Proving a Linear Transformation if onto/one-to-one for vector spaces
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Given two vector spaces $V$ and $W$ of finite dimensions and a linear transformation $T: V rightarrow W$, let $m = dim(V)$ and $n = dim(W)$. Show that if $T$ is one to one, then $m leq n$ and if $T$ is onto then $m geq n$.
So I am thinking this is best approached by contradiction. Can I say that the number of vectors in the basis of $V$ is $m$ and the number of vectors in the basis of $W$ is $n$ and to assume $m geq n$. Let $V = {v_1,v_2,...,v_m}$ and $W = {w_1,w_2,...,w_n}$, if we consider the function $T: V rightarrow W$ then the number of vectors in the $span(V) leq span(W)$ so then you wouldn't have unique mapping and $T$ isn't one to one.
I'm not sure where to even begin for the onto part of the question.
linear-algebra proof-verification vector-spaces vectors linear-transformations
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
$endgroup$
add a comment |
$begingroup$
Given two vector spaces $V$ and $W$ of finite dimensions and a linear transformation $T: V rightarrow W$, let $m = dim(V)$ and $n = dim(W)$. Show that if $T$ is one to one, then $m leq n$ and if $T$ is onto then $m geq n$.
So I am thinking this is best approached by contradiction. Can I say that the number of vectors in the basis of $V$ is $m$ and the number of vectors in the basis of $W$ is $n$ and to assume $m geq n$. Let $V = {v_1,v_2,...,v_m}$ and $W = {w_1,w_2,...,w_n}$, if we consider the function $T: V rightarrow W$ then the number of vectors in the $span(V) leq span(W)$ so then you wouldn't have unique mapping and $T$ isn't one to one.
I'm not sure where to even begin for the onto part of the question.
linear-algebra proof-verification vector-spaces vectors linear-transformations
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
$endgroup$
$begingroup$
If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 6:04
add a comment |
$begingroup$
Given two vector spaces $V$ and $W$ of finite dimensions and a linear transformation $T: V rightarrow W$, let $m = dim(V)$ and $n = dim(W)$. Show that if $T$ is one to one, then $m leq n$ and if $T$ is onto then $m geq n$.
So I am thinking this is best approached by contradiction. Can I say that the number of vectors in the basis of $V$ is $m$ and the number of vectors in the basis of $W$ is $n$ and to assume $m geq n$. Let $V = {v_1,v_2,...,v_m}$ and $W = {w_1,w_2,...,w_n}$, if we consider the function $T: V rightarrow W$ then the number of vectors in the $span(V) leq span(W)$ so then you wouldn't have unique mapping and $T$ isn't one to one.
I'm not sure where to even begin for the onto part of the question.
linear-algebra proof-verification vector-spaces vectors linear-transformations
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
$endgroup$
Given two vector spaces $V$ and $W$ of finite dimensions and a linear transformation $T: V rightarrow W$, let $m = dim(V)$ and $n = dim(W)$. Show that if $T$ is one to one, then $m leq n$ and if $T$ is onto then $m geq n$.
So I am thinking this is best approached by contradiction. Can I say that the number of vectors in the basis of $V$ is $m$ and the number of vectors in the basis of $W$ is $n$ and to assume $m geq n$. Let $V = {v_1,v_2,...,v_m}$ and $W = {w_1,w_2,...,w_n}$, if we consider the function $T: V rightarrow W$ then the number of vectors in the $span(V) leq span(W)$ so then you wouldn't have unique mapping and $T$ isn't one to one.
I'm not sure where to even begin for the onto part of the question.
linear-algebra proof-verification vector-spaces vectors linear-transformations
linear-algebra proof-verification vector-spaces vectors linear-transformations
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
asked Dec 6 '18 at 5:56
FundementalJTheoremFundementalJTheorem
424
424
$begingroup$
If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 6:04
add a comment |
$begingroup$
If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 6:04
$begingroup$
If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 6:04
$begingroup$
If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 6:04
add a comment |
1 Answer
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We have $m= dim V = dim ker(T)+ dim Im(T).$
If $T$ is one-to-one, then $dim ker(T)=0$, thus $m = dim Im(T) le dim W =n.$
If $T$ is onto, then...... your turn !
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
$endgroup$
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
add a comment |
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1 Answer
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$begingroup$
We have $m= dim V = dim ker(T)+ dim Im(T).$
If $T$ is one-to-one, then $dim ker(T)=0$, thus $m = dim Im(T) le dim W =n.$
If $T$ is onto, then...... your turn !
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
$endgroup$
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
add a comment |
$begingroup$
We have $m= dim V = dim ker(T)+ dim Im(T).$
If $T$ is one-to-one, then $dim ker(T)=0$, thus $m = dim Im(T) le dim W =n.$
If $T$ is onto, then...... your turn !
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
$endgroup$
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
add a comment |
$begingroup$
We have $m= dim V = dim ker(T)+ dim Im(T).$
If $T$ is one-to-one, then $dim ker(T)=0$, thus $m = dim Im(T) le dim W =n.$
If $T$ is onto, then...... your turn !
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
$endgroup$
We have $m= dim V = dim ker(T)+ dim Im(T).$
If $T$ is one-to-one, then $dim ker(T)=0$, thus $m = dim Im(T) le dim W =n.$
If $T$ is onto, then...... your turn !
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
answered Dec 6 '18 at 6:05
FredFred
45.2k1847
45.2k1847
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
add a comment |
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $leq$ relationships?
$endgroup$
– FundementalJTheorem
Dec 6 '18 at 6:29
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
It is not correct, that if $T$ is onto, then $dim(V)=dim(W)$. Example : let $T mathbb R^2 to mathbb R$ be defined by $T(x,y)=x$.
$endgroup$
– Fred
Dec 6 '18 at 11:08
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
$begingroup$
If $T$ is onto, then $Im(T)=W$, hence $dim Im(T)=n$, thus $m= dim ker(T)+ dim Im(T)= dim ker(T)+n ge n.$
$endgroup$
– Fred
Dec 6 '18 at 11:10
add a comment |
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$begingroup$
If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 6:04