A question on linear independence and dependence
$begingroup$
I have been trying to solve the second half of the following problem but so far I have not succeeded so far:
Let $alpha_{1},ldots,alpha_{n}$ be $n$ elements in a vector space
$V$ over $mathbb{F}$. If they are linearly dependent and if any $n-1$
vectors among them are linearly dependent then there exists
$c_1,c_2,ldots,c_ninmathbb{F}$ , none of which is $0$, such that
$sum_{i=1}^{n}c_{i}alpha_{i}=0$. If $d_1,ldots,d_{n}inmathbb{F}$
satisfy the same requirements then there is a $kne 0inmathbb{F}$
such that $d_{i}=kc_{i}, 1le ile n$.
Here's my attempt:
If $alpha_{1},ldots,alpha_{n}$ are linearly dependent then there are $c_1,c_2,ldots,c_ninmathbb{F}$, not all zero, such that
$$sum_{i=1}^{n} c_{i}alpha_{i} =0$$
If some $c_j=0$ we can assume after rearranging the indices that $c_1=0$ thus
$$sum_{i=2}^{n} c_{i}alpha_{i} =0$$
Since any $n-1$ vectors are independent we must have $c_1=c_2=ldots c_{n}=0$.
Thus $alpha_{1},ldots,alpha_{n}$ are linearly independent which is a contradiction thus all of $c_i$'s are nonzero. This completes the first half of the proof.
Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Then we can write
$$sum_{i=1}^{n}d_{i}alpha_{i}=0=sum_{i=1}^{n}c_{i}alpha_{i}$$
that is
$sum_{i=1}^{n}(d_{i}-c_{i})alpha_{i}=0$.
I'm not sure what can follow from here; How do I bring out the existence of such a $k$? Hints would be appreciated.
UPDATE: Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Since $d_1 ne 0$ we can write
$$sum_{i=1}^{n}d_{i}d_{1}^{-1}alpha_{i}=0$$
And thus like previously,
$$sum_{i=1}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
i.e.
$$sum_{i=2}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
Since any $(n-1)$ vectors are independent, we have that $d_{i}=frac{d_1}{c_1} c_{i}$ for all $2le ile n$. This completes the second part.
linear-algebra vector-spaces
asked Jan 20 at 8:53
Ashish KAshish K
910613
$endgroup$
add a comment |
$begingroup$
I have been trying to solve the second half of the following problem but so far I have not succeeded so far:
Let $alpha_{1},ldots,alpha_{n}$ be $n$ elements in a vector space
$V$ over $mathbb{F}$. If they are linearly dependent and if any $n-1$
vectors among them are linearly dependent then there exists
$c_1,c_2,ldots,c_ninmathbb{F}$ , none of which is $0$, such that
$sum_{i=1}^{n}c_{i}alpha_{i}=0$. If $d_1,ldots,d_{n}inmathbb{F}$
satisfy the same requirements then there is a $kne 0inmathbb{F}$
such that $d_{i}=kc_{i}, 1le ile n$.
Here's my attempt:
If $alpha_{1},ldots,alpha_{n}$ are linearly dependent then there are $c_1,c_2,ldots,c_ninmathbb{F}$, not all zero, such that
$$sum_{i=1}^{n} c_{i}alpha_{i} =0$$
If some $c_j=0$ we can assume after rearranging the indices that $c_1=0$ thus
$$sum_{i=2}^{n} c_{i}alpha_{i} =0$$
Since any $n-1$ vectors are independent we must have $c_1=c_2=ldots c_{n}=0$.
Thus $alpha_{1},ldots,alpha_{n}$ are linearly independent which is a contradiction thus all of $c_i$'s are nonzero. This completes the first half of the proof.
Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Then we can write
$$sum_{i=1}^{n}d_{i}alpha_{i}=0=sum_{i=1}^{n}c_{i}alpha_{i}$$
that is
$sum_{i=1}^{n}(d_{i}-c_{i})alpha_{i}=0$.
I'm not sure what can follow from here; How do I bring out the existence of such a $k$? Hints would be appreciated.
UPDATE: Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Since $d_1 ne 0$ we can write
$$sum_{i=1}^{n}d_{i}d_{1}^{-1}alpha_{i}=0$$
And thus like previously,
$$sum_{i=1}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
i.e.
$$sum_{i=2}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
Since any $(n-1)$ vectors are independent, we have that $d_{i}=frac{d_1}{c_1} c_{i}$ for all $2le ile n$. This completes the second part.
linear-algebra vector-spaces
asked Jan 20 at 8:53
Ashish KAshish K
910613
$endgroup$
$begingroup$
Indirect hint: you are given that $c_j, d_j$'s are all nonzero. Then could you do something to reduce the "degree of freedom"? Also note that we are working on a field.
$endgroup$
– xbh
Jan 20 at 9:02
$begingroup$
@xbh hey I am not sure if I follow you; can you break it down even more?
$endgroup$
– Ashish K
Jan 20 at 9:13
$begingroup$
Okay, my apologies. Directly, I claim that we could assume $c_1 = 1$, is it correct? If it is, then the unknown coefficients are only $c_2$ thru $c_n$, i.e. we reduce "# of free variables" from $n$ to $n-1$, then we might be easier to proceed.
$endgroup$
– xbh
Jan 20 at 9:23
$begingroup$
@xbh thanks for the hint! I completed the proof. Can you please check it?
$endgroup$
– Ashish K
Jan 20 at 9:32
1
$begingroup$
Yeah, that's pretty much of it. Well done!
$endgroup$
– xbh
Jan 20 at 9:34
add a comment |
$begingroup$
I have been trying to solve the second half of the following problem but so far I have not succeeded so far:
Let $alpha_{1},ldots,alpha_{n}$ be $n$ elements in a vector space
$V$ over $mathbb{F}$. If they are linearly dependent and if any $n-1$
vectors among them are linearly dependent then there exists
$c_1,c_2,ldots,c_ninmathbb{F}$ , none of which is $0$, such that
$sum_{i=1}^{n}c_{i}alpha_{i}=0$. If $d_1,ldots,d_{n}inmathbb{F}$
satisfy the same requirements then there is a $kne 0inmathbb{F}$
such that $d_{i}=kc_{i}, 1le ile n$.
Here's my attempt:
If $alpha_{1},ldots,alpha_{n}$ are linearly dependent then there are $c_1,c_2,ldots,c_ninmathbb{F}$, not all zero, such that
$$sum_{i=1}^{n} c_{i}alpha_{i} =0$$
If some $c_j=0$ we can assume after rearranging the indices that $c_1=0$ thus
$$sum_{i=2}^{n} c_{i}alpha_{i} =0$$
Since any $n-1$ vectors are independent we must have $c_1=c_2=ldots c_{n}=0$.
Thus $alpha_{1},ldots,alpha_{n}$ are linearly independent which is a contradiction thus all of $c_i$'s are nonzero. This completes the first half of the proof.
Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Then we can write
$$sum_{i=1}^{n}d_{i}alpha_{i}=0=sum_{i=1}^{n}c_{i}alpha_{i}$$
that is
$sum_{i=1}^{n}(d_{i}-c_{i})alpha_{i}=0$.
I'm not sure what can follow from here; How do I bring out the existence of such a $k$? Hints would be appreciated.
UPDATE: Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Since $d_1 ne 0$ we can write
$$sum_{i=1}^{n}d_{i}d_{1}^{-1}alpha_{i}=0$$
And thus like previously,
$$sum_{i=1}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
i.e.
$$sum_{i=2}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
Since any $(n-1)$ vectors are independent, we have that $d_{i}=frac{d_1}{c_1} c_{i}$ for all $2le ile n$. This completes the second part.
linear-algebra vector-spaces
asked Jan 20 at 8:53
Ashish KAshish K
910613
$endgroup$
I have been trying to solve the second half of the following problem but so far I have not succeeded so far:
Let $alpha_{1},ldots,alpha_{n}$ be $n$ elements in a vector space
$V$ over $mathbb{F}$. If they are linearly dependent and if any $n-1$
vectors among them are linearly dependent then there exists
$c_1,c_2,ldots,c_ninmathbb{F}$ , none of which is $0$, such that
$sum_{i=1}^{n}c_{i}alpha_{i}=0$. If $d_1,ldots,d_{n}inmathbb{F}$
satisfy the same requirements then there is a $kne 0inmathbb{F}$
such that $d_{i}=kc_{i}, 1le ile n$.
Here's my attempt:
If $alpha_{1},ldots,alpha_{n}$ are linearly dependent then there are $c_1,c_2,ldots,c_ninmathbb{F}$, not all zero, such that
$$sum_{i=1}^{n} c_{i}alpha_{i} =0$$
If some $c_j=0$ we can assume after rearranging the indices that $c_1=0$ thus
$$sum_{i=2}^{n} c_{i}alpha_{i} =0$$
Since any $n-1$ vectors are independent we must have $c_1=c_2=ldots c_{n}=0$.
Thus $alpha_{1},ldots,alpha_{n}$ are linearly independent which is a contradiction thus all of $c_i$'s are nonzero. This completes the first half of the proof.
Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Then we can write
$$sum_{i=1}^{n}d_{i}alpha_{i}=0=sum_{i=1}^{n}c_{i}alpha_{i}$$
that is
$sum_{i=1}^{n}(d_{i}-c_{i})alpha_{i}=0$.
I'm not sure what can follow from here; How do I bring out the existence of such a $k$? Hints would be appreciated.
UPDATE: Suppose that there are some $d_{1},ldots,d_{n} inmathbb{F}$ where none of which are zero such that
$$sum_{i=1}^{n}d_{i}alpha_{i}=0$$
Since $d_1 ne 0$ we can write
$$sum_{i=1}^{n}d_{i}d_{1}^{-1}alpha_{i}=0$$
And thus like previously,
$$sum_{i=1}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
i.e.
$$sum_{i=2}^{n}(d_{i}d_{1}^{-1}-c_{i}c_{1}^{-1})alpha_{i}=0$$
Since any $(n-1)$ vectors are independent, we have that $d_{i}=frac{d_1}{c_1} c_{i}$ for all $2le ile n$. This completes the second part.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 20 at 8:53
Ashish KAshish K
910613
asked Jan 20 at 8:53
Ashish KAshish K
910613
edited Jan 20 at 9:39
Ashish K
asked Jan 20 at 8:53
Ashish KAshish K
910613
asked Jan 20 at 8:53
Ashish KAshish K
910613
asked Jan 20 at 8:53
Ashish KAshish K
910613
910613
$begingroup$
Indirect hint: you are given that $c_j, d_j$'s are all nonzero. Then could you do something to reduce the "degree of freedom"? Also note that we are working on a field.
$endgroup$
– xbh
Jan 20 at 9:02
$begingroup$
@xbh hey I am not sure if I follow you; can you break it down even more?
$endgroup$
– Ashish K
Jan 20 at 9:13
$begingroup$
Okay, my apologies. Directly, I claim that we could assume $c_1 = 1$, is it correct? If it is, then the unknown coefficients are only $c_2$ thru $c_n$, i.e. we reduce "# of free variables" from $n$ to $n-1$, then we might be easier to proceed.
$endgroup$
– xbh
Jan 20 at 9:23
$begingroup$
@xbh thanks for the hint! I completed the proof. Can you please check it?
$endgroup$
– Ashish K
Jan 20 at 9:32
1
$begingroup$
Yeah, that's pretty much of it. Well done!
$endgroup$
– xbh
Jan 20 at 9:34
add a comment |
$begingroup$
Indirect hint: you are given that $c_j, d_j$'s are all nonzero. Then could you do something to reduce the "degree of freedom"? Also note that we are working on a field.
$endgroup$
– xbh
Jan 20 at 9:02
$begingroup$
@xbh hey I am not sure if I follow you; can you break it down even more?
$endgroup$
– Ashish K
Jan 20 at 9:13
$begingroup$
Okay, my apologies. Directly, I claim that we could assume $c_1 = 1$, is it correct? If it is, then the unknown coefficients are only $c_2$ thru $c_n$, i.e. we reduce "# of free variables" from $n$ to $n-1$, then we might be easier to proceed.
$endgroup$
– xbh
Jan 20 at 9:23
$begingroup$
@xbh thanks for the hint! I completed the proof. Can you please check it?
$endgroup$
– Ashish K
Jan 20 at 9:32
1
$begingroup$
Yeah, that's pretty much of it. Well done!
$endgroup$
– xbh
Jan 20 at 9:34
$begingroup$
Indirect hint: you are given that $c_j, d_j$'s are all nonzero. Then could you do something to reduce the "degree of freedom"? Also note that we are working on a field.
$endgroup$
– xbh
Jan 20 at 9:02
$begingroup$
Indirect hint: you are given that $c_j, d_j$'s are all nonzero. Then could you do something to reduce the "degree of freedom"? Also note that we are working on a field.
$endgroup$
– xbh
Jan 20 at 9:02
$begingroup$
@xbh hey I am not sure if I follow you; can you break it down even more?
$endgroup$
– Ashish K
Jan 20 at 9:13
$begingroup$
@xbh hey I am not sure if I follow you; can you break it down even more?
$endgroup$
– Ashish K
Jan 20 at 9:13
$begingroup$
Okay, my apologies. Directly, I claim that we could assume $c_1 = 1$, is it correct? If it is, then the unknown coefficients are only $c_2$ thru $c_n$, i.e. we reduce "# of free variables" from $n$ to $n-1$, then we might be easier to proceed.
$endgroup$
– xbh
Jan 20 at 9:23
$begingroup$
Okay, my apologies. Directly, I claim that we could assume $c_1 = 1$, is it correct? If it is, then the unknown coefficients are only $c_2$ thru $c_n$, i.e. we reduce "# of free variables" from $n$ to $n-1$, then we might be easier to proceed.
$endgroup$
– xbh
Jan 20 at 9:23
$begingroup$
@xbh thanks for the hint! I completed the proof. Can you please check it?
$endgroup$
– Ashish K
Jan 20 at 9:32
$begingroup$
@xbh thanks for the hint! I completed the proof. Can you please check it?
$endgroup$
– Ashish K
Jan 20 at 9:32
1
1
$begingroup$
Yeah, that's pretty much of it. Well done!
$endgroup$
– xbh
Jan 20 at 9:34
$begingroup$
Yeah, that's pretty much of it. Well done!
$endgroup$
– xbh
Jan 20 at 9:34
add a comment |
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$begingroup$
Indirect hint: you are given that $c_j, d_j$'s are all nonzero. Then could you do something to reduce the "degree of freedom"? Also note that we are working on a field.
$endgroup$
– xbh
Jan 20 at 9:02
$begingroup$
@xbh hey I am not sure if I follow you; can you break it down even more?
$endgroup$
– Ashish K
Jan 20 at 9:13
$begingroup$
Okay, my apologies. Directly, I claim that we could assume $c_1 = 1$, is it correct? If it is, then the unknown coefficients are only $c_2$ thru $c_n$, i.e. we reduce "# of free variables" from $n$ to $n-1$, then we might be easier to proceed.
$endgroup$
– xbh
Jan 20 at 9:23
$begingroup$
@xbh thanks for the hint! I completed the proof. Can you please check it?
$endgroup$
– Ashish K
Jan 20 at 9:32
1
$begingroup$
Yeah, that's pretty much of it. Well done!
$endgroup$
– xbh
Jan 20 at 9:34