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Prove that a set of elements in a vector space are linearly dependent

What exactly do we mean when say “linear” combination?Check if two 3D vectors are linearly dependentProve that vectors x,y are linearly dependent exactly when …Determining properties of solution of a second order ODEDescribe explicitly the linear transformation T from $F^2$ to $F^2$ such that $Tepsilon_1=(a,b),Tepsilon_2=(c,d)$How to determine if set of linear mappings are linearly independent or dependent?Linear Independence- How do I show that the vectors are linearly independent?Find if $N=langle y_1,y_2,y_3rangle$ is linearly dependent/independent, given $y_1=x_1+x_2, y_2=x_1+x_3, y_3=x_2+x_3$

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3

$begingroup$

I have this question on my homework and its as follows:

Given $X_1,X_2,X_3$ which belongs to a vector space $V$ and $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$. If $Y_1,Y_2$ is linearly dependent prove that $X_1,X_2,X_3$ is linearly dependent.

What I thought to do was first show that $c_1Y_1+c_2Y_2=0$ then substitute in $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ into the equation such that $c_1Y_1+c_2Y_2=c_2(X_1+X_2-X_3)+c_1(X_1+X_2+2X_3)=0$
then manipulate the equation as follows:

$X_1(c_1+c_2)+X_2(c_1+c_2)+X_3(2c_1-c_2)=0$ and that would show that the set $X_1,X_2,X_3$ is linearly dependent because they are non zero constants. But now I am considering what if $c_1=1$ and $c_2=-1$ which would make the constants $(c_1+c_2)=0$ for $X_1$ and $X_2$. Any advice on where to go from here?

thanks!

linear-algebra

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asked 8 hours ago

Zach LedermanZach Lederman

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3

$begingroup$

I have this question on my homework and its as follows:

Given $X_1,X_2,X_3$ which belongs to a vector space $V$ and $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$. If $Y_1,Y_2$ is linearly dependent prove that $X_1,X_2,X_3$ is linearly dependent.

What I thought to do was first show that $c_1Y_1+c_2Y_2=0$ then substitute in $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ into the equation such that $c_1Y_1+c_2Y_2=c_2(X_1+X_2-X_3)+c_1(X_1+X_2+2X_3)=0$
then manipulate the equation as follows:

$X_1(c_1+c_2)+X_2(c_1+c_2)+X_3(2c_1-c_2)=0$ and that would show that the set $X_1,X_2,X_3$ is linearly dependent because they are non zero constants. But now I am considering what if $c_1=1$ and $c_2=-1$ which would make the constants $(c_1+c_2)=0$ for $X_1$ and $X_2$. Any advice on where to go from here?

thanks!

linear-algebra

share|cite|improve this question

asked 8 hours ago

Zach LedermanZach Lederman

4155 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I have this question on my homework and its as follows:

Given $X_1,X_2,X_3$ which belongs to a vector space $V$ and $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$. If $Y_1,Y_2$ is linearly dependent prove that $X_1,X_2,X_3$ is linearly dependent.

What I thought to do was first show that $c_1Y_1+c_2Y_2=0$ then substitute in $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ into the equation such that $c_1Y_1+c_2Y_2=c_2(X_1+X_2-X_3)+c_1(X_1+X_2+2X_3)=0$
then manipulate the equation as follows:

$X_1(c_1+c_2)+X_2(c_1+c_2)+X_3(2c_1-c_2)=0$ and that would show that the set $X_1,X_2,X_3$ is linearly dependent because they are non zero constants. But now I am considering what if $c_1=1$ and $c_2=-1$ which would make the constants $(c_1+c_2)=0$ for $X_1$ and $X_2$. Any advice on where to go from here?

thanks!

linear-algebra

share|cite|improve this question

asked 8 hours ago

Zach LedermanZach Lederman

4155 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

Zach LedermanZach Lederman

4155 bronze badges

share|cite|improve this question

asked 8 hours ago

Zach LedermanZach Lederman

4155 bronze badges

asked 8 hours ago

Zach LedermanZach Lederman

4155 bronze badges

asked 8 hours ago

Zach LedermanZach Lederman

4155 bronze badges

3 Answers
3

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3

$begingroup$

In this case I think it would be easiest and cleanest to prove the contrapositive, namely, that if $X_1,dotsc, X_3$ is a linearly independent set of vectors, then so is $Y_1, Y_2$. To this end suppose that
$$
c_1Y_1+c_2Y_2=0tag1
$$
for some $c_iinmathbbR$. Since $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ we can rewrite (1) as
$$
(c_1+c_2)X_1+(c_1+c_2)X_2+(2c_1-c_2)X_3=0
$$
whence $c_1+c_2=0$ and $2c_1=c_2$ by linear independence of the $X_i$. In particular $3c_1=0$ whence $c_1=0$ and so $c_2=0$. It follows that the $Y_i$ are linearly independent.

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answered 7 hours ago

Foobaz JohnFoobaz John

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2

$begingroup$

You're on the right track: you have a linear combination
$$
d_1 X_1 + d_2 X_2 + d_3 X_3 = 0
$$
where $d_1 = d_2 = c_1 + c_2$ and $d_3 = 2c_1 - c_2$.

What you need to show is that not all of $d_1,d_2,d_3$ are $0$. So, suppose $d_1 = d_2 = 0$. This means $c_1 = -c_2$. In this case, what can you say about $d_3$? (If you can conclude it is non-zero, you are done).

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answered 8 hours ago

BaronVTBaronVT

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2

$begingroup$

Note that you only need to prove that $c_1 + c_2$ and $2c_1 - c_2$ are not both zero to show that $X_1,X_2,X_3$ is linearly dependent.

To see this, suppose that $c_1 + c_2 = 2c_1 - c_2 = 0$. This would imply that $c_1 = c_2 = 0$. Do you see how this contradicts your assumption that $ Y_1, Y_2$ is linearly dependent?

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answered 8 hours ago

Theoretical EconomistTheoretical Economist

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3

$begingroup$

In this case I think it would be easiest and cleanest to prove the contrapositive, namely, that if $X_1,dotsc, X_3$ is a linearly independent set of vectors, then so is $Y_1, Y_2$. To this end suppose that
$$
c_1Y_1+c_2Y_2=0tag1
$$
for some $c_iinmathbbR$. Since $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ we can rewrite (1) as
$$
(c_1+c_2)X_1+(c_1+c_2)X_2+(2c_1-c_2)X_3=0
$$
whence $c_1+c_2=0$ and $2c_1=c_2$ by linear independence of the $X_i$. In particular $3c_1=0$ whence $c_1=0$ and so $c_2=0$. It follows that the $Y_i$ are linearly independent.

share|cite|improve this answer

answered 7 hours ago

Foobaz JohnFoobaz John

25.2k44 gold badges1515 silver badges5555 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

In this case I think it would be easiest and cleanest to prove the contrapositive, namely, that if $X_1,dotsc, X_3$ is a linearly independent set of vectors, then so is $Y_1, Y_2$. To this end suppose that
$$
c_1Y_1+c_2Y_2=0tag1
$$
for some $c_iinmathbbR$. Since $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ we can rewrite (1) as
$$
(c_1+c_2)X_1+(c_1+c_2)X_2+(2c_1-c_2)X_3=0
$$
whence $c_1+c_2=0$ and $2c_1=c_2$ by linear independence of the $X_i$. In particular $3c_1=0$ whence $c_1=0$ and so $c_2=0$. It follows that the $Y_i$ are linearly independent.

share|cite|improve this answer

answered 7 hours ago

Foobaz JohnFoobaz John

25.2k44 gold badges1515 silver badges5555 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

In this case I think it would be easiest and cleanest to prove the contrapositive, namely, that if $X_1,dotsc, X_3$ is a linearly independent set of vectors, then so is $Y_1, Y_2$. To this end suppose that
$$
c_1Y_1+c_2Y_2=0tag1
$$
for some $c_iinmathbbR$. Since $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ we can rewrite (1) as
$$
(c_1+c_2)X_1+(c_1+c_2)X_2+(2c_1-c_2)X_3=0
$$
whence $c_1+c_2=0$ and $2c_1=c_2$ by linear independence of the $X_i$. In particular $3c_1=0$ whence $c_1=0$ and so $c_2=0$. It follows that the $Y_i$ are linearly independent.

share|cite|improve this answer

answered 7 hours ago

Foobaz JohnFoobaz John

25.2k44 gold badges1515 silver badges5555 bronze badges

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share|cite|improve this answer

answered 7 hours ago

Foobaz JohnFoobaz John

25.2k44 gold badges1515 silver badges5555 bronze badges

answered 7 hours ago

Foobaz JohnFoobaz John

25.2k44 gold badges1515 silver badges5555 bronze badges

answered 7 hours ago

Foobaz JohnFoobaz John

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2

$begingroup$

You're on the right track: you have a linear combination
$$
d_1 X_1 + d_2 X_2 + d_3 X_3 = 0
$$
where $d_1 = d_2 = c_1 + c_2$ and $d_3 = 2c_1 - c_2$.

What you need to show is that not all of $d_1,d_2,d_3$ are $0$. So, suppose $d_1 = d_2 = 0$. This means $c_1 = -c_2$. In this case, what can you say about $d_3$? (If you can conclude it is non-zero, you are done).

share|cite|improve this answer

answered 8 hours ago

BaronVTBaronVT

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add a comment
 | 

2

$begingroup$

You're on the right track: you have a linear combination
$$
d_1 X_1 + d_2 X_2 + d_3 X_3 = 0
$$
where $d_1 = d_2 = c_1 + c_2$ and $d_3 = 2c_1 - c_2$.

What you need to show is that not all of $d_1,d_2,d_3$ are $0$. So, suppose $d_1 = d_2 = 0$. This means $c_1 = -c_2$. In this case, what can you say about $d_3$? (If you can conclude it is non-zero, you are done).

share|cite|improve this answer

answered 8 hours ago

BaronVTBaronVT

11.8k11 gold badge1414 silver badges3838 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You're on the right track: you have a linear combination
$$
d_1 X_1 + d_2 X_2 + d_3 X_3 = 0
$$
where $d_1 = d_2 = c_1 + c_2$ and $d_3 = 2c_1 - c_2$.

What you need to show is that not all of $d_1,d_2,d_3$ are $0$. So, suppose $d_1 = d_2 = 0$. This means $c_1 = -c_2$. In this case, what can you say about $d_3$? (If you can conclude it is non-zero, you are done).

share|cite|improve this answer

answered 8 hours ago

BaronVTBaronVT

11.8k11 gold badge1414 silver badges3838 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

BaronVTBaronVT

11.8k11 gold badge1414 silver badges3838 bronze badges

answered 8 hours ago

BaronVTBaronVT

11.8k11 gold badge1414 silver badges3838 bronze badges

answered 8 hours ago

BaronVTBaronVT

11.8k11 gold badge1414 silver badges3838 bronze badges

2

$begingroup$

Note that you only need to prove that $c_1 + c_2$ and $2c_1 - c_2$ are not both zero to show that $X_1,X_2,X_3$ is linearly dependent.

To see this, suppose that $c_1 + c_2 = 2c_1 - c_2 = 0$. This would imply that $c_1 = c_2 = 0$. Do you see how this contradicts your assumption that $ Y_1, Y_2$ is linearly dependent?

share|cite|improve this answer

answered 8 hours ago

Theoretical EconomistTheoretical Economist

4,06722 gold badges99 silver badges3131 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Note that you only need to prove that $c_1 + c_2$ and $2c_1 - c_2$ are not both zero to show that $X_1,X_2,X_3$ is linearly dependent.

To see this, suppose that $c_1 + c_2 = 2c_1 - c_2 = 0$. This would imply that $c_1 = c_2 = 0$. Do you see how this contradicts your assumption that $ Y_1, Y_2$ is linearly dependent?

share|cite|improve this answer

answered 8 hours ago

Theoretical EconomistTheoretical Economist

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$endgroup$

add a comment
 | 

$begingroup$

Note that you only need to prove that $c_1 + c_2$ and $2c_1 - c_2$ are not both zero to show that $X_1,X_2,X_3$ is linearly dependent.

To see this, suppose that $c_1 + c_2 = 2c_1 - c_2 = 0$. This would imply that $c_1 = c_2 = 0$. Do you see how this contradicts your assumption that $ Y_1, Y_2$ is linearly dependent?

share|cite|improve this answer

answered 8 hours ago

Theoretical EconomistTheoretical Economist

4,06722 gold badges99 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Theoretical EconomistTheoretical Economist

4,06722 gold badges99 silver badges3131 bronze badges

answered 8 hours ago

Theoretical EconomistTheoretical Economist

4,06722 gold badges99 silver badges3131 bronze badges

answered 8 hours ago

Theoretical EconomistTheoretical Economist

4,06722 gold badges99 silver badges3131 bronze badges