What exactly are bases vectors
I have watched 3 blue one brown Essence of Linear Algebra tutorials enter link description here. I want to ask one question. Let suppose we have a 3x4 transformation matrix as follows
A =
7 3 -9 8
6 2 -8 7
8 4 0 7
The basis of vector A are
{ (7,3,-9,8), (6, 2, -8, 7), (8, 4, 0, 7) }
And here is the vector which is to be transformed
B =
1 2 4 5
When we will multiply the matrix with vector it will take it from R4 --> R3.
as
17 13 51
Please tell me, when we write the multiplication in terms of linear combination it is like that.
1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]
But as the video of 3blue one brown suggested for R2 --> R2 transformation.
a( transformed vector i ) + b (transormed vector j) for vector to be transformed [a, b]
But in the case defined above it is streching its coloumns which are not vectors.
Please tell me the solution of this, or any wrong concept I have.
Thanks
linear-algebra vector-spaces vectors linear-transformations
asked Nov 26 at 15:17
Muhammad Rafique
11
add a comment |
I have watched 3 blue one brown Essence of Linear Algebra tutorials enter link description here. I want to ask one question. Let suppose we have a 3x4 transformation matrix as follows
A =
7 3 -9 8
6 2 -8 7
8 4 0 7
The basis of vector A are
{ (7,3,-9,8), (6, 2, -8, 7), (8, 4, 0, 7) }
And here is the vector which is to be transformed
B =
1 2 4 5
When we will multiply the matrix with vector it will take it from R4 --> R3.
as
17 13 51
Please tell me, when we write the multiplication in terms of linear combination it is like that.
1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]
But as the video of 3blue one brown suggested for R2 --> R2 transformation.
a( transformed vector i ) + b (transormed vector j) for vector to be transformed [a, b]
But in the case defined above it is streching its coloumns which are not vectors.
Please tell me the solution of this, or any wrong concept I have.
Thanks
linear-algebra vector-spaces vectors linear-transformations
asked Nov 26 at 15:17
Muhammad Rafique
11
When we talk about a basis, it is with respect to a vector space. $A$ is not a vector space here. Rather, the rows of $A$ form a basis for a subspace of $mathbb{R}^4.$
– Sean Roberson
Nov 26 at 15:21
Okay, yes we have 3 vectors from R4 which span R3. But my question is How exactly [ 1 2 4 5 ] are stretching the vectors of A to reach R3. In this case it is stretching in this way 1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]. But [ 7 6 8] is not a vector
– Muhammad Rafique
Nov 26 at 15:29
No. They do not span $mathbb{R}^3.$
– Sean Roberson
Nov 26 at 16:26
add a comment |
I have watched 3 blue one brown Essence of Linear Algebra tutorials enter link description here. I want to ask one question. Let suppose we have a 3x4 transformation matrix as follows
A =
7 3 -9 8
6 2 -8 7
8 4 0 7
The basis of vector A are
{ (7,3,-9,8), (6, 2, -8, 7), (8, 4, 0, 7) }
And here is the vector which is to be transformed
B =
1 2 4 5
When we will multiply the matrix with vector it will take it from R4 --> R3.
as
17 13 51
Please tell me, when we write the multiplication in terms of linear combination it is like that.
1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]
But as the video of 3blue one brown suggested for R2 --> R2 transformation.
a( transformed vector i ) + b (transormed vector j) for vector to be transformed [a, b]
But in the case defined above it is streching its coloumns which are not vectors.
Please tell me the solution of this, or any wrong concept I have.
Thanks
linear-algebra vector-spaces vectors linear-transformations
asked Nov 26 at 15:17
Muhammad Rafique
11
I have watched 3 blue one brown Essence of Linear Algebra tutorials enter link description here. I want to ask one question. Let suppose we have a 3x4 transformation matrix as follows
A =
7 3 -9 8
6 2 -8 7
8 4 0 7
The basis of vector A are
{ (7,3,-9,8), (6, 2, -8, 7), (8, 4, 0, 7) }
And here is the vector which is to be transformed
B =
1 2 4 5
When we will multiply the matrix with vector it will take it from R4 --> R3.
as
17 13 51
Please tell me, when we write the multiplication in terms of linear combination it is like that.
1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]
But as the video of 3blue one brown suggested for R2 --> R2 transformation.
a( transformed vector i ) + b (transormed vector j) for vector to be transformed [a, b]
But in the case defined above it is streching its coloumns which are not vectors.
Please tell me the solution of this, or any wrong concept I have.
Thanks
linear-algebra vector-spaces vectors linear-transformations
linear-algebra vector-spaces vectors linear-transformations
asked Nov 26 at 15:17
Muhammad Rafique
11
asked Nov 26 at 15:17
Muhammad Rafique
11
asked Nov 26 at 15:17
Muhammad Rafique
11
asked Nov 26 at 15:17
Muhammad Rafique
11
asked Nov 26 at 15:17
Muhammad Rafique
11
11
When we talk about a basis, it is with respect to a vector space. $A$ is not a vector space here. Rather, the rows of $A$ form a basis for a subspace of $mathbb{R}^4.$
– Sean Roberson
Nov 26 at 15:21
Okay, yes we have 3 vectors from R4 which span R3. But my question is How exactly [ 1 2 4 5 ] are stretching the vectors of A to reach R3. In this case it is stretching in this way 1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]. But [ 7 6 8] is not a vector
– Muhammad Rafique
Nov 26 at 15:29
No. They do not span $mathbb{R}^3.$
– Sean Roberson
Nov 26 at 16:26
add a comment |
When we talk about a basis, it is with respect to a vector space. $A$ is not a vector space here. Rather, the rows of $A$ form a basis for a subspace of $mathbb{R}^4.$
– Sean Roberson
Nov 26 at 15:21
Okay, yes we have 3 vectors from R4 which span R3. But my question is How exactly [ 1 2 4 5 ] are stretching the vectors of A to reach R3. In this case it is stretching in this way 1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]. But [ 7 6 8] is not a vector
– Muhammad Rafique
Nov 26 at 15:29
No. They do not span $mathbb{R}^3.$
– Sean Roberson
Nov 26 at 16:26
When we talk about a basis, it is with respect to a vector space. $A$ is not a vector space here. Rather, the rows of $A$ form a basis for a subspace of $mathbb{R}^4.$
– Sean Roberson
Nov 26 at 15:21
When we talk about a basis, it is with respect to a vector space. $A$ is not a vector space here. Rather, the rows of $A$ form a basis for a subspace of $mathbb{R}^4.$
– Sean Roberson
Nov 26 at 15:21
Okay, yes we have 3 vectors from R4 which span R3. But my question is How exactly [ 1 2 4 5 ] are stretching the vectors of A to reach R3. In this case it is stretching in this way 1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]. But [ 7 6 8] is not a vector
– Muhammad Rafique
Nov 26 at 15:29
Okay, yes we have 3 vectors from R4 which span R3. But my question is How exactly [ 1 2 4 5 ] are stretching the vectors of A to reach R3. In this case it is stretching in this way 1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]. But [ 7 6 8] is not a vector
– Muhammad Rafique
Nov 26 at 15:29
No. They do not span $mathbb{R}^3.$
– Sean Roberson
Nov 26 at 16:26
No. They do not span $mathbb{R}^3.$
– Sean Roberson
Nov 26 at 16:26
add a comment |
1 Answer
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The columns of the matrix are vectors too, though in this case in $mathbb R^3$ (not $mathbb R^4$ like the rows).
It is a fact that the image of a linear transformation is always the span of the $mathbf{columns}$ of any matrix representing it.
answered Nov 26 at 15:30
Chris Custer
10.8k3724
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
|
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1 Answer
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active
oldest
votes
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oldest
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active
oldest
votes
The columns of the matrix are vectors too, though in this case in $mathbb R^3$ (not $mathbb R^4$ like the rows).
It is a fact that the image of a linear transformation is always the span of the $mathbf{columns}$ of any matrix representing it.
answered Nov 26 at 15:30
Chris Custer
10.8k3724
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
|
show 1 more comment
The columns of the matrix are vectors too, though in this case in $mathbb R^3$ (not $mathbb R^4$ like the rows).
It is a fact that the image of a linear transformation is always the span of the $mathbf{columns}$ of any matrix representing it.
answered Nov 26 at 15:30
Chris Custer
10.8k3724
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
|
show 1 more comment
The columns of the matrix are vectors too, though in this case in $mathbb R^3$ (not $mathbb R^4$ like the rows).
It is a fact that the image of a linear transformation is always the span of the $mathbf{columns}$ of any matrix representing it.
answered Nov 26 at 15:30
Chris Custer
10.8k3724
The columns of the matrix are vectors too, though in this case in $mathbb R^3$ (not $mathbb R^4$ like the rows).
It is a fact that the image of a linear transformation is always the span of the $mathbf{columns}$ of any matrix representing it.
answered Nov 26 at 15:30
Chris Custer
10.8k3724
edited Nov 26 at 15:43
answered Nov 26 at 15:30
Chris Custer
10.8k3724
answered Nov 26 at 15:30
Chris Custer
10.8k3724
answered Nov 26 at 15:30
Chris Custer
10.8k3724
10.8k3724
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
|
show 1 more comment
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
But how can columns be a vector. They do not have any magnitude or direction ? Please explain me
– Muhammad Rafique
Nov 26 at 15:31
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
Any $n$-tuple can be thought of as a vector.
– Chris Custer
Nov 26 at 15:33
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
@MuhammadRafique The "definition" of a vector being something that has magnitude and direction is incorrect. See a more complete definition of vector here. That being said, $left[begin{smallmatrix}7\6\8end{smallmatrix}right]$ does happen to have magnitude and direction when treated as an element in $Bbb R^3$ with the standard euclidean bases and metric... It is $7$ units in the $i$ direction, $6$ units in the $j$ direction and $8$ units in the $k$ direction and has magnitude $sqrt{7^2+6^2+8^2}$
– JMoravitz
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
But what will be its geometric interpretation. Like if we consider the rows of this vector we can understand that they have been created using the linear combination of basis vectors of that space. But how can we explain the origin of the column vectors. It means there will be whole different geometry for row vector and column vectors ?
– Muhammad Rafique
Nov 26 at 15:35
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
@MuhammadRafique read about row spaces and column spaces, two of the fundamental subspaces related to a matrix.
– JMoravitz
Nov 26 at 15:37
|
show 1 more comment
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When we talk about a basis, it is with respect to a vector space. $A$ is not a vector space here. Rather, the rows of $A$ form a basis for a subspace of $mathbb{R}^4.$
– Sean Roberson
Nov 26 at 15:21
Okay, yes we have 3 vectors from R4 which span R3. But my question is How exactly [ 1 2 4 5 ] are stretching the vectors of A to reach R3. In this case it is stretching in this way 1 [ 7 6 8 ] + 2 [ 3 2 4] + 4 [ -9 8 0 ] + 5 [ 8 7 7]. But [ 7 6 8] is not a vector
– Muhammad Rafique
Nov 26 at 15:29
No. They do not span $mathbb{R}^3.$
– Sean Roberson
Nov 26 at 16:26