Mixing
Question regarding orthogonality and linear independence
$begingroup$
I am new to linear algebra and was a bit confused regarding the following… Any feedback would be really appreciated...
True or false?
- If $K$ is a non-empty set of vectors in $R^n$, then $(K^bot)^bot=K$.
$A={v_1,v_2}$ is a set of vectors in $R^4$. If $(Sp(A))^bot=Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$, then A is linearly dependent.
My answer for 1 is true, per definition.
My answer for 2 is:
The vectors in $Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$ are linearly dependent, and can be reduced to $Sp{(1,2,1,1),(2,2,2,2)}$. This means that the dimension of $Sp(A)^bot$ is 2 and the same for $Sp(A)$. Based on $A^bot=Sp{(1,2,1,1),(2,2,2,2)}$, $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$, which I worked out via a matrix based on $(x,y,z,w)·(1,2,1,1)=0$ and $(x,y,z,w)·(2,2,2,2)=0$. The vectors in $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$ are linearly independent, so the statement is incorrect.
Thank you!
linear-algebra orthogonality
asked Jan 18 at 9:47
daltadalta
1168
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra and was a bit confused regarding the following… Any feedback would be really appreciated...
True or false?
- If $K$ is a non-empty set of vectors in $R^n$, then $(K^bot)^bot=K$.
$A={v_1,v_2}$ is a set of vectors in $R^4$. If $(Sp(A))^bot=Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$, then A is linearly dependent.
My answer for 1 is true, per definition.
My answer for 2 is:
The vectors in $Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$ are linearly dependent, and can be reduced to $Sp{(1,2,1,1),(2,2,2,2)}$. This means that the dimension of $Sp(A)^bot$ is 2 and the same for $Sp(A)$. Based on $A^bot=Sp{(1,2,1,1),(2,2,2,2)}$, $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$, which I worked out via a matrix based on $(x,y,z,w)·(1,2,1,1)=0$ and $(x,y,z,w)·(2,2,2,2)=0$. The vectors in $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$ are linearly independent, so the statement is incorrect.
Thank you!
linear-algebra orthogonality
asked Jan 18 at 9:47
daltadalta
1168
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra and was a bit confused regarding the following… Any feedback would be really appreciated...
True or false?
- If $K$ is a non-empty set of vectors in $R^n$, then $(K^bot)^bot=K$.
$A={v_1,v_2}$ is a set of vectors in $R^4$. If $(Sp(A))^bot=Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$, then A is linearly dependent.
My answer for 1 is true, per definition.
My answer for 2 is:
The vectors in $Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$ are linearly dependent, and can be reduced to $Sp{(1,2,1,1),(2,2,2,2)}$. This means that the dimension of $Sp(A)^bot$ is 2 and the same for $Sp(A)$. Based on $A^bot=Sp{(1,2,1,1),(2,2,2,2)}$, $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$, which I worked out via a matrix based on $(x,y,z,w)·(1,2,1,1)=0$ and $(x,y,z,w)·(2,2,2,2)=0$. The vectors in $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$ are linearly independent, so the statement is incorrect.
Thank you!
linear-algebra orthogonality
asked Jan 18 at 9:47
daltadalta
1168
$endgroup$
I am new to linear algebra and was a bit confused regarding the following… Any feedback would be really appreciated...
True or false?
- If $K$ is a non-empty set of vectors in $R^n$, then $(K^bot)^bot=K$.
$A={v_1,v_2}$ is a set of vectors in $R^4$. If $(Sp(A))^bot=Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$, then A is linearly dependent.
My answer for 1 is true, per definition.
My answer for 2 is:
The vectors in $Sp{(1,2,1,1),(2,2,2,2),(2,1,2,2)}$ are linearly dependent, and can be reduced to $Sp{(1,2,1,1),(2,2,2,2)}$. This means that the dimension of $Sp(A)^bot$ is 2 and the same for $Sp(A)$. Based on $A^bot=Sp{(1,2,1,1),(2,2,2,2)}$, $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$, which I worked out via a matrix based on $(x,y,z,w)·(1,2,1,1)=0$ and $(x,y,z,w)·(2,2,2,2)=0$. The vectors in $(A^bot)^bot=Sp{(-1,0,1,0),(-1,0,0,1)$ are linearly independent, so the statement is incorrect.
Thank you!
linear-algebra orthogonality
linear-algebra orthogonality
asked Jan 18 at 9:47
daltadalta
1168
asked Jan 18 at 9:47
daltadalta
1168
asked Jan 18 at 9:47
daltadalta
1168
asked Jan 18 at 9:47
daltadalta
1168
asked Jan 18 at 9:47
daltadalta
1168
1168
add a comment |
add a comment |
1 Answer
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$begingroup$
- Be careful, $(K^{perp})^{perp}$ is always a subspace. So what if $K$ is just a set and not a subspace?
- You reasoning seems correct.
answered Jan 18 at 9:59
KlausKlaus
2,0099
$endgroup$
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
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– Klaus
Jan 18 at 10:08
add a comment |
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1 Answer
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$begingroup$
- Be careful, $(K^{perp})^{perp}$ is always a subspace. So what if $K$ is just a set and not a subspace?
- You reasoning seems correct.
answered Jan 18 at 9:59
KlausKlaus
2,0099
$endgroup$
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
$endgroup$
– Klaus
Jan 18 at 10:08
add a comment |
$begingroup$
- Be careful, $(K^{perp})^{perp}$ is always a subspace. So what if $K$ is just a set and not a subspace?
- You reasoning seems correct.
answered Jan 18 at 9:59
KlausKlaus
2,0099
$endgroup$
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
$endgroup$
– Klaus
Jan 18 at 10:08
add a comment |
$begingroup$
- Be careful, $(K^{perp})^{perp}$ is always a subspace. So what if $K$ is just a set and not a subspace?
- You reasoning seems correct.
answered Jan 18 at 9:59
KlausKlaus
2,0099
$endgroup$
- Be careful, $(K^{perp})^{perp}$ is always a subspace. So what if $K$ is just a set and not a subspace?
- You reasoning seems correct.
answered Jan 18 at 9:59
KlausKlaus
2,0099
answered Jan 18 at 9:59
KlausKlaus
2,0099
answered Jan 18 at 9:59
KlausKlaus
2,0099
answered Jan 18 at 9:59
KlausKlaus
2,0099
2,0099
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
$endgroup$
– Klaus
Jan 18 at 10:08
add a comment |
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
$endgroup$
– Klaus
Jan 18 at 10:08
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Thank you! We were not taught this in class, nor can I find it in the textbook. So the answer is "false", because it was not stated that K is a subspace? Just trying to understand...
$endgroup$
– dalta
Jan 18 at 10:01
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
$endgroup$
– Klaus
Jan 18 at 10:08
$begingroup$
Yes, 1. is false if $K$ is not a subspace. That the orthogonal complement of any set is a subspace is probably mentioned in every textbook. However, you can easily prove this simple fact yourself. Take $x,y$ in the complement and $lambda in mathbb{R}$. Try to show that $x+lambda y$ is in the complement as well.
$endgroup$
– Klaus
Jan 18 at 10:08
add a comment |
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