Mixing
Ranks of matrix
$begingroup$
Find the rank of the following matrix
$$begin{bmatrix}1&-1&2\2&1&3end{bmatrix}$$
My approach:
The row space exists in $R^3$ and is spanned by two vectors. Since the vectors are independent of each other(because they are not scalar multiples of each other). Therefore, the row rank of the matrix which is the rank of this is two, which is the correct answer.
However, I'm still confused as to why the answer is the answer. If the row space exists in R^3 doesn't it have be be spanned by at least three vectors. For example, the unit vectors $u_1, u_2, u_3$ span row space and are independent of each other so the rank of the space should be 3.
Can someone please tell me the flaw in my logic/understanding?
linear-algebra matrices vectors
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
$endgroup$
add a comment |
$begingroup$
Find the rank of the following matrix
$$begin{bmatrix}1&-1&2\2&1&3end{bmatrix}$$
My approach:
The row space exists in $R^3$ and is spanned by two vectors. Since the vectors are independent of each other(because they are not scalar multiples of each other). Therefore, the row rank of the matrix which is the rank of this is two, which is the correct answer.
However, I'm still confused as to why the answer is the answer. If the row space exists in R^3 doesn't it have be be spanned by at least three vectors. For example, the unit vectors $u_1, u_2, u_3$ span row space and are independent of each other so the rank of the space should be 3.
Can someone please tell me the flaw in my logic/understanding?
linear-algebra matrices vectors
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
$endgroup$
1
$begingroup$
The row space of this matrix is not the same as the space spanned by $u_1,u_2,u_3$ (which is $Bbb R^n$).
$endgroup$
– Lord Shark the Unknown
Jan 15 at 20:12
$begingroup$
@LordSharktheUnknown is the row space a subspace of $R^3$. Does that mean the space $R^2$ is a subspace of $R^3$ ?
$endgroup$
– Wilson Guo
Jan 15 at 20:17
$begingroup$
The principal component vectors do not span the row space. You cannot form the vector $(1,0,0), $ for example, as a linear combination of the two row vectors.
$endgroup$
– Doug M
Jan 15 at 20:25
2
$begingroup$
You seem to have misunderstood what “span” means. Review that definition.
$endgroup$
– amd
Jan 15 at 20:26
add a comment |
$begingroup$
Find the rank of the following matrix
$$begin{bmatrix}1&-1&2\2&1&3end{bmatrix}$$
My approach:
The row space exists in $R^3$ and is spanned by two vectors. Since the vectors are independent of each other(because they are not scalar multiples of each other). Therefore, the row rank of the matrix which is the rank of this is two, which is the correct answer.
However, I'm still confused as to why the answer is the answer. If the row space exists in R^3 doesn't it have be be spanned by at least three vectors. For example, the unit vectors $u_1, u_2, u_3$ span row space and are independent of each other so the rank of the space should be 3.
Can someone please tell me the flaw in my logic/understanding?
linear-algebra matrices vectors
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
$endgroup$
Find the rank of the following matrix
$$begin{bmatrix}1&-1&2\2&1&3end{bmatrix}$$
My approach:
The row space exists in $R^3$ and is spanned by two vectors. Since the vectors are independent of each other(because they are not scalar multiples of each other). Therefore, the row rank of the matrix which is the rank of this is two, which is the correct answer.
However, I'm still confused as to why the answer is the answer. If the row space exists in R^3 doesn't it have be be spanned by at least three vectors. For example, the unit vectors $u_1, u_2, u_3$ span row space and are independent of each other so the rank of the space should be 3.
Can someone please tell me the flaw in my logic/understanding?
linear-algebra matrices vectors
linear-algebra matrices vectors
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
asked Jan 15 at 20:11
Wilson GuoWilson Guo
273
273
1
$begingroup$
The row space of this matrix is not the same as the space spanned by $u_1,u_2,u_3$ (which is $Bbb R^n$).
$endgroup$
– Lord Shark the Unknown
Jan 15 at 20:12
$begingroup$
@LordSharktheUnknown is the row space a subspace of $R^3$. Does that mean the space $R^2$ is a subspace of $R^3$ ?
$endgroup$
– Wilson Guo
Jan 15 at 20:17
$begingroup$
The principal component vectors do not span the row space. You cannot form the vector $(1,0,0), $ for example, as a linear combination of the two row vectors.
$endgroup$
– Doug M
Jan 15 at 20:25
2
$begingroup$
You seem to have misunderstood what “span” means. Review that definition.
$endgroup$
– amd
Jan 15 at 20:26
add a comment |
1
$begingroup$
The row space of this matrix is not the same as the space spanned by $u_1,u_2,u_3$ (which is $Bbb R^n$).
$endgroup$
– Lord Shark the Unknown
Jan 15 at 20:12
$begingroup$
@LordSharktheUnknown is the row space a subspace of $R^3$. Does that mean the space $R^2$ is a subspace of $R^3$ ?
$endgroup$
– Wilson Guo
Jan 15 at 20:17
$begingroup$
The principal component vectors do not span the row space. You cannot form the vector $(1,0,0), $ for example, as a linear combination of the two row vectors.
$endgroup$
– Doug M
Jan 15 at 20:25
2
$begingroup$
You seem to have misunderstood what “span” means. Review that definition.
$endgroup$
– amd
Jan 15 at 20:26
1
1
$begingroup$
The row space of this matrix is not the same as the space spanned by $u_1,u_2,u_3$ (which is $Bbb R^n$).
$endgroup$
– Lord Shark the Unknown
Jan 15 at 20:12
$begingroup$
The row space of this matrix is not the same as the space spanned by $u_1,u_2,u_3$ (which is $Bbb R^n$).
$endgroup$
– Lord Shark the Unknown
Jan 15 at 20:12
$begingroup$
@LordSharktheUnknown is the row space a subspace of $R^3$. Does that mean the space $R^2$ is a subspace of $R^3$ ?
$endgroup$
– Wilson Guo
Jan 15 at 20:17
$begingroup$
@LordSharktheUnknown is the row space a subspace of $R^3$. Does that mean the space $R^2$ is a subspace of $R^3$ ?
$endgroup$
– Wilson Guo
Jan 15 at 20:17
$begingroup$
The principal component vectors do not span the row space. You cannot form the vector $(1,0,0), $ for example, as a linear combination of the two row vectors.
$endgroup$
– Doug M
Jan 15 at 20:25
$begingroup$
The principal component vectors do not span the row space. You cannot form the vector $(1,0,0), $ for example, as a linear combination of the two row vectors.
$endgroup$
– Doug M
Jan 15 at 20:25
2
2
$begingroup$
You seem to have misunderstood what “span” means. Review that definition.
$endgroup$
– amd
Jan 15 at 20:26
$begingroup$
You seem to have misunderstood what “span” means. Review that definition.
$endgroup$
– amd
Jan 15 at 20:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
With the case of $mathbb{R}^3$, the dimension is 3, since it has a basis that contains 3 elements.
The row space of your matrix lives as a "subspace" of the bigger structure $mathbb{R}^3$. That is, you don't view it as $mathbb{R}^3$, but rather as its own entity within $mathbb{R}^3$. It's a nicely structured chunk of $mathbb{R}^3$, if you will.
Being its own entity, it must have its own basis! The row space is filled with linear combinations of the two rows of your matrix, and since the two rows are linearly independent (as you rightfully pointed out), its basis contains only 2 elements, so its dimension is 2!
I think an example is more enlightening. Consider this simpler matrix instead:
$$begin{bmatrix}1&0&0\0&1&0end{bmatrix}$$
Its rows are linearly independent, but are elements of $mathbb{R}^3$. In this case, your intuition wouldn't have told you that the dimension is $3$ simply because they are elements of $mathbb{R}^3$, right? If you look at its row space (i.e., the linear combination of its rows), I'm sure you can see the plane, which is of dimension 2.
answered Jan 16 at 3:49
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
The rank of a matrix is simply the number of nonzero rows in reduced row echelon form (rref). If you find the Reduced row echelon form of this given matrix it will yield:
(1, 0, 5/3)
(0, 1, -1/3)
Clearly, there are 2 nonzero rows in the reduced row echelon form of the given matrix. Thus, the rank is thereby 2.
answered Jan 15 at 20:25
Taffies1Taffies1
52
$endgroup$
add a comment |
$begingroup$
The definition of rank is the number of linearly independent row vectors of a matrix. For a matrix with $n$ linearly independent col, the max of rank is $n$.
Span means the linear combination of these vectors includes all vectors in this space, which means at least there are $n$ vectors(for matrix with col $n$).
edited Jan 16 at 0:21
dantopa
6,50942244
answered Jan 15 at 20:21
eason 曲eason 曲
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With the case of $mathbb{R}^3$, the dimension is 3, since it has a basis that contains 3 elements.
The row space of your matrix lives as a "subspace" of the bigger structure $mathbb{R}^3$. That is, you don't view it as $mathbb{R}^3$, but rather as its own entity within $mathbb{R}^3$. It's a nicely structured chunk of $mathbb{R}^3$, if you will.
Being its own entity, it must have its own basis! The row space is filled with linear combinations of the two rows of your matrix, and since the two rows are linearly independent (as you rightfully pointed out), its basis contains only 2 elements, so its dimension is 2!
I think an example is more enlightening. Consider this simpler matrix instead:
$$begin{bmatrix}1&0&0\0&1&0end{bmatrix}$$
Its rows are linearly independent, but are elements of $mathbb{R}^3$. In this case, your intuition wouldn't have told you that the dimension is $3$ simply because they are elements of $mathbb{R}^3$, right? If you look at its row space (i.e., the linear combination of its rows), I'm sure you can see the plane, which is of dimension 2.
answered Jan 16 at 3:49
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
With the case of $mathbb{R}^3$, the dimension is 3, since it has a basis that contains 3 elements.
The row space of your matrix lives as a "subspace" of the bigger structure $mathbb{R}^3$. That is, you don't view it as $mathbb{R}^3$, but rather as its own entity within $mathbb{R}^3$. It's a nicely structured chunk of $mathbb{R}^3$, if you will.
Being its own entity, it must have its own basis! The row space is filled with linear combinations of the two rows of your matrix, and since the two rows are linearly independent (as you rightfully pointed out), its basis contains only 2 elements, so its dimension is 2!
I think an example is more enlightening. Consider this simpler matrix instead:
$$begin{bmatrix}1&0&0\0&1&0end{bmatrix}$$
Its rows are linearly independent, but are elements of $mathbb{R}^3$. In this case, your intuition wouldn't have told you that the dimension is $3$ simply because they are elements of $mathbb{R}^3$, right? If you look at its row space (i.e., the linear combination of its rows), I'm sure you can see the plane, which is of dimension 2.
answered Jan 16 at 3:49
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
With the case of $mathbb{R}^3$, the dimension is 3, since it has a basis that contains 3 elements.
The row space of your matrix lives as a "subspace" of the bigger structure $mathbb{R}^3$. That is, you don't view it as $mathbb{R}^3$, but rather as its own entity within $mathbb{R}^3$. It's a nicely structured chunk of $mathbb{R}^3$, if you will.
Being its own entity, it must have its own basis! The row space is filled with linear combinations of the two rows of your matrix, and since the two rows are linearly independent (as you rightfully pointed out), its basis contains only 2 elements, so its dimension is 2!
I think an example is more enlightening. Consider this simpler matrix instead:
$$begin{bmatrix}1&0&0\0&1&0end{bmatrix}$$
Its rows are linearly independent, but are elements of $mathbb{R}^3$. In this case, your intuition wouldn't have told you that the dimension is $3$ simply because they are elements of $mathbb{R}^3$, right? If you look at its row space (i.e., the linear combination of its rows), I'm sure you can see the plane, which is of dimension 2.
answered Jan 16 at 3:49
MetricMetric
1,23649
$endgroup$
With the case of $mathbb{R}^3$, the dimension is 3, since it has a basis that contains 3 elements.
The row space of your matrix lives as a "subspace" of the bigger structure $mathbb{R}^3$. That is, you don't view it as $mathbb{R}^3$, but rather as its own entity within $mathbb{R}^3$. It's a nicely structured chunk of $mathbb{R}^3$, if you will.
Being its own entity, it must have its own basis! The row space is filled with linear combinations of the two rows of your matrix, and since the two rows are linearly independent (as you rightfully pointed out), its basis contains only 2 elements, so its dimension is 2!
I think an example is more enlightening. Consider this simpler matrix instead:
$$begin{bmatrix}1&0&0\0&1&0end{bmatrix}$$
Its rows are linearly independent, but are elements of $mathbb{R}^3$. In this case, your intuition wouldn't have told you that the dimension is $3$ simply because they are elements of $mathbb{R}^3$, right? If you look at its row space (i.e., the linear combination of its rows), I'm sure you can see the plane, which is of dimension 2.
answered Jan 16 at 3:49
MetricMetric
1,23649
edited Jan 16 at 4:17
answered Jan 16 at 3:49
MetricMetric
1,23649
answered Jan 16 at 3:49
MetricMetric
1,23649
answered Jan 16 at 3:49
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
$begingroup$
The rank of a matrix is simply the number of nonzero rows in reduced row echelon form (rref). If you find the Reduced row echelon form of this given matrix it will yield:
(1, 0, 5/3)
(0, 1, -1/3)
Clearly, there are 2 nonzero rows in the reduced row echelon form of the given matrix. Thus, the rank is thereby 2.
answered Jan 15 at 20:25
Taffies1Taffies1
52
$endgroup$
add a comment |
$begingroup$
The rank of a matrix is simply the number of nonzero rows in reduced row echelon form (rref). If you find the Reduced row echelon form of this given matrix it will yield:
(1, 0, 5/3)
(0, 1, -1/3)
Clearly, there are 2 nonzero rows in the reduced row echelon form of the given matrix. Thus, the rank is thereby 2.
answered Jan 15 at 20:25
Taffies1Taffies1
52
$endgroup$
add a comment |
$begingroup$
The rank of a matrix is simply the number of nonzero rows in reduced row echelon form (rref). If you find the Reduced row echelon form of this given matrix it will yield:
(1, 0, 5/3)
(0, 1, -1/3)
Clearly, there are 2 nonzero rows in the reduced row echelon form of the given matrix. Thus, the rank is thereby 2.
answered Jan 15 at 20:25
Taffies1Taffies1
52
$endgroup$
The rank of a matrix is simply the number of nonzero rows in reduced row echelon form (rref). If you find the Reduced row echelon form of this given matrix it will yield:
(1, 0, 5/3)
(0, 1, -1/3)
Clearly, there are 2 nonzero rows in the reduced row echelon form of the given matrix. Thus, the rank is thereby 2.
answered Jan 15 at 20:25
Taffies1Taffies1
52
answered Jan 15 at 20:25
Taffies1Taffies1
52
answered Jan 15 at 20:25
Taffies1Taffies1
52
answered Jan 15 at 20:25
Taffies1Taffies1
52
52
add a comment |
add a comment |
$begingroup$
The definition of rank is the number of linearly independent row vectors of a matrix. For a matrix with $n$ linearly independent col, the max of rank is $n$.
Span means the linear combination of these vectors includes all vectors in this space, which means at least there are $n$ vectors(for matrix with col $n$).
edited Jan 16 at 0:21
dantopa
6,50942244
answered Jan 15 at 20:21
eason 曲eason 曲
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
add a comment |
$begingroup$
The definition of rank is the number of linearly independent row vectors of a matrix. For a matrix with $n$ linearly independent col, the max of rank is $n$.
Span means the linear combination of these vectors includes all vectors in this space, which means at least there are $n$ vectors(for matrix with col $n$).
edited Jan 16 at 0:21
dantopa
6,50942244
answered Jan 15 at 20:21
eason 曲eason 曲
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
add a comment |
$begingroup$
The definition of rank is the number of linearly independent row vectors of a matrix. For a matrix with $n$ linearly independent col, the max of rank is $n$.
Span means the linear combination of these vectors includes all vectors in this space, which means at least there are $n$ vectors(for matrix with col $n$).
edited Jan 16 at 0:21
dantopa
6,50942244
answered Jan 15 at 20:21
eason 曲eason 曲
1
$endgroup$
The definition of rank is the number of linearly independent row vectors of a matrix. For a matrix with $n$ linearly independent col, the max of rank is $n$.
Span means the linear combination of these vectors includes all vectors in this space, which means at least there are $n$ vectors(for matrix with col $n$).
edited Jan 16 at 0:21
dantopa
6,50942244
answered Jan 15 at 20:21
eason 曲eason 曲
1
edited Jan 16 at 0:21
dantopa
6,50942244
edited Jan 16 at 0:21
dantopa
6,50942244
edited Jan 16 at 0:21
dantopa
6,50942244
6,50942244
answered Jan 15 at 20:21
eason 曲eason 曲
1
answered Jan 15 at 20:21
eason 曲eason 曲
1
answered Jan 15 at 20:21
eason 曲eason 曲
1
1
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 16 at 0:20
add a comment |
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$begingroup$
The row space of this matrix is not the same as the space spanned by $u_1,u_2,u_3$ (which is $Bbb R^n$).
$endgroup$
– Lord Shark the Unknown
Jan 15 at 20:12
$begingroup$
@LordSharktheUnknown is the row space a subspace of $R^3$. Does that mean the space $R^2$ is a subspace of $R^3$ ?
$endgroup$
– Wilson Guo
Jan 15 at 20:17
$begingroup$
The principal component vectors do not span the row space. You cannot form the vector $(1,0,0), $ for example, as a linear combination of the two row vectors.
$endgroup$
– Doug M
Jan 15 at 20:25
2
$begingroup$
You seem to have misunderstood what “span” means. Review that definition.
$endgroup$
– amd
Jan 15 at 20:26