Showing a set of subspaces are subspaces of a homomorphism?












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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







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    1














    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







    share|cite|improve this question



























      1












      1








      1







      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







      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 5:54









      André 3000

      12.5k22042




      12.5k22042










      asked Jun 5 '17 at 11:55









      Vivek KatialVivek Katial

      113




      113






















          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:




          1. If $S_1,S_2 in K$, then $S_1 + S_2 in K$.


          2. 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!)






          share|cite|improve this answer























          • 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, $Tin K$
            – Hagen von Eitzen
            Jun 5 '17 at 12:04












          • @HagenvonEitzen Yes, thank you for pointing this out.
            – астон вілла олоф мэллбэрг
            Jun 5 '17 at 12:06











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          Let us call the given set K. You will be done if you can show the following two steps:




          1. If $S_1,S_2 in K$, then $S_1 + S_2 in K$.


          2. 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!)






          share|cite|improve this answer























          • 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, $Tin K$
            – Hagen von Eitzen
            Jun 5 '17 at 12:04












          • @HagenvonEitzen Yes, thank you for pointing this out.
            – астон вілла олоф мэллбэрг
            Jun 5 '17 at 12:06
















          1














          Let us call the given set K. You will be done if you can show the following two steps:




          1. If $S_1,S_2 in K$, then $S_1 + S_2 in K$.


          2. 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!)






          share|cite|improve this answer























          • 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, $Tin K$
            – Hagen von Eitzen
            Jun 5 '17 at 12:04












          • @HagenvonEitzen Yes, thank you for pointing this out.
            – астон вілла олоф мэллбэрг
            Jun 5 '17 at 12:06














          1












          1








          1






          Let us call the given set K. You will be done if you can show the following two steps:




          1. If $S_1,S_2 in K$, then $S_1 + S_2 in K$.


          2. 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!)






          share|cite|improve this answer














          Let us call the given set K. You will be done if you can show the following two steps:




          1. If $S_1,S_2 in K$, then $S_1 + S_2 in K$.


          2. 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!)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 5:45

























          answered Jun 5 '17 at 12:01









          астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

          37.4k33376




          37.4k33376












          • 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, $Tin K$
            – Hagen von Eitzen
            Jun 5 '17 at 12:04












          • @HagenvonEitzen Yes, thank you for pointing this out.
            – астон вілла олоф мэллбэрг
            Jun 5 '17 at 12:06


















          • 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, $Tin K$
            – Hagen von Eitzen
            Jun 5 '17 at 12:04












          • @HagenvonEitzen Yes, thank you for pointing this out.
            – астон вілла олоф мэллбэрг
            Jun 5 '17 at 12:06
















          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, $Tin K$
          – Hagen von Eitzen
          Jun 5 '17 at 12:04






          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, $Tin K$
          – Hagen von Eitzen
          Jun 5 '17 at 12:04














          @HagenvonEitzen Yes, thank you for pointing this out.
          – астон вілла олоф мэллбэрг
          Jun 5 '17 at 12:06




          @HagenvonEitzen Yes, thank you for pointing this out.
          – астон вілла олоф мэллбэрг
          Jun 5 '17 at 12:06


















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