependent (you can multiply it by any scalar you want to get the zero vector) so it is not a basis. The only basis of the trivial subspace is the empty set. Its span is indeed the trivial space (the smallest vector subspace which contains the empty set) and it is obviously linearly independent.
answered Jan 31 at 21:29
MarkMark
10.5k1622
$endgroup$
1
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
$begingroup$
Recall the following:
Theorem 1: Let V be an arbitrary finite-dimensional vector space and define a morphism $phi : V to V$. Then the following holds when $V = kerV oplus ImV$ : $$ text{dimV} = text{dim(kerV))} + text{dim(ImV))} $$.
This implies that, when we have full rank, $V = kerV oplus ImV = emptyset oplus ImV = ImV$
So, in the case of full rank, i.e. when $text{dimV} = text{dim(ImV)}$, the dimension of the kernel is zero. Since, $dim(kerV) = 0$, $kerV = emptyset$. Clearly, the zero vector is not a basis for the kernel in the case of full rank.
answered Jan 31 at 22:37
Victoria MVictoria M
42618
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Indeed the kernel equals to the trivial space ${0}$. However the zero vector is linearly dependent (you can multiply it by any scalar you want to get the zero vector) so it is not a basis. The only basis of the trivial subspace is the empty set. Its span is indeed the trivial space (the smallest vector subspace which contains the empty set) and it is obviously linearly independent.
answered Jan 31 at 21:29
MarkMark
10.5k1622
$endgroup$
1
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
$begingroup$
Indeed the kernel equals to the trivial space ${0}$. However the zero vector is linearly dependent (you can multiply it by any scalar you want to get the zero vector) so it is not a basis. The only basis of the trivial subspace is the empty set. Its span is indeed the trivial space (the smallest vector subspace which contains the empty set) and it is obviously linearly independent.
answered Jan 31 at 21:29
MarkMark
10.5k1622
$endgroup$
1
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
$begingroup$
Indeed the kernel equals to the trivial space ${0}$. However the zero vector is linearly dependent (you can multiply it by any scalar you want to get the zero vector) so it is not a basis. The only basis of the trivial subspace is the empty set. Its span is indeed the trivial space (the smallest vector subspace which contains the empty set) and it is obviously linearly independent.
answered Jan 31 at 21:29
MarkMark
10.5k1622
$endgroup$
Indeed the kernel equals to the trivial space ${0}$. However the zero vector is linearly dependent (you can multiply it by any scalar you want to get the zero vector) so it is not a basis. The only basis of the trivial subspace is the empty set. Its span is indeed the trivial space (the smallest vector subspace which contains the empty set) and it is obviously linearly independent.
answered Jan 31 at 21:29
MarkMark
10.5k1622
answered Jan 31 at 21:29
MarkMark
10.5k1622
answered Jan 31 at 21:29
MarkMark
10.5k1622
answered Jan 31 at 21:29
MarkMark
10.5k1622
10.5k1622
1
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
1
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
1
1
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
$begingroup$
That makes sense. Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35
add a comment |
$begingroup$
Recall the following:
Theorem 1: Let V be an arbitrary finite-dimensional vector space and define a morphism $phi : V to V$. Then the following holds when $V = kerV oplus ImV$ : $$ text{dimV} = text{dim(kerV))} + text{dim(ImV))} $$.
This implies that, when we have full rank, $V = kerV oplus ImV = emptyset oplus ImV = ImV$
So, in the case of full rank, i.e. when $text{dimV} = text{dim(ImV)}$, the dimension of the kernel is zero. Since, $dim(kerV) = 0$, $kerV = emptyset$. Clearly, the zero vector is not a basis for the kernel in the case of full rank.
answered Jan 31 at 22:37
Victoria MVictoria M
42618
$endgroup$
add a comment |
$begingroup$
Recall the following:
Theorem 1: Let V be an arbitrary finite-dimensional vector space and define a morphism $phi : V to V$. Then the following holds when $V = kerV oplus ImV$ : $$ text{dimV} = text{dim(kerV))} + text{dim(ImV))} $$.
This implies that, when we have full rank, $V = kerV oplus ImV = emptyset oplus ImV = ImV$
So, in the case of full rank, i.e. when $text{dimV} = text{dim(ImV)}$, the dimension of the kernel is zero. Since, $dim(kerV) = 0$, $kerV = emptyset$. Clearly, the zero vector is not a basis for the kernel in the case of full rank.
answered Jan 31 at 22:37
Victoria MVictoria M
42618
$endgroup$
add a comment |
$begingroup$
Recall the following:
Theorem 1: Let V be an arbitrary finite-dimensional vector space and define a morphism $phi : V to V$. Then the following holds when $V = kerV oplus ImV$ : $$ text{dimV} = text{dim(kerV))} + text{dim(ImV))} $$.
This implies that, when we have full rank, $V = kerV oplus ImV = emptyset oplus ImV = ImV$
So, in the case of full rank, i.e. when $text{dimV} = text{dim(ImV)}$, the dimension of the kernel is zero. Since, $dim(kerV) = 0$, $kerV = emptyset$. Clearly, the zero vector is not a basis for the kernel in the case of full rank.
answered Jan 31 at 22:37
Victoria MVictoria M
42618
$endgroup$
Recall the following:
Theorem 1: Let V be an arbitrary finite-dimensional vector space and define a morphism $phi : V to V$. Then the following holds when $V = kerV oplus ImV$ : $$ text{dimV} = text{dim(kerV))} + text{dim(ImV))} $$.
This implies that, when we have full rank, $V = kerV oplus ImV = emptyset oplus ImV = ImV$
So, in the case of full rank, i.e. when $text{dimV} = text{dim(ImV)}$, the dimension of the kernel is zero. Since, $dim(kerV) = 0$, $kerV = emptyset$. Clearly, the zero vector is not a basis for the kernel in the case of full rank.
answered Jan 31 at 22:37
Victoria MVictoria M
42618
answered Jan 31 at 22:37
Victoria MVictoria M
42618
answered Jan 31 at 22:37
Victoria MVictoria M
42618
answered Jan 31 at 22:37
Victoria MVictoria M
42618
42618
add a comment |
add a comment |
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1
$begingroup$
Basis sets must be linearly independent. That isn't.
$endgroup$
– user3482749
Jan 31 at 21:28
$begingroup$
Check this answer: math.stackexchange.com/questions/1909645/…
$endgroup$
– Shirish Kulhari
Jan 31 at 21:29
1
$begingroup$
Yes, the zero vector is in the kernel. In fact, a full-rank matrix has only the zero vector in its kernel. However, a vector space that only is the zero vector has dimension zero. The zero vector can't be part of any basis for any vector space as it is linearly dependent. All that's left is to realize that the span of the empty set is the set containing only the zero vector by virtue of the empty linear combination.
$endgroup$
– Thorgott
Jan 31 at 21:30
1
$begingroup$
@Thorgott Thanks for your help!
$endgroup$
– Nullspace
Jan 31 at 21:35