Linear dependence after a linear transformation
$begingroup$
Find an example for each of the following, or explain why no example can be found:
a) a linear map $phi_3 : M_{2times 2}(mathbf{R}) to mathbf{R}_4$ and matrices $ell_1, ldots, ell_4 in M_{2times2}(mathbf{R})$ that are linearly dependent such that $phi_3(ell_1), ldots, phi_3(ell_4)$ are linearly independent.
I believe an example cannot be found for this statement, but what would a proof for this be?
b) a linear map $phi_4 : M_{2times2}(mathbf{R}) to mathbf{R}^4$ and matrices $k_1, ldots, k_4 in M_{2times2}(mathbf{R})$ that are linearly independent such that $phi_4(k_1), ldots, phi_4(k_4)$ are linearly dependent.
I believe an example for this can be found, my example is
begin{align*}
m_1 &= left( begin{smallmatrix} {1} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_2 &= left( begin{smallmatrix} {0} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_3 &= left( begin{smallmatrix} {0} & {0}\ {1} & {0} \end{smallmatrix} right) \
m_4 &= left( begin{smallmatrix} {0} & {0}\ {0} & {1} \end{smallmatrix} right)
end{align*}
This set of matrices is clearly linearly independent, and
$$phi_4: left( begin{smallmatrix} {a} & {b}\ {c} & {d} \end{smallmatrix} right) mapsto (a-b,c+d, 2c, 2d)$$
gives $phi_4(m_1) = (0,0,0,0)$.
So as we already have a 0 vector, it does not matter what $phi_4(m_2),phi_4(m_3), phi_4(m_4)$ are, as we have already attained a 0 vector so any integer multiplied by ϕ(m1) satisfies the condition for linear dependence.
Is this example sufficient?
vector-analysis
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
asked Dec 3 '18 at 18:20
conot conot
1
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add a comment |
$begingroup$
Find an example for each of the following, or explain why no example can be found:
a) a linear map $phi_3 : M_{2times 2}(mathbf{R}) to mathbf{R}_4$ and matrices $ell_1, ldots, ell_4 in M_{2times2}(mathbf{R})$ that are linearly dependent such that $phi_3(ell_1), ldots, phi_3(ell_4)$ are linearly independent.
I believe an example cannot be found for this statement, but what would a proof for this be?
b) a linear map $phi_4 : M_{2times2}(mathbf{R}) to mathbf{R}^4$ and matrices $k_1, ldots, k_4 in M_{2times2}(mathbf{R})$ that are linearly independent such that $phi_4(k_1), ldots, phi_4(k_4)$ are linearly dependent.
I believe an example for this can be found, my example is
begin{align*}
m_1 &= left( begin{smallmatrix} {1} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_2 &= left( begin{smallmatrix} {0} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_3 &= left( begin{smallmatrix} {0} & {0}\ {1} & {0} \end{smallmatrix} right) \
m_4 &= left( begin{smallmatrix} {0} & {0}\ {0} & {1} \end{smallmatrix} right)
end{align*}
This set of matrices is clearly linearly independent, and
$$phi_4: left( begin{smallmatrix} {a} & {b}\ {c} & {d} \end{smallmatrix} right) mapsto (a-b,c+d, 2c, 2d)$$
gives $phi_4(m_1) = (0,0,0,0)$.
So as we already have a 0 vector, it does not matter what $phi_4(m_2),phi_4(m_3), phi_4(m_4)$ are, as we have already attained a 0 vector so any integer multiplied by ϕ(m1) satisfies the condition for linear dependence.
Is this example sufficient?
vector-analysis
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
asked Dec 3 '18 at 18:20
conot conot
1
$endgroup$
add a comment |
$begingroup$
Find an example for each of the following, or explain why no example can be found:
a) a linear map $phi_3 : M_{2times 2}(mathbf{R}) to mathbf{R}_4$ and matrices $ell_1, ldots, ell_4 in M_{2times2}(mathbf{R})$ that are linearly dependent such that $phi_3(ell_1), ldots, phi_3(ell_4)$ are linearly independent.
I believe an example cannot be found for this statement, but what would a proof for this be?
b) a linear map $phi_4 : M_{2times2}(mathbf{R}) to mathbf{R}^4$ and matrices $k_1, ldots, k_4 in M_{2times2}(mathbf{R})$ that are linearly independent such that $phi_4(k_1), ldots, phi_4(k_4)$ are linearly dependent.
I believe an example for this can be found, my example is
begin{align*}
m_1 &= left( begin{smallmatrix} {1} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_2 &= left( begin{smallmatrix} {0} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_3 &= left( begin{smallmatrix} {0} & {0}\ {1} & {0} \end{smallmatrix} right) \
m_4 &= left( begin{smallmatrix} {0} & {0}\ {0} & {1} \end{smallmatrix} right)
end{align*}
This set of matrices is clearly linearly independent, and
$$phi_4: left( begin{smallmatrix} {a} & {b}\ {c} & {d} \end{smallmatrix} right) mapsto (a-b,c+d, 2c, 2d)$$
gives $phi_4(m_1) = (0,0,0,0)$.
So as we already have a 0 vector, it does not matter what $phi_4(m_2),phi_4(m_3), phi_4(m_4)$ are, as we have already attained a 0 vector so any integer multiplied by ϕ(m1) satisfies the condition for linear dependence.
Is this example sufficient?
vector-analysis
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
asked Dec 3 '18 at 18:20
conot conot
1
$endgroup$
Find an example for each of the following, or explain why no example can be found:
a) a linear map $phi_3 : M_{2times 2}(mathbf{R}) to mathbf{R}_4$ and matrices $ell_1, ldots, ell_4 in M_{2times2}(mathbf{R})$ that are linearly dependent such that $phi_3(ell_1), ldots, phi_3(ell_4)$ are linearly independent.
I believe an example cannot be found for this statement, but what would a proof for this be?
b) a linear map $phi_4 : M_{2times2}(mathbf{R}) to mathbf{R}^4$ and matrices $k_1, ldots, k_4 in M_{2times2}(mathbf{R})$ that are linearly independent such that $phi_4(k_1), ldots, phi_4(k_4)$ are linearly dependent.
I believe an example for this can be found, my example is
begin{align*}
m_1 &= left( begin{smallmatrix} {1} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_2 &= left( begin{smallmatrix} {0} & {1}\ {0} & {0} \end{smallmatrix} right) \
m_3 &= left( begin{smallmatrix} {0} & {0}\ {1} & {0} \end{smallmatrix} right) \
m_4 &= left( begin{smallmatrix} {0} & {0}\ {0} & {1} \end{smallmatrix} right)
end{align*}
This set of matrices is clearly linearly independent, and
$$phi_4: left( begin{smallmatrix} {a} & {b}\ {c} & {d} \end{smallmatrix} right) mapsto (a-b,c+d, 2c, 2d)$$
gives $phi_4(m_1) = (0,0,0,0)$.
So as we already have a 0 vector, it does not matter what $phi_4(m_2),phi_4(m_3), phi_4(m_4)$ are, as we have already attained a 0 vector so any integer multiplied by ϕ(m1) satisfies the condition for linear dependence.
Is this example sufficient?
vector-analysis
vector-analysis
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
asked Dec 3 '18 at 18:20
conot conot
1
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
asked Dec 3 '18 at 18:20
conot conot
1
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
edited Dec 3 '18 at 18:34
Connor Harris
4,355724
4,355724
asked Dec 3 '18 at 18:20
conot conot
1
asked Dec 3 '18 at 18:20
conot conot
1
asked Dec 3 '18 at 18:20
conot conot
1
1
add a comment |
add a comment |
2 Answers
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$begingroup$
$l_1,l_2,l_3, l_4$ are linearly dependent
there exists some not trivial set of scalars $(c_1,c_2,c_3,c_4)$ such that $c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4 = 0$
$phi(l)$ is a linear map
$phi(c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4) = c_1phi(l_1) + c_2phi(l_2) + c_3phi(l_3) + c_4phi(l_4) = 0$
$phi(l_1),phi(l_2),phi(l_3),phi(l_4)$ are linearly dependent.
2)
Your example seems fine, but you could probably find something even simpler.
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
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$begingroup$
A: In general, if $T: V to W$ is a linear map and ${v_1, ldots, v_n} subset V$ is a linearly dependent set of vectors—i.e., if there exist $c_1, ldots, c_n$ such that $c_1 v_1 + cdots + c_n v_n = 0$—then $c_1 T(v_1) + cdots + c_n T(v_n) = T(c_1 v_1 + cdots + c_n v_n) = T(0) = 0$, so ${c_1 T(v_1), ldots, c_n T(v_n)}$ is linearly dependent. So no example can be found.
B: Any set of vectors including the zero vector is trivially linearly dependent, so your example works fine.
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
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2 Answers
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$begingroup$
$l_1,l_2,l_3, l_4$ are linearly dependent
there exists some not trivial set of scalars $(c_1,c_2,c_3,c_4)$ such that $c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4 = 0$
$phi(l)$ is a linear map
$phi(c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4) = c_1phi(l_1) + c_2phi(l_2) + c_3phi(l_3) + c_4phi(l_4) = 0$
$phi(l_1),phi(l_2),phi(l_3),phi(l_4)$ are linearly dependent.
2)
Your example seems fine, but you could probably find something even simpler.
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
$endgroup$
add a comment |
$begingroup$
$l_1,l_2,l_3, l_4$ are linearly dependent
there exists some not trivial set of scalars $(c_1,c_2,c_3,c_4)$ such that $c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4 = 0$
$phi(l)$ is a linear map
$phi(c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4) = c_1phi(l_1) + c_2phi(l_2) + c_3phi(l_3) + c_4phi(l_4) = 0$
$phi(l_1),phi(l_2),phi(l_3),phi(l_4)$ are linearly dependent.
2)
Your example seems fine, but you could probably find something even simpler.
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
$endgroup$
add a comment |
$begingroup$
$l_1,l_2,l_3, l_4$ are linearly dependent
there exists some not trivial set of scalars $(c_1,c_2,c_3,c_4)$ such that $c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4 = 0$
$phi(l)$ is a linear map
$phi(c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4) = c_1phi(l_1) + c_2phi(l_2) + c_3phi(l_3) + c_4phi(l_4) = 0$
$phi(l_1),phi(l_2),phi(l_3),phi(l_4)$ are linearly dependent.
2)
Your example seems fine, but you could probably find something even simpler.
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
$endgroup$
$l_1,l_2,l_3, l_4$ are linearly dependent
there exists some not trivial set of scalars $(c_1,c_2,c_3,c_4)$ such that $c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4 = 0$
$phi(l)$ is a linear map
$phi(c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4) = c_1phi(l_1) + c_2phi(l_2) + c_3phi(l_3) + c_4phi(l_4) = 0$
$phi(l_1),phi(l_2),phi(l_3),phi(l_4)$ are linearly dependent.
2)
Your example seems fine, but you could probably find something even simpler.
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
answered Dec 3 '18 at 18:34
Doug MDoug M
44.4k31854
44.4k31854
add a comment |
add a comment |
$begingroup$
A: In general, if $T: V to W$ is a linear map and ${v_1, ldots, v_n} subset V$ is a linearly dependent set of vectors—i.e., if there exist $c_1, ldots, c_n$ such that $c_1 v_1 + cdots + c_n v_n = 0$—then $c_1 T(v_1) + cdots + c_n T(v_n) = T(c_1 v_1 + cdots + c_n v_n) = T(0) = 0$, so ${c_1 T(v_1), ldots, c_n T(v_n)}$ is linearly dependent. So no example can be found.
B: Any set of vectors including the zero vector is trivially linearly dependent, so your example works fine.
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
$endgroup$
add a comment |
$begingroup$
A: In general, if $T: V to W$ is a linear map and ${v_1, ldots, v_n} subset V$ is a linearly dependent set of vectors—i.e., if there exist $c_1, ldots, c_n$ such that $c_1 v_1 + cdots + c_n v_n = 0$—then $c_1 T(v_1) + cdots + c_n T(v_n) = T(c_1 v_1 + cdots + c_n v_n) = T(0) = 0$, so ${c_1 T(v_1), ldots, c_n T(v_n)}$ is linearly dependent. So no example can be found.
B: Any set of vectors including the zero vector is trivially linearly dependent, so your example works fine.
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
$endgroup$
add a comment |
$begingroup$
A: In general, if $T: V to W$ is a linear map and ${v_1, ldots, v_n} subset V$ is a linearly dependent set of vectors—i.e., if there exist $c_1, ldots, c_n$ such that $c_1 v_1 + cdots + c_n v_n = 0$—then $c_1 T(v_1) + cdots + c_n T(v_n) = T(c_1 v_1 + cdots + c_n v_n) = T(0) = 0$, so ${c_1 T(v_1), ldots, c_n T(v_n)}$ is linearly dependent. So no example can be found.
B: Any set of vectors including the zero vector is trivially linearly dependent, so your example works fine.
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
$endgroup$
A: In general, if $T: V to W$ is a linear map and ${v_1, ldots, v_n} subset V$ is a linearly dependent set of vectors—i.e., if there exist $c_1, ldots, c_n$ such that $c_1 v_1 + cdots + c_n v_n = 0$—then $c_1 T(v_1) + cdots + c_n T(v_n) = T(c_1 v_1 + cdots + c_n v_n) = T(0) = 0$, so ${c_1 T(v_1), ldots, c_n T(v_n)}$ is linearly dependent. So no example can be found.
B: Any set of vectors including the zero vector is trivially linearly dependent, so your example works fine.
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
answered Dec 3 '18 at 18:39
Connor HarrisConnor Harris
4,355724
4,355724
add a comment |
add a comment |
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