find the rank of the matrix if, is my answer right?
$begingroup$
if $$ B =
pmatrix{-3&-2&1 \ 1&0&1\0&-4&8\0&1&-2 \3&0&3}
$$
if $C(B)$=$N(A)$ then find the rank of matrix A ?
the answer will be as follow
$C(B)=2$
Rank of $B=2$
$N(B)=3-2=1$
$N(A)=2$
the rank $=$ number of columns $-$ the nullity
then
rank of A $=$ number of columns $-$ 2
the problem is that iam not sure about the number of A columns is it 3?
if it is then the rank of $A = 1$
if it not then any help on that and thank you in advance
linear-algebra matrices
asked Dec 12 '18 at 23:16
zolmanzolman
34
$endgroup$
|
show 1 more comment
$begingroup$
if $$ B =
pmatrix{-3&-2&1 \ 1&0&1\0&-4&8\0&1&-2 \3&0&3}
$$
if $C(B)$=$N(A)$ then find the rank of matrix A ?
the answer will be as follow
$C(B)=2$
Rank of $B=2$
$N(B)=3-2=1$
$N(A)=2$
the rank $=$ number of columns $-$ the nullity
then
rank of A $=$ number of columns $-$ 2
the problem is that iam not sure about the number of A columns is it 3?
if it is then the rank of $A = 1$
if it not then any help on that and thank you in advance
linear-algebra matrices
asked Dec 12 '18 at 23:16
zolmanzolman
34
$endgroup$
$begingroup$
What does $C(B)$ mean? How are $A$ and $B$ related?
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:20
$begingroup$
@Theo Bendit $C(B)$ all linear combination of of columns of B, they related by the relation $C(B)=N(A)$
$endgroup$
– zolman
Dec 12 '18 at 23:24
$begingroup$
In that case, I would refrain from referring to $C(B)$ as a number, as it is a space of vectors. You could talk about $operatorname{dim} C(B)$ as a number, but this is precisely $operatorname{rank} B$.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:29
$begingroup$
@Theo Bendit $C(B)$ as number it is the number of independent columns of B which is 2 and rank also is 2
$endgroup$
– zolman
Dec 12 '18 at 23:32
$begingroup$
These are different concepts! If $C(B)$ is the set of all linear combinations of columns of $B$, then it is a space of vectors. If it is the number of linearly independent columns of $B$, then it is a number (the dimension of this subspace). The latter number is the definition of "rank". There's nothing wrong with using alternative names for the rank, but it's confusing when you refer both to $C(B)$ and $operatorname{rank} B$ in the same argument!
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:35
|
show 1 more comment
$begingroup$
if $$ B =
pmatrix{-3&-2&1 \ 1&0&1\0&-4&8\0&1&-2 \3&0&3}
$$
if $C(B)$=$N(A)$ then find the rank of matrix A ?
the answer will be as follow
$C(B)=2$
Rank of $B=2$
$N(B)=3-2=1$
$N(A)=2$
the rank $=$ number of columns $-$ the nullity
then
rank of A $=$ number of columns $-$ 2
the problem is that iam not sure about the number of A columns is it 3?
if it is then the rank of $A = 1$
if it not then any help on that and thank you in advance
linear-algebra matrices
asked Dec 12 '18 at 23:16
zolmanzolman
34
$endgroup$
if $$ B =
pmatrix{-3&-2&1 \ 1&0&1\0&-4&8\0&1&-2 \3&0&3}
$$
if $C(B)$=$N(A)$ then find the rank of matrix A ?
the answer will be as follow
$C(B)=2$
Rank of $B=2$
$N(B)=3-2=1$
$N(A)=2$
the rank $=$ number of columns $-$ the nullity
then
rank of A $=$ number of columns $-$ 2
the problem is that iam not sure about the number of A columns is it 3?
if it is then the rank of $A = 1$
if it not then any help on that and thank you in advance
linear-algebra matrices
linear-algebra matrices
asked Dec 12 '18 at 23:16
zolmanzolman
34
asked Dec 12 '18 at 23:16
zolmanzolman
34
edited Dec 12 '18 at 23:28
zolman
asked Dec 12 '18 at 23:16
zolmanzolman
34
asked Dec 12 '18 at 23:16
zolmanzolman
34
asked Dec 12 '18 at 23:16
zolmanzolman
34
34
$begingroup$
What does $C(B)$ mean? How are $A$ and $B$ related?
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:20
$begingroup$
@Theo Bendit $C(B)$ all linear combination of of columns of B, they related by the relation $C(B)=N(A)$
$endgroup$
– zolman
Dec 12 '18 at 23:24
$begingroup$
In that case, I would refrain from referring to $C(B)$ as a number, as it is a space of vectors. You could talk about $operatorname{dim} C(B)$ as a number, but this is precisely $operatorname{rank} B$.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:29
$begingroup$
@Theo Bendit $C(B)$ as number it is the number of independent columns of B which is 2 and rank also is 2
$endgroup$
– zolman
Dec 12 '18 at 23:32
$begingroup$
These are different concepts! If $C(B)$ is the set of all linear combinations of columns of $B$, then it is a space of vectors. If it is the number of linearly independent columns of $B$, then it is a number (the dimension of this subspace). The latter number is the definition of "rank". There's nothing wrong with using alternative names for the rank, but it's confusing when you refer both to $C(B)$ and $operatorname{rank} B$ in the same argument!
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:35
|
show 1 more comment
$begingroup$
What does $C(B)$ mean? How are $A$ and $B$ related?
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:20
$begingroup$
@Theo Bendit $C(B)$ all linear combination of of columns of B, they related by the relation $C(B)=N(A)$
$endgroup$
– zolman
Dec 12 '18 at 23:24
$begingroup$
In that case, I would refrain from referring to $C(B)$ as a number, as it is a space of vectors. You could talk about $operatorname{dim} C(B)$ as a number, but this is precisely $operatorname{rank} B$.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:29
$begingroup$
@Theo Bendit $C(B)$ as number it is the number of independent columns of B which is 2 and rank also is 2
$endgroup$
– zolman
Dec 12 '18 at 23:32
$begingroup$
These are different concepts! If $C(B)$ is the set of all linear combinations of columns of $B$, then it is a space of vectors. If it is the number of linearly independent columns of $B$, then it is a number (the dimension of this subspace). The latter number is the definition of "rank". There's nothing wrong with using alternative names for the rank, but it's confusing when you refer both to $C(B)$ and $operatorname{rank} B$ in the same argument!
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:35
$begingroup$
What does $C(B)$ mean? How are $A$ and $B$ related?
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:20
$begingroup$
What does $C(B)$ mean? How are $A$ and $B$ related?
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:20
$begingroup$
@Theo Bendit $C(B)$ all linear combination of of columns of B, they related by the relation $C(B)=N(A)$
$endgroup$
– zolman
Dec 12 '18 at 23:24
$begingroup$
@Theo Bendit $C(B)$ all linear combination of of columns of B, they related by the relation $C(B)=N(A)$
$endgroup$
– zolman
Dec 12 '18 at 23:24
$begingroup$
In that case, I would refrain from referring to $C(B)$ as a number, as it is a space of vectors. You could talk about $operatorname{dim} C(B)$ as a number, but this is precisely $operatorname{rank} B$.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:29
$begingroup$
In that case, I would refrain from referring to $C(B)$ as a number, as it is a space of vectors. You could talk about $operatorname{dim} C(B)$ as a number, but this is precisely $operatorname{rank} B$.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:29
$begingroup$
@Theo Bendit $C(B)$ as number it is the number of independent columns of B which is 2 and rank also is 2
$endgroup$
– zolman
Dec 12 '18 at 23:32
$begingroup$
@Theo Bendit $C(B)$ as number it is the number of independent columns of B which is 2 and rank also is 2
$endgroup$
– zolman
Dec 12 '18 at 23:32
$begingroup$
These are different concepts! If $C(B)$ is the set of all linear combinations of columns of $B$, then it is a space of vectors. If it is the number of linearly independent columns of $B$, then it is a number (the dimension of this subspace). The latter number is the definition of "rank". There's nothing wrong with using alternative names for the rank, but it's confusing when you refer both to $C(B)$ and $operatorname{rank} B$ in the same argument!
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:35
$begingroup$
These are different concepts! If $C(B)$ is the set of all linear combinations of columns of $B$, then it is a space of vectors. If it is the number of linearly independent columns of $B$, then it is a number (the dimension of this subspace). The latter number is the definition of "rank". There's nothing wrong with using alternative names for the rank, but it's confusing when you refer both to $C(B)$ and $operatorname{rank} B$ in the same argument!
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:35
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The number of columns of $A$ must be $5$. You can tell this from the nullspace of $A$, which is the columnspace of $B$. Note that the columnspace of $B$ consists of vectors with $5$ components, and $A$ must multiply to each of these vectors (to produce the $0$ vector). In order for this multiplication to be well-defined, the number of columns of $A$ must be $5$.
Hence,
$$5 = operatorname{rank} A + operatorname{nullity} A = operatorname{rank} A + 2 implies operatorname{rank} A = 3.$$
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
$endgroup$
1
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
The number of columns of $A$ must be $5$. You can tell this from the nullspace of $A$, which is the columnspace of $B$. Note that the columnspace of $B$ consists of vectors with $5$ components, and $A$ must multiply to each of these vectors (to produce the $0$ vector). In order for this multiplication to be well-defined, the number of columns of $A$ must be $5$.
Hence,
$$5 = operatorname{rank} A + operatorname{nullity} A = operatorname{rank} A + 2 implies operatorname{rank} A = 3.$$
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
$endgroup$
1
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
add a comment |
$begingroup$
The number of columns of $A$ must be $5$. You can tell this from the nullspace of $A$, which is the columnspace of $B$. Note that the columnspace of $B$ consists of vectors with $5$ components, and $A$ must multiply to each of these vectors (to produce the $0$ vector). In order for this multiplication to be well-defined, the number of columns of $A$ must be $5$.
Hence,
$$5 = operatorname{rank} A + operatorname{nullity} A = operatorname{rank} A + 2 implies operatorname{rank} A = 3.$$
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
$endgroup$
1
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
add a comment |
$begingroup$
The number of columns of $A$ must be $5$. You can tell this from the nullspace of $A$, which is the columnspace of $B$. Note that the columnspace of $B$ consists of vectors with $5$ components, and $A$ must multiply to each of these vectors (to produce the $0$ vector). In order for this multiplication to be well-defined, the number of columns of $A$ must be $5$.
Hence,
$$5 = operatorname{rank} A + operatorname{nullity} A = operatorname{rank} A + 2 implies operatorname{rank} A = 3.$$
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
$endgroup$
The number of columns of $A$ must be $5$. You can tell this from the nullspace of $A$, which is the columnspace of $B$. Note that the columnspace of $B$ consists of vectors with $5$ components, and $A$ must multiply to each of these vectors (to produce the $0$ vector). In order for this multiplication to be well-defined, the number of columns of $A$ must be $5$.
Hence,
$$5 = operatorname{rank} A + operatorname{nullity} A = operatorname{rank} A + 2 implies operatorname{rank} A = 3.$$
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
answered Dec 12 '18 at 23:38
Theo BenditTheo Bendit
19k12353
19k12353
1
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
add a comment |
1
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
1
1
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
$begingroup$
thank you very much for explaining to me i was very confused about that and now it is clear thanks again
$endgroup$
– zolman
Dec 12 '18 at 23:42
add a comment |
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$begingroup$
What does $C(B)$ mean? How are $A$ and $B$ related?
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:20
$begingroup$
@Theo Bendit $C(B)$ all linear combination of of columns of B, they related by the relation $C(B)=N(A)$
$endgroup$
– zolman
Dec 12 '18 at 23:24
$begingroup$
In that case, I would refrain from referring to $C(B)$ as a number, as it is a space of vectors. You could talk about $operatorname{dim} C(B)$ as a number, but this is precisely $operatorname{rank} B$.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:29
$begingroup$
@Theo Bendit $C(B)$ as number it is the number of independent columns of B which is 2 and rank also is 2
$endgroup$
– zolman
Dec 12 '18 at 23:32
$begingroup$
These are different concepts! If $C(B)$ is the set of all linear combinations of columns of $B$, then it is a space of vectors. If it is the number of linearly independent columns of $B$, then it is a number (the dimension of this subspace). The latter number is the definition of "rank". There's nothing wrong with using alternative names for the rank, but it's confusing when you refer both to $C(B)$ and $operatorname{rank} B$ in the same argument!
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:35