Deciding which properties of linear transformations are true
Clash Royale CLAN TAG#URR8PPP
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For a linear transformation $T colon V to W$, choose which of the following are correct:
a) $T$ preserves vector space operations.
b) $T$ maps zero vector of $V$ to that of $W$.
c) When $dim V = m$, $dim W = n$ and $T$ is invertible then $m$ equals $n$.
d) When $dim V = m$ and $dim W = n$ then $T$ is invertible iff $m$ equals $n$.
I wonder if I'm thinking correctly:
About a), I think it means scalar multiplication and addition, so it's true.
About b), $T$ is linear, so it's true.
About c) and d), if a transformation maps $mathbbR^n=m$ to $mathbbR^n=m$ then it's invertible, so it's true.
Please correct me if I'm wrong.
linear-transformations
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
asked Aug 26 at 13:06
nik
1045
 |Â
show 2 more comments
up vote
1
down vote
favorite
For a linear transformation $T colon V to W$, choose which of the following are correct:
a) $T$ preserves vector space operations.
b) $T$ maps zero vector of $V$ to that of $W$.
c) When $dim V = m$, $dim W = n$ and $T$ is invertible then $m$ equals $n$.
d) When $dim V = m$ and $dim W = n$ then $T$ is invertible iff $m$ equals $n$.
I wonder if I'm thinking correctly:
About a), I think it means scalar multiplication and addition, so it's true.
About b), $T$ is linear, so it's true.
About c) and d), if a transformation maps $mathbbR^n=m$ to $mathbbR^n=m$ then it's invertible, so it's true.
Please correct me if I'm wrong.
linear-transformations
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
asked Aug 26 at 13:06
nik
1045
(a), (b), (c): correct; (d): false
– amsmath
Aug 26 at 13:13
" a 4 by 3 matrix A and Rank(A)=3 " Would it be a counter example?
– nik
Aug 26 at 14:13
In (d) you have $m=n$. So how can your example be a counterexample???
– amsmath
Aug 26 at 14:16
For $[M]_4 times 3 cdot x$, assume Rank(M)=3, and $x$ to span $mathbbR^3$. then wouldn't m and n equal to 3?
– nik
Aug 26 at 16:32
I thought assuming x to span R3 means n=3, and the result of transformation will have form as mx+ny+lz and since m, n, and l are L.D., it has 3 basis which means m=3. Am I going wrong way?
– nik
Aug 26 at 16:36
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For a linear transformation $T colon V to W$, choose which of the following are correct:
a) $T$ preserves vector space operations.
b) $T$ maps zero vector of $V$ to that of $W$.
c) When $dim V = m$, $dim W = n$ and $T$ is invertible then $m$ equals $n$.
d) When $dim V = m$ and $dim W = n$ then $T$ is invertible iff $m$ equals $n$.
I wonder if I'm thinking correctly:
About a), I think it means scalar multiplication and addition, so it's true.
About b), $T$ is linear, so it's true.
About c) and d), if a transformation maps $mathbbR^n=m$ to $mathbbR^n=m$ then it's invertible, so it's true.
Please correct me if I'm wrong.
linear-transformations
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
asked Aug 26 at 13:06
nik
1045
For a linear transformation $T colon V to W$, choose which of the following are correct:
a) $T$ preserves vector space operations.
b) $T$ maps zero vector of $V$ to that of $W$.
c) When $dim V = m$, $dim W = n$ and $T$ is invertible then $m$ equals $n$.
d) When $dim V = m$ and $dim W = n$ then $T$ is invertible iff $m$ equals $n$.
I wonder if I'm thinking correctly:
About a), I think it means scalar multiplication and addition, so it's true.
About b), $T$ is linear, so it's true.
About c) and d), if a transformation maps $mathbbR^n=m$ to $mathbbR^n=m$ then it's invertible, so it's true.
Please correct me if I'm wrong.
linear-transformations
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
asked Aug 26 at 13:06
nik
1045
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
edited Aug 26 at 13:13
Jendrik Stelzner
7,58221037
7,58221037
asked Aug 26 at 13:06
nik
1045
asked Aug 26 at 13:06
nik
1045
asked Aug 26 at 13:06
nik
1045
1045
(a), (b), (c): correct; (d): false
– amsmath
Aug 26 at 13:13
" a 4 by 3 matrix A and Rank(A)=3 " Would it be a counter example?
– nik
Aug 26 at 14:13
In (d) you have $m=n$. So how can your example be a counterexample???
– amsmath
Aug 26 at 14:16
For $[M]_4 times 3 cdot x$, assume Rank(M)=3, and $x$ to span $mathbbR^3$. then wouldn't m and n equal to 3?
– nik
Aug 26 at 16:32
I thought assuming x to span R3 means n=3, and the result of transformation will have form as mx+ny+lz and since m, n, and l are L.D., it has 3 basis which means m=3. Am I going wrong way?
– nik
Aug 26 at 16:36
 |Â
show 2 more comments
(a), (b), (c): correct; (d): false
– amsmath
Aug 26 at 13:13
" a 4 by 3 matrix A and Rank(A)=3 " Would it be a counter example?
– nik
Aug 26 at 14:13
In (d) you have $m=n$. So how can your example be a counterexample???
– amsmath
Aug 26 at 14:16
For $[M]_4 times 3 cdot x$, assume Rank(M)=3, and $x$ to span $mathbbR^3$. then wouldn't m and n equal to 3?
– nik
Aug 26 at 16:32
I thought assuming x to span R3 means n=3, and the result of transformation will have form as mx+ny+lz and since m, n, and l are L.D., it has 3 basis which means m=3. Am I going wrong way?
– nik
Aug 26 at 16:36
(a), (b), (c): correct; (d): false
– amsmath
Aug 26 at 13:13
(a), (b), (c): correct; (d): false
– amsmath
Aug 26 at 13:13
" a 4 by 3 matrix A and Rank(A)=3 " Would it be a counter example?
– nik
Aug 26 at 14:13
" a 4 by 3 matrix A and Rank(A)=3 " Would it be a counter example?
– nik
Aug 26 at 14:13
In (d) you have $m=n$. So how can your example be a counterexample???
– amsmath
Aug 26 at 14:16
In (d) you have $m=n$. So how can your example be a counterexample???
– amsmath
Aug 26 at 14:16
For $[M]_4 times 3 cdot x$, assume Rank(M)=3, and $x$ to span $mathbbR^3$. then wouldn't m and n equal to 3?
– nik
Aug 26 at 16:32
For $[M]_4 times 3 cdot x$, assume Rank(M)=3, and $x$ to span $mathbbR^3$. then wouldn't m and n equal to 3?
– nik
Aug 26 at 16:32
I thought assuming x to span R3 means n=3, and the result of transformation will have form as mx+ny+lz and since m, n, and l are L.D., it has 3 basis which means m=3. Am I going wrong way?
– nik
Aug 26 at 16:36
I thought assuming x to span R3 means n=3, and the result of transformation will have form as mx+ny+lz and since m, n, and l are L.D., it has 3 basis which means m=3. Am I going wrong way?
– nik
Aug 26 at 16:36
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
If a transformation maps from $mathbbR^m$ to $mathbb R^m$, it need not be invertible. For example, consider the map which takes any element in $mathbbR^m$ to the zero vector. This is neither one-one nor onto.
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
 |Â
show 6 more comments
up vote
1
down vote
a),b) and c) are true.
The iff in d) is false. Any map that has nonzero kernel, for instance, won't be invertible. (For a map to be invertible it must be one-one.) Not every map between spaces of the same dimension is invertible. For a trivial example, consider the zero map (kernel is $V$), as @Sudheesh points out... For other examples, consider any map whose rank is less than $n$ (of course, we only need one counterexample)...
answered Aug 26 at 15:33
Chris Custer
6,2752622
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If a transformation maps from $mathbbR^m$ to $mathbb R^m$, it need not be invertible. For example, consider the map which takes any element in $mathbbR^m$ to the zero vector. This is neither one-one nor onto.
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
 |Â
show 6 more comments
up vote
0
down vote
accepted
If a transformation maps from $mathbbR^m$ to $mathbb R^m$, it need not be invertible. For example, consider the map which takes any element in $mathbbR^m$ to the zero vector. This is neither one-one nor onto.
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
 |Â
show 6 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If a transformation maps from $mathbbR^m$ to $mathbb R^m$, it need not be invertible. For example, consider the map which takes any element in $mathbbR^m$ to the zero vector. This is neither one-one nor onto.
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
If a transformation maps from $mathbbR^m$ to $mathbb R^m$, it need not be invertible. For example, consider the map which takes any element in $mathbbR^m$ to the zero vector. This is neither one-one nor onto.
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
answered Aug 26 at 13:13
Sudheesh Surendranath
17417
17417
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
 |Â
show 6 more comments
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
I'm not sure how your example fits into the mentioned transformation.
– nik
Aug 26 at 13:45
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik: It does..
– amsmath
Aug 26 at 14:17
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
@nik This map is not invertible, even though it's from $mathbbR^m$ to $mathbbR^m$. Hence d) is false.
– Sudheesh Surendranath
Aug 26 at 15:27
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
Could you explain how zero vector is considered to have a dim=m ?
– nik
Aug 26 at 15:59
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
The zero vector is the image, not the codomain. The dimension of the codomain is $m$ and the dimension of the image (zero vector) is $0$. When the dimension of the image and the codomain are the same, the map is onto.
– Sudheesh Surendranath
Aug 26 at 16:36
 |Â
show 6 more comments
up vote
1
down vote
a),b) and c) are true.
The iff in d) is false. Any map that has nonzero kernel, for instance, won't be invertible. (For a map to be invertible it must be one-one.) Not every map between spaces of the same dimension is invertible. For a trivial example, consider the zero map (kernel is $V$), as @Sudheesh points out... For other examples, consider any map whose rank is less than $n$ (of course, we only need one counterexample)...
answered Aug 26 at 15:33
Chris Custer
6,2752622
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
 |Â
show 2 more comments
up vote
1
down vote
a),b) and c) are true.
The iff in d) is false. Any map that has nonzero kernel, for instance, won't be invertible. (For a map to be invertible it must be one-one.) Not every map between spaces of the same dimension is invertible. For a trivial example, consider the zero map (kernel is $V$), as @Sudheesh points out... For other examples, consider any map whose rank is less than $n$ (of course, we only need one counterexample)...
answered Aug 26 at 15:33
Chris Custer
6,2752622
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
a),b) and c) are true.
The iff in d) is false. Any map that has nonzero kernel, for instance, won't be invertible. (For a map to be invertible it must be one-one.) Not every map between spaces of the same dimension is invertible. For a trivial example, consider the zero map (kernel is $V$), as @Sudheesh points out... For other examples, consider any map whose rank is less than $n$ (of course, we only need one counterexample)...
answered Aug 26 at 15:33
Chris Custer
6,2752622
a),b) and c) are true.
The iff in d) is false. Any map that has nonzero kernel, for instance, won't be invertible. (For a map to be invertible it must be one-one.) Not every map between spaces of the same dimension is invertible. For a trivial example, consider the zero map (kernel is $V$), as @Sudheesh points out... For other examples, consider any map whose rank is less than $n$ (of course, we only need one counterexample)...
answered Aug 26 at 15:33
Chris Custer
6,2752622
answered Aug 26 at 15:33
Chris Custer
6,2752622
answered Aug 26 at 15:33
Chris Custer
6,2752622
answered Aug 26 at 15:33
Chris Custer
6,2752622
6,2752622
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
 |Â
show 2 more comments
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
Thanks. 'kernel is V' gave me some hint. I'll think about it more.
– nik
Aug 26 at 17:26
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
$underbrace(0,0,dots,0)_textm-times$ is $vec0inmathbb R^m$, btw
– Chris Custer
Aug 26 at 17:58
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
I can understand it does belong to $mathbbR^m$, but how does it alone have dimension of m? Shouldn't there be m L.D. vectors?
– nik
Aug 27 at 6:33
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
It has $m$ coordinates. Thus it is the $m$-dimensional zero vector. Linear dependence is a different notion. Good question, incidentally...
– Chris Custer
Aug 27 at 8:25
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
Sometimes just $0$ is written.
– Chris Custer
Aug 27 at 8:27
 |Â
show 2 more comments
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(a), (b), (c): correct; (d): false
– amsmath
Aug 26 at 13:13
" a 4 by 3 matrix A and Rank(A)=3 " Would it be a counter example?
– nik
Aug 26 at 14:13
In (d) you have $m=n$. So how can your example be a counterexample???
– amsmath
Aug 26 at 14:16
For $[M]_4 times 3 cdot x$, assume Rank(M)=3, and $x$ to span $mathbbR^3$. then wouldn't m and n equal to 3?
– nik
Aug 26 at 16:32
I thought assuming x to span R3 means n=3, and the result of transformation will have form as mx+ny+lz and since m, n, and l are L.D., it has 3 basis which means m=3. Am I going wrong way?
– nik
Aug 26 at 16:36