Showing subspace is a vector space. Why is this step necessary?

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Why do we need to show the transpose of the sum is equal to the sum? Isn't it enough to just show that W, the set of of all 2x2 symmetric matrices, is closed under addition and multiplication? I realize symmetric means that the transpose is equal to the original matrix, but we don't need to show this right to show closure under addition right?




linear-algebra matrices symmetric-matrices




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    How else would you do it?
    – Sobi
    Aug 28 at 14:39














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I am reading this text:



enter image description here



Why do we need to show the transpose of the sum is equal to the sum? Isn't it enough to just show that W, the set of of all 2x2 symmetric matrices, is closed under addition and multiplication? I realize symmetric means that the transpose is equal to the original matrix, but we don't need to show this right to show closure under addition right?




linear-algebra matrices symmetric-matrices




share|cite|improve this question


















  • 2




    How else would you do it?
    – Sobi
    Aug 28 at 14:39












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am reading this text:



enter image description here



Why do we need to show the transpose of the sum is equal to the sum? Isn't it enough to just show that W, the set of of all 2x2 symmetric matrices, is closed under addition and multiplication? I realize symmetric means that the transpose is equal to the original matrix, but we don't need to show this right to show closure under addition right?




linear-algebra matrices symmetric-matrices




share|cite|improve this question














I am reading this text:



enter image description here



Why do we need to show the transpose of the sum is equal to the sum? Isn't it enough to just show that W, the set of of all 2x2 symmetric matrices, is closed under addition and multiplication? I realize symmetric means that the transpose is equal to the original matrix, but we don't need to show this right to show closure under addition right?





linear-algebra matrices symmetric-matrices





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edited Aug 28 at 18:43









José Carlos Santos

120k16101182




120k16101182










asked Aug 28 at 14:38









Jwan622

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  • 2




    How else would you do it?
    – Sobi
    Aug 28 at 14:39












  • 2




    How else would you do it?
    – Sobi
    Aug 28 at 14:39







2




2




How else would you do it?
– Sobi
Aug 28 at 14:39




How else would you do it?
– Sobi
Aug 28 at 14:39










2 Answers
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The author of that text needed to show that $A_1+A_2$ is symmetric, assuming that $A_1$ and $A_2$ are symmetric. So, since $A$ being symmetric means that $A^T=A$, he or she proved that $(A_1+A_2)^T=A_1+A_2$. That's all. The same thing applies to the product by a scalar.






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    That is exactly how you prove the closure under addition and scalar multiplication.j



    You show that if you add two symmetric matrices, the sum is symmetric, that is $$(A+B)^T= A+B$$ and $$(lambda A)^T=lambda A $$ and that is what the author is doing.






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      The author of that text needed to show that $A_1+A_2$ is symmetric, assuming that $A_1$ and $A_2$ are symmetric. So, since $A$ being symmetric means that $A^T=A$, he or she proved that $(A_1+A_2)^T=A_1+A_2$. That's all. The same thing applies to the product by a scalar.






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        The author of that text needed to show that $A_1+A_2$ is symmetric, assuming that $A_1$ and $A_2$ are symmetric. So, since $A$ being symmetric means that $A^T=A$, he or she proved that $(A_1+A_2)^T=A_1+A_2$. That's all. The same thing applies to the product by a scalar.






        share|cite|improve this answer






















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          up vote
          1
          down vote









          The author of that text needed to show that $A_1+A_2$ is symmetric, assuming that $A_1$ and $A_2$ are symmetric. So, since $A$ being symmetric means that $A^T=A$, he or she proved that $(A_1+A_2)^T=A_1+A_2$. That's all. The same thing applies to the product by a scalar.






          share|cite|improve this answer












          The author of that text needed to show that $A_1+A_2$ is symmetric, assuming that $A_1$ and $A_2$ are symmetric. So, since $A$ being symmetric means that $A^T=A$, he or she proved that $(A_1+A_2)^T=A_1+A_2$. That's all. The same thing applies to the product by a scalar.







          share|cite|improve this answer












          share|cite|improve this answer



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          answered Aug 28 at 14:42









          José Carlos Santos

          120k16101182




          120k16101182




















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              That is exactly how you prove the closure under addition and scalar multiplication.j



              You show that if you add two symmetric matrices, the sum is symmetric, that is $$(A+B)^T= A+B$$ and $$(lambda A)^T=lambda A $$ and that is what the author is doing.






              share|cite|improve this answer
























                up vote
                0
                down vote













                That is exactly how you prove the closure under addition and scalar multiplication.j



                You show that if you add two symmetric matrices, the sum is symmetric, that is $$(A+B)^T= A+B$$ and $$(lambda A)^T=lambda A $$ and that is what the author is doing.






                share|cite|improve this answer






















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                  up vote
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                  down vote









                  That is exactly how you prove the closure under addition and scalar multiplication.j



                  You show that if you add two symmetric matrices, the sum is symmetric, that is $$(A+B)^T= A+B$$ and $$(lambda A)^T=lambda A $$ and that is what the author is doing.






                  share|cite|improve this answer












                  That is exactly how you prove the closure under addition and scalar multiplication.j



                  You show that if you add two symmetric matrices, the sum is symmetric, that is $$(A+B)^T= A+B$$ and $$(lambda A)^T=lambda A $$ and that is what the author is doing.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 28 at 14:50









                  Mohammad Riazi-Kermani

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