Showing subspace is a vector space. Why is this step necessary?
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I am reading this text:
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
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
asked Aug 28 at 14:38
Jwan622
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up vote
0
down vote
favorite
I am reading this text:
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
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
asked Aug 28 at 14:38
Jwan622
1,75711224
2
How else would you do it?
– Sobi
Aug 28 at 14:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading this text:
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
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
asked Aug 28 at 14:38
Jwan622
1,75711224
I am reading this text:
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
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
asked Aug 28 at 14:38
Jwan622
1,75711224
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
edited Aug 28 at 18:43
José Carlos Santos
120k16101182
120k16101182
asked Aug 28 at 14:38
Jwan622
1,75711224
asked Aug 28 at 14:38
Jwan622
1,75711224
asked Aug 28 at 14:38
Jwan622
1,75711224
1,75711224
2
How else would you do it?
– Sobi
Aug 28 at 14:39
add a comment |Â
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
add a comment |Â
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.
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
add a comment |Â
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.
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
add a comment |Â
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.
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
add a comment |Â
up vote
1
down vote
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.
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
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.
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
answered Aug 28 at 14:42
José Carlos Santos
120k16101182
120k16101182
add a comment |Â
add a comment |Â
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.
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
add a comment |Â
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.
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
add a comment |Â
up vote
0
down vote
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.
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
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.
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
answered Aug 28 at 14:50
Mohammad Riazi-Kermani
30.7k41852
30.7k41852
add a comment |Â
add a comment |Â
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2
How else would you do it?
– Sobi
Aug 28 at 14:39