Does a family of linearly independent injective maps have a vector with linearly independent images?
$begingroup$
Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.
Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?
This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.
In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.
linear-algebra linear-transformations
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
asked Dec 17 '18 at 5:18
user277182user277182
456212
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.
Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?
This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.
In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.
linear-algebra linear-transformations
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
asked Dec 17 '18 at 5:18
user277182user277182
456212
$endgroup$
1
$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57
add a comment |
$begingroup$
Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.
Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?
This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.
In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.
linear-algebra linear-transformations
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
asked Dec 17 '18 at 5:18
user277182user277182
456212
$endgroup$
Let $V,W$ be finite dimensional vector spaces over an infinite field $k$. Fix a positive integer $n leq dim(W), dim(V)$.
Given $n$ injective linear maps $f_i:Vrightarrow W$, such that the $f_i$ are linearly independent as elements of $operatorname{Hom}(V,W)$, is there necessarily a vector $vin V$ with $f_i(v)$ linearly independent in $W$?
This holds for $n=1,2$ and if the $f_i$ commute and are diagonalisable, since then one may simultaneously diagonalise.
In terms of the conditions, we clearly require linear independence of the maps for the conclusion, and injectivity is required to exclude the case of $dim(V) > dim(W)$, where the images will never be linearly independent.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
asked Dec 17 '18 at 5:18
user277182user277182
456212
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
asked Dec 17 '18 at 5:18
user277182user277182
456212
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
edited Dec 19 '18 at 14:21
Brahadeesh
6,46942363
6,46942363
asked Dec 17 '18 at 5:18
user277182user277182
456212
asked Dec 17 '18 at 5:18
user277182user277182
456212
asked Dec 17 '18 at 5:18
user277182user277182
456212
456212
1
$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57
add a comment |
1
$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57
1
1
$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57
$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This fails for $n=3.$ Consider $f_i$ with matrix representations
$$
begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
$$
Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$
answered Dec 19 '18 at 14:03
DapDap
17.5k841
$endgroup$
add a comment |
$begingroup$
Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.
When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This fails for $n=3.$ Consider $f_i$ with matrix representations
$$
begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
$$
Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$
answered Dec 19 '18 at 14:03
DapDap
17.5k841
$endgroup$
add a comment |
$begingroup$
This fails for $n=3.$ Consider $f_i$ with matrix representations
$$
begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
$$
Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$
answered Dec 19 '18 at 14:03
DapDap
17.5k841
$endgroup$
add a comment |
$begingroup$
This fails for $n=3.$ Consider $f_i$ with matrix representations
$$
begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
$$
Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$
answered Dec 19 '18 at 14:03
DapDap
17.5k841
$endgroup$
This fails for $n=3.$ Consider $f_i$ with matrix representations
$$
begin{pmatrix}1&0&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&1&0\0&1&0\0&0&1end{pmatrix},
begin{pmatrix}1&0&1\0&1&0\0&0&1end{pmatrix}.
$$
Or in other words: $mathrm{id}, mathrm{id}+e_1e_2^*, mathrm{id}+e_1e_3^*.$ Then for any $v,$ all three vectors $f_i(v)$ lies in the space spanned by ${v,e_1}.$
answered Dec 19 '18 at 14:03
DapDap
17.5k841
answered Dec 19 '18 at 14:03
DapDap
17.5k841
answered Dec 19 '18 at 14:03
DapDap
17.5k841
answered Dec 19 '18 at 14:03
DapDap
17.5k841
17.5k841
add a comment |
add a comment |
$begingroup$
Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.
When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
$endgroup$
add a comment |
$begingroup$
Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.
When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
$endgroup$
add a comment |
$begingroup$
Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.
When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
$endgroup$
Dap's answer can be generalized as follows. First we pick a linear injection $f_{n}$, and then pick $n-1$ linearly independent maps $g_{i}$, each with the image a subspace of some $U subseteq W$ of dimension $m < n-1$. Basically, we are going to choose $f_{i} = f_{n} + lambda g_{i}$ for sufficiently small $lambda > 0$. That is, using that the determinant function is continuous (restricted to submatrices), the choice of $lambda$ ensures that the rank of $f_{i}$ is the same as $f_{n}$. Then note that the image of any $v in V$ under $f_{i}$ is in $U + mathrm{span}(f_{n}(v))$ which has dimension $m + 1 < n$. Thus we have that the $f_{i}(v)$ are linearly dependent.
When can we form this construction? It is sufficient to have $m^{2} geq n-2$, because for any basis $beta = { u_{j} : j }$ of $U$, we can let $g_{i}$ be the map that sends some $u_{j}$ to another $u_{j'}$ and everything else to zero (similarly to $e_{1}e_{2}^{ast}$), and there are $m^{2}$ such possible maps. If we take $m = n-2$ (the least restrictive case), then $m^{2} geq n-2$ for any $m geq 0$. Since we also need $m geq 1$, the construction thus works in general for any $n geq 3$.
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
answered Dec 20 '18 at 0:42
Jacob MaibachJacob Maibach
1,4802917
1,4802917
add a comment |
add a comment |
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$begingroup$
Injectivity also excludes the case $f_i = wu_i^*$ for $n$ linearly independent $u_i^*in V^*$ and an arbitrary nonzero $win W$.
$endgroup$
– Rahul
Dec 18 '18 at 5:57