Showing a set of subspaces are subspaces of a homomorphism?
$begingroup$
another question from a problem set that I am struggling to answer. Any guidance or help would be appreciated. I don't even really understand what the question is asking.
Let $V$ be a nonzero vector space and let $T in text{Hom}(V, V )$. Show that ${S in Hom(V, V ) : S circ T = T circ S }$ is a subspace of $ text{Hom}(V, V )$.
linear-algebra vector-spaces proof-writing dual-spaces
edited Jan 4 at 5:54
André 3000
12.8k22243
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
$endgroup$
add a comment |
$begingroup$
another question from a problem set that I am struggling to answer. Any guidance or help would be appreciated. I don't even really understand what the question is asking.
Let $V$ be a nonzero vector space and let $T in text{Hom}(V, V )$. Show that ${S in Hom(V, V ) : S circ T = T circ S }$ is a subspace of $ text{Hom}(V, V )$.
linear-algebra vector-spaces proof-writing dual-spaces
edited Jan 4 at 5:54
André 3000
12.8k22243
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
$endgroup$
add a comment |
$begingroup$
another question from a problem set that I am struggling to answer. Any guidance or help would be appreciated. I don't even really understand what the question is asking.
Let $V$ be a nonzero vector space and let $T in text{Hom}(V, V )$. Show that ${S in Hom(V, V ) : S circ T = T circ S }$ is a subspace of $ text{Hom}(V, V )$.
linear-algebra vector-spaces proof-writing dual-spaces
edited Jan 4 at 5:54
André 3000
12.8k22243
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
$endgroup$
another question from a problem set that I am struggling to answer. Any guidance or help would be appreciated. I don't even really understand what the question is asking.
Let $V$ be a nonzero vector space and let $T in text{Hom}(V, V )$. Show that ${S in Hom(V, V ) : S circ T = T circ S }$ is a subspace of $ text{Hom}(V, V )$.
linear-algebra vector-spaces proof-writing dual-spaces
linear-algebra vector-spaces proof-writing dual-spaces
edited Jan 4 at 5:54
André 3000
12.8k22243
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
edited Jan 4 at 5:54
André 3000
12.8k22243
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
edited Jan 4 at 5:54
André 3000
12.8k22243
edited Jan 4 at 5:54
André 3000
12.8k22243
edited Jan 4 at 5:54
André 3000
12.8k22243
12.8k22243
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
asked Jun 5 '17 at 11:55
Vivek KatialVivek Katial
113
113
add a comment |
add a comment |
1 Answer
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Let us call the given set K. You will be done if you can show the following two steps:
If $S_1,S_2 in K$, then $S_1 + S_2 in K$.
If $S in K$, then $kS in K$ for all $k$.
To show the first, note that $(S_1+S_2) circ T = S_1 circ T + S_2 circ T = T circ S_1 + T circ S_2 = T circ (S_1 + S_2)$.
The second simply follows: $(kS) circ T = k(S circ T) = k(T circ S) = T circ (kS)$.
You can verify these steps yourself, but do ask if doubts persist. The above two steps show that $K$ is closed under addition, and scalar multiplication respectively, so $K$ is a subspace of $operatorname{Hom} (V,V)$.
ADDENDUM : I should mention, thanks to the comment below, that $K$ is non-empty , since $T$ commutes with itself, hence is in $K$. (The identity map is in $K$ as well, as is the zero map!)
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
$endgroup$
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Let us call the given set K. You will be done if you can show the following two steps:
If $S_1,S_2 in K$, then $S_1 + S_2 in K$.
If $S in K$, then $kS in K$ for all $k$.
To show the first, note that $(S_1+S_2) circ T = S_1 circ T + S_2 circ T = T circ S_1 + T circ S_2 = T circ (S_1 + S_2)$.
The second simply follows: $(kS) circ T = k(S circ T) = k(T circ S) = T circ (kS)$.
You can verify these steps yourself, but do ask if doubts persist. The above two steps show that $K$ is closed under addition, and scalar multiplication respectively, so $K$ is a subspace of $operatorname{Hom} (V,V)$.
ADDENDUM : I should mention, thanks to the comment below, that $K$ is non-empty , since $T$ commutes with itself, hence is in $K$. (The identity map is in $K$ as well, as is the zero map!)
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
$endgroup$
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
add a comment |
$begingroup$
Let us call the given set K. You will be done if you can show the following two steps:
If $S_1,S_2 in K$, then $S_1 + S_2 in K$.
If $S in K$, then $kS in K$ for all $k$.
To show the first, note that $(S_1+S_2) circ T = S_1 circ T + S_2 circ T = T circ S_1 + T circ S_2 = T circ (S_1 + S_2)$.
The second simply follows: $(kS) circ T = k(S circ T) = k(T circ S) = T circ (kS)$.
You can verify these steps yourself, but do ask if doubts persist. The above two steps show that $K$ is closed under addition, and scalar multiplication respectively, so $K$ is a subspace of $operatorname{Hom} (V,V)$.
ADDENDUM : I should mention, thanks to the comment below, that $K$ is non-empty , since $T$ commutes with itself, hence is in $K$. (The identity map is in $K$ as well, as is the zero map!)
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
$endgroup$
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
add a comment |
$begingroup$
Let us call the given set K. You will be done if you can show the following two steps:
If $S_1,S_2 in K$, then $S_1 + S_2 in K$.
If $S in K$, then $kS in K$ for all $k$.
To show the first, note that $(S_1+S_2) circ T = S_1 circ T + S_2 circ T = T circ S_1 + T circ S_2 = T circ (S_1 + S_2)$.
The second simply follows: $(kS) circ T = k(S circ T) = k(T circ S) = T circ (kS)$.
You can verify these steps yourself, but do ask if doubts persist. The above two steps show that $K$ is closed under addition, and scalar multiplication respectively, so $K$ is a subspace of $operatorname{Hom} (V,V)$.
ADDENDUM : I should mention, thanks to the comment below, that $K$ is non-empty , since $T$ commutes with itself, hence is in $K$. (The identity map is in $K$ as well, as is the zero map!)
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
$endgroup$
Let us call the given set K. You will be done if you can show the following two steps:
If $S_1,S_2 in K$, then $S_1 + S_2 in K$.
If $S in K$, then $kS in K$ for all $k$.
To show the first, note that $(S_1+S_2) circ T = S_1 circ T + S_2 circ T = T circ S_1 + T circ S_2 = T circ (S_1 + S_2)$.
The second simply follows: $(kS) circ T = k(S circ T) = k(T circ S) = T circ (kS)$.
You can verify these steps yourself, but do ask if doubts persist. The above two steps show that $K$ is closed under addition, and scalar multiplication respectively, so $K$ is a subspace of $operatorname{Hom} (V,V)$.
ADDENDUM : I should mention, thanks to the comment below, that $K$ is non-empty , since $T$ commutes with itself, hence is in $K$. (The identity map is in $K$ as well, as is the zero map!)
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
edited Jan 4 at 5:45
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
answered Jun 5 '17 at 12:01
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.2k33678
40.2k33678
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
add a comment |
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
Strictly speaking, the two conditions are not sufficient for $K$ to be a subspace. With only these conditions, it might happen that $K=emptyset$. Fortunately, $Tin K$
$endgroup$
– Hagen von Eitzen
Jun 5 '17 at 12:04
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
$begingroup$
@HagenvonEitzen Yes, thank you for pointing this out.
$endgroup$
– астон вілла олоф мэллбэрг
Jun 5 '17 at 12:06
add a comment |
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