Help for finding Bases of Sets
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I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
asked yesterday
Danny
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up vote
-1
down vote
favorite
I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
asked yesterday
Danny
32
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
yesterday
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
yesterday
Thank you for the advice.
– Danny
yesterday
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
asked yesterday
Danny
32
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
linear-algebra
asked yesterday
Danny
32
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Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Danny
32
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
Danny
32
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Danny
32
asked yesterday
Danny
32
32
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
yesterday
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
yesterday
Thank you for the advice.
– Danny
yesterday
add a comment |
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
yesterday
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
yesterday
Thank you for the advice.
– Danny
yesterday
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
yesterday
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
yesterday
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
yesterday
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
yesterday
Thank you for the advice.
– Danny
yesterday
Thank you for the advice.
– Danny
yesterday
add a comment |
1 Answer
1
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0
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There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered yesterday
gandalf61
7,102523
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered yesterday
gandalf61
7,102523
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
add a comment |
up vote
0
down vote
accepted
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered yesterday
gandalf61
7,102523
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered yesterday
gandalf61
7,102523
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered yesterday
gandalf61
7,102523
answered yesterday
gandalf61
7,102523
answered yesterday
gandalf61
7,102523
answered yesterday
gandalf61
7,102523
7,102523
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
add a comment |
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
yesterday
add a comment |
Danny is a new contributor. Be nice, and check out our Code of Conduct.
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Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
yesterday
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
yesterday
Thank you for the advice.
– Danny
yesterday