Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Prove that...
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Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Prove that $mathcal{L}(V,W)$ is infinite dimensional.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Suppose $V$ is finite-dimensional and $T_1,T_2∈L(V,W)$. Prove that $rangeT_1 ⊂ rangeT_2$ if and only if there exists $S∈L(V,V)$ such that $T_1=T_2S$Prob. 9, Sec. 3.9 in Erwin Kreyszig's INTRODUCTORY FUNCTIONAL ANALYSIS WITH APPLICATIONS: Finite-dimensional range and the form of imagesSuppose $V, $and $W$ are both finite dimensional . Prove that there exits an injective linear map from $V$ to $W$ if and only if $dimV leq dim W.$Prove that $dim range T = 1$ if and only if there is a basis of $V$ and a basis of $W$ all entries of $M(T)$ equal $1$Proof of $mathcal{L}(V,W)$ being infinite dimensionalInequality $dim(F)le K$ for all finite dimensional vector spaces of $E$ implies $dim(E)le K$Show that a set of linear maps that are not surjective, is not a subspace of all linear maps from $V$ to $W$How to prove that $text{null} ;T_1 subset text{null} ;T_2$ implies $text{dim(range}; T_1) geq text{dim(range} T_2)$?How can I prove that two null spaces are equal by using the existence of an invertible operator?Can I say dim range $T = $ dim null $S$ + range $S$ if $S in mathcal{L}(V,W)$, and $T in mathcal{L}(U,V)$?
$begingroup$
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Prove that $mathcal{L}(V,W)$ is infinite dimensional. (The question is from Linear Algebra Done Right)
My trial (I tried to prove by contradiction. But I am not sure whether this is correct):
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Suppose $mathcal{L}(V,W)$ is finite dimensional. Let $T_1, T_2, ..., T_n$ be the basis of $mathcal{L}(V,W)$. Then $Tv = span(T_1,...,T_n)$.
Define $T_j(v_j) = w_j $ for $j = 1,...,n$. Then $Tv = W = span(w_1,...,w_n)$. That contradicts with the assumption that $W$ is infinite dimensional.
I am not sure whether this is correct. I am not sure whether $Tv = W$ is correct. Should it be range $T$ but not $W$?
linear-algebra linear-transformations
asked Mar 24 at 11:27
JOHN JOHN
4589
$endgroup$
add a comment |
$begingroup$
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Prove that $mathcal{L}(V,W)$ is infinite dimensional. (The question is from Linear Algebra Done Right)
My trial (I tried to prove by contradiction. But I am not sure whether this is correct):
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Suppose $mathcal{L}(V,W)$ is finite dimensional. Let $T_1, T_2, ..., T_n$ be the basis of $mathcal{L}(V,W)$. Then $Tv = span(T_1,...,T_n)$.
Define $T_j(v_j) = w_j $ for $j = 1,...,n$. Then $Tv = W = span(w_1,...,w_n)$. That contradicts with the assumption that $W$ is infinite dimensional.
I am not sure whether this is correct. I am not sure whether $Tv = W$ is correct. Should it be range $T$ but not $W$?
linear-algebra linear-transformations
asked Mar 24 at 11:27
JOHN JOHN
4589
$endgroup$
add a comment |
$begingroup$
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Prove that $mathcal{L}(V,W)$ is infinite dimensional. (The question is from Linear Algebra Done Right)
My trial (I tried to prove by contradiction. But I am not sure whether this is correct):
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Suppose $mathcal{L}(V,W)$ is finite dimensional. Let $T_1, T_2, ..., T_n$ be the basis of $mathcal{L}(V,W)$. Then $Tv = span(T_1,...,T_n)$.
Define $T_j(v_j) = w_j $ for $j = 1,...,n$. Then $Tv = W = span(w_1,...,w_n)$. That contradicts with the assumption that $W$ is infinite dimensional.
I am not sure whether this is correct. I am not sure whether $Tv = W$ is correct. Should it be range $T$ but not $W$?
linear-algebra linear-transformations
asked Mar 24 at 11:27
JOHN JOHN
4589
$endgroup$
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Prove that $mathcal{L}(V,W)$ is infinite dimensional. (The question is from Linear Algebra Done Right)
My trial (I tried to prove by contradiction. But I am not sure whether this is correct):
Suppose $V$ is finite-dimensional with dim $V gt 0$, and $W$ is infinite dimensional. Suppose $mathcal{L}(V,W)$ is finite dimensional. Let $T_1, T_2, ..., T_n$ be the basis of $mathcal{L}(V,W)$. Then $Tv = span(T_1,...,T_n)$.
Define $T_j(v_j) = w_j $ for $j = 1,...,n$. Then $Tv = W = span(w_1,...,w_n)$. That contradicts with the assumption that $W$ is infinite dimensional.
I am not sure whether this is correct. I am not sure whether $Tv = W$ is correct. Should it be range $T$ but not $W$?
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Mar 24 at 11:27
JOHN JOHN
4589
asked Mar 24 at 11:27
JOHN JOHN
4589
asked Mar 24 at 11:27
JOHN JOHN
4589
asked Mar 24 at 11:27
JOHN JOHN
4589
asked Mar 24 at 11:27
JOHN JOHN
4589
4589
add a comment |
add a comment |
3 Answers
3
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$begingroup$
Apart from the choice of words (a basis of $mathcal{L}(V,W)$, not the basis), the proof is not good.
The idea is good, though. Suppose $mathcal{L}(V,W)$ is finite dimensional. We want to show that also $W$ is. To this end, choose a basis ${T_1,dots,T_n}$ of $mathcal{L}(V,W)$. Set
$$
W_0=operatorname{span}{T_i(v_j)mid 1le ile n, 1le jle m},
$$
where ${v_1,dots,v_m}$ is a basis of $V$. We want to show that $W_0=W$. If not, take $win W$, $wnotin W_0$. Then we can define a linear map $Tcolon Vto W$ by
$$
T(v_j)=w
$$
Since ${T_1,dots,T_n}$ is supposed to be a basis of $mathcal{L}(V,W)$, we have
$$
T=a_1T_1+a_2T_2+dots+a_nT_n
$$
which easily leads to a contradiction.
answered Mar 24 at 12:00
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
There is no such thing as the basis of $mathcal{L}(V,W)$ and you don't say what $Tv$ is.
Let $S$ be an infinite linearly independent subset of $W$ and let ${e_1,ldots,e_n}$ be a basis of $V$. For each $vin S$, let $f_vcolon Vlongrightarrow W$ be the linear map such that $f_v(e_j)=v$, for each $jin{1,ldots,n}$. It follows from the fact that $S$ is lenarly independent that ${f_v,|,vin S}$ is linearly independent too. Since $mathcal{L}(V,W)$ contains an infinite linearly independent subset, it is infinite-dimensional.
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
Let $n=dim V;(>0)$. If $K$ is the base field, we have an isomorphism $varphi: V stackrel{sim:}longrightarrow K^n$, whence an isomorphism $;mathscr L(K^n,W)stackrel{sim:}{longrightarrow}mathscr L(V,W)$.
Now we have canonical isomorphisms:
$$mathscr L(K^n,W)stackrel{sim:}{longrightarrow}bigl(mathscr L(K,W)bigr)^nquadtext{and}quadmathscr L(K,W)stackrel{sim:}{longrightarrow} W, $$
so that ultimately
$$mathscr L(V,W)stackrel{sim:}{longrightarrow} W^n.$$
answered Mar 24 at 11:46
BernardBernard
124k742117
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Apart from the choice of words (a basis of $mathcal{L}(V,W)$, not the basis), the proof is not good.
The idea is good, though. Suppose $mathcal{L}(V,W)$ is finite dimensional. We want to show that also $W$ is. To this end, choose a basis ${T_1,dots,T_n}$ of $mathcal{L}(V,W)$. Set
$$
W_0=operatorname{span}{T_i(v_j)mid 1le ile n, 1le jle m},
$$
where ${v_1,dots,v_m}$ is a basis of $V$. We want to show that $W_0=W$. If not, take $win W$, $wnotin W_0$. Then we can define a linear map $Tcolon Vto W$ by
$$
T(v_j)=w
$$
Since ${T_1,dots,T_n}$ is supposed to be a basis of $mathcal{L}(V,W)$, we have
$$
T=a_1T_1+a_2T_2+dots+a_nT_n
$$
which easily leads to a contradiction.
answered Mar 24 at 12:00
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Apart from the choice of words (a basis of $mathcal{L}(V,W)$, not the basis), the proof is not good.
The idea is good, though. Suppose $mathcal{L}(V,W)$ is finite dimensional. We want to show that also $W$ is. To this end, choose a basis ${T_1,dots,T_n}$ of $mathcal{L}(V,W)$. Set
$$
W_0=operatorname{span}{T_i(v_j)mid 1le ile n, 1le jle m},
$$
where ${v_1,dots,v_m}$ is a basis of $V$. We want to show that $W_0=W$. If not, take $win W$, $wnotin W_0$. Then we can define a linear map $Tcolon Vto W$ by
$$
T(v_j)=w
$$
Since ${T_1,dots,T_n}$ is supposed to be a basis of $mathcal{L}(V,W)$, we have
$$
T=a_1T_1+a_2T_2+dots+a_nT_n
$$
which easily leads to a contradiction.
answered Mar 24 at 12:00
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Apart from the choice of words (a basis of $mathcal{L}(V,W)$, not the basis), the proof is not good.
The idea is good, though. Suppose $mathcal{L}(V,W)$ is finite dimensional. We want to show that also $W$ is. To this end, choose a basis ${T_1,dots,T_n}$ of $mathcal{L}(V,W)$. Set
$$
W_0=operatorname{span}{T_i(v_j)mid 1le ile n, 1le jle m},
$$
where ${v_1,dots,v_m}$ is a basis of $V$. We want to show that $W_0=W$. If not, take $win W$, $wnotin W_0$. Then we can define a linear map $Tcolon Vto W$ by
$$
T(v_j)=w
$$
Since ${T_1,dots,T_n}$ is supposed to be a basis of $mathcal{L}(V,W)$, we have
$$
T=a_1T_1+a_2T_2+dots+a_nT_n
$$
which easily leads to a contradiction.
answered Mar 24 at 12:00
egregegreg
186k1486208
$endgroup$
Apart from the choice of words (a basis of $mathcal{L}(V,W)$, not the basis), the proof is not good.
The idea is good, though. Suppose $mathcal{L}(V,W)$ is finite dimensional. We want to show that also $W$ is. To this end, choose a basis ${T_1,dots,T_n}$ of $mathcal{L}(V,W)$. Set
$$
W_0=operatorname{span}{T_i(v_j)mid 1le ile n, 1le jle m},
$$
where ${v_1,dots,v_m}$ is a basis of $V$. We want to show that $W_0=W$. If not, take $win W$, $wnotin W_0$. Then we can define a linear map $Tcolon Vto W$ by
$$
T(v_j)=w
$$
Since ${T_1,dots,T_n}$ is supposed to be a basis of $mathcal{L}(V,W)$, we have
$$
T=a_1T_1+a_2T_2+dots+a_nT_n
$$
which easily leads to a contradiction.
answered Mar 24 at 12:00
egregegreg
186k1486208
answered Mar 24 at 12:00
egregegreg
186k1486208
answered Mar 24 at 12:00
egregegreg
186k1486208
answered Mar 24 at 12:00
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
$begingroup$
There is no such thing as the basis of $mathcal{L}(V,W)$ and you don't say what $Tv$ is.
Let $S$ be an infinite linearly independent subset of $W$ and let ${e_1,ldots,e_n}$ be a basis of $V$. For each $vin S$, let $f_vcolon Vlongrightarrow W$ be the linear map such that $f_v(e_j)=v$, for each $jin{1,ldots,n}$. It follows from the fact that $S$ is lenarly independent that ${f_v,|,vin S}$ is linearly independent too. Since $mathcal{L}(V,W)$ contains an infinite linearly independent subset, it is infinite-dimensional.
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
There is no such thing as the basis of $mathcal{L}(V,W)$ and you don't say what $Tv$ is.
Let $S$ be an infinite linearly independent subset of $W$ and let ${e_1,ldots,e_n}$ be a basis of $V$. For each $vin S$, let $f_vcolon Vlongrightarrow W$ be the linear map such that $f_v(e_j)=v$, for each $jin{1,ldots,n}$. It follows from the fact that $S$ is lenarly independent that ${f_v,|,vin S}$ is linearly independent too. Since $mathcal{L}(V,W)$ contains an infinite linearly independent subset, it is infinite-dimensional.
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
There is no such thing as the basis of $mathcal{L}(V,W)$ and you don't say what $Tv$ is.
Let $S$ be an infinite linearly independent subset of $W$ and let ${e_1,ldots,e_n}$ be a basis of $V$. For each $vin S$, let $f_vcolon Vlongrightarrow W$ be the linear map such that $f_v(e_j)=v$, for each $jin{1,ldots,n}$. It follows from the fact that $S$ is lenarly independent that ${f_v,|,vin S}$ is linearly independent too. Since $mathcal{L}(V,W)$ contains an infinite linearly independent subset, it is infinite-dimensional.
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
There is no such thing as the basis of $mathcal{L}(V,W)$ and you don't say what $Tv$ is.
Let $S$ be an infinite linearly independent subset of $W$ and let ${e_1,ldots,e_n}$ be a basis of $V$. For each $vin S$, let $f_vcolon Vlongrightarrow W$ be the linear map such that $f_v(e_j)=v$, for each $jin{1,ldots,n}$. It follows from the fact that $S$ is lenarly independent that ${f_v,|,vin S}$ is linearly independent too. Since $mathcal{L}(V,W)$ contains an infinite linearly independent subset, it is infinite-dimensional.
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 24 at 11:33
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
add a comment |
add a comment |
$begingroup$
Let $n=dim V;(>0)$. If $K$ is the base field, we have an isomorphism $varphi: V stackrel{sim:}longrightarrow K^n$, whence an isomorphism $;mathscr L(K^n,W)stackrel{sim:}{longrightarrow}mathscr L(V,W)$.
Now we have canonical isomorphisms:
$$mathscr L(K^n,W)stackrel{sim:}{longrightarrow}bigl(mathscr L(K,W)bigr)^nquadtext{and}quadmathscr L(K,W)stackrel{sim:}{longrightarrow} W, $$
so that ultimately
$$mathscr L(V,W)stackrel{sim:}{longrightarrow} W^n.$$
answered Mar 24 at 11:46
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
Let $n=dim V;(>0)$. If $K$ is the base field, we have an isomorphism $varphi: V stackrel{sim:}longrightarrow K^n$, whence an isomorphism $;mathscr L(K^n,W)stackrel{sim:}{longrightarrow}mathscr L(V,W)$.
Now we have canonical isomorphisms:
$$mathscr L(K^n,W)stackrel{sim:}{longrightarrow}bigl(mathscr L(K,W)bigr)^nquadtext{and}quadmathscr L(K,W)stackrel{sim:}{longrightarrow} W, $$
so that ultimately
$$mathscr L(V,W)stackrel{sim:}{longrightarrow} W^n.$$
answered Mar 24 at 11:46
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
Let $n=dim V;(>0)$. If $K$ is the base field, we have an isomorphism $varphi: V stackrel{sim:}longrightarrow K^n$, whence an isomorphism $;mathscr L(K^n,W)stackrel{sim:}{longrightarrow}mathscr L(V,W)$.
Now we have canonical isomorphisms:
$$mathscr L(K^n,W)stackrel{sim:}{longrightarrow}bigl(mathscr L(K,W)bigr)^nquadtext{and}quadmathscr L(K,W)stackrel{sim:}{longrightarrow} W, $$
so that ultimately
$$mathscr L(V,W)stackrel{sim:}{longrightarrow} W^n.$$
answered Mar 24 at 11:46
BernardBernard
124k742117
$endgroup$
Let $n=dim V;(>0)$. If $K$ is the base field, we have an isomorphism $varphi: V stackrel{sim:}longrightarrow K^n$, whence an isomorphism $;mathscr L(K^n,W)stackrel{sim:}{longrightarrow}mathscr L(V,W)$.
Now we have canonical isomorphisms:
$$mathscr L(K^n,W)stackrel{sim:}{longrightarrow}bigl(mathscr L(K,W)bigr)^nquadtext{and}quadmathscr L(K,W)stackrel{sim:}{longrightarrow} W, $$
so that ultimately
$$mathscr L(V,W)stackrel{sim:}{longrightarrow} W^n.$$
answered Mar 24 at 11:46
BernardBernard
124k742117
answered Mar 24 at 11:46
BernardBernard
124k742117
answered Mar 24 at 11:46
BernardBernard
124k742117
answered Mar 24 at 11:46
BernardBernard
124k742117
124k742117
add a comment |
add a comment |
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