Mixing
Question about proof regarding span
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I was working through Linear Algebra Done Right and had a question about whether a proof I gave for a question is valid or not. I've seen solutions to this question online but I wanted to see if this method worked.
Prove that if $(v_1,...,v_n)$ spans $V$, then so does the list
$(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
My attempt:
Let $L$ denote $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$ and $L_k$ be the $k$'th element of $L$.
Note that: $$v_k = sum_{i=k}^{n} L_i$$
Let $U$ denote the vector space spanned by $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
From the sum above we can see that $v_1,...v_nin U$. Since $U$ is a vector space every linear combination of its elements is in $U$. It follows that $Vsubseteq U$, and thus $L$ spans $V$.
Does this method work or are there any tips?
linear-algebra proof-verification
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
$endgroup$
add a comment |
$begingroup$
I was working through Linear Algebra Done Right and had a question about whether a proof I gave for a question is valid or not. I've seen solutions to this question online but I wanted to see if this method worked.
Prove that if $(v_1,...,v_n)$ spans $V$, then so does the list
$(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
My attempt:
Let $L$ denote $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$ and $L_k$ be the $k$'th element of $L$.
Note that: $$v_k = sum_{i=k}^{n} L_i$$
Let $U$ denote the vector space spanned by $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
From the sum above we can see that $v_1,...v_nin U$. Since $U$ is a vector space every linear combination of its elements is in $U$. It follows that $Vsubseteq U$, and thus $L$ spans $V$.
Does this method work or are there any tips?
linear-algebra proof-verification
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
$endgroup$
3
$begingroup$
Yep that works.
$endgroup$
– John Doe
Jan 3 at 4:33
1
$begingroup$
Nice proof, Jack.
$endgroup$
– Chris Custer
Jan 3 at 4:41
add a comment |
$begingroup$
I was working through Linear Algebra Done Right and had a question about whether a proof I gave for a question is valid or not. I've seen solutions to this question online but I wanted to see if this method worked.
Prove that if $(v_1,...,v_n)$ spans $V$, then so does the list
$(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
My attempt:
Let $L$ denote $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$ and $L_k$ be the $k$'th element of $L$.
Note that: $$v_k = sum_{i=k}^{n} L_i$$
Let $U$ denote the vector space spanned by $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
From the sum above we can see that $v_1,...v_nin U$. Since $U$ is a vector space every linear combination of its elements is in $U$. It follows that $Vsubseteq U$, and thus $L$ spans $V$.
Does this method work or are there any tips?
linear-algebra proof-verification
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
$endgroup$
I was working through Linear Algebra Done Right and had a question about whether a proof I gave for a question is valid or not. I've seen solutions to this question online but I wanted to see if this method worked.
Prove that if $(v_1,...,v_n)$ spans $V$, then so does the list
$(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
My attempt:
Let $L$ denote $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$ and $L_k$ be the $k$'th element of $L$.
Note that: $$v_k = sum_{i=k}^{n} L_i$$
Let $U$ denote the vector space spanned by $(v_1-v_2,v_2-v_3,...,v_{n-1}-v_n,v_n)$.
From the sum above we can see that $v_1,...v_nin U$. Since $U$ is a vector space every linear combination of its elements is in $U$. It follows that $Vsubseteq U$, and thus $L$ spans $V$.
Does this method work or are there any tips?
linear-algebra proof-verification
linear-algebra proof-verification
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
asked Jan 3 at 4:31
Jack PfaffingerJack Pfaffinger
885
885
3
$begingroup$
Yep that works.
$endgroup$
– John Doe
Jan 3 at 4:33
1
$begingroup$
Nice proof, Jack.
$endgroup$
– Chris Custer
Jan 3 at 4:41
add a comment |
3
$begingroup$
Yep that works.
$endgroup$
– John Doe
Jan 3 at 4:33
1
$begingroup$
Nice proof, Jack.
$endgroup$
– Chris Custer
Jan 3 at 4:41
3
3
$begingroup$
Yep that works.
$endgroup$
– John Doe
Jan 3 at 4:33
$begingroup$
Yep that works.
$endgroup$
– John Doe
Jan 3 at 4:33
1
1
$begingroup$
Nice proof, Jack.
$endgroup$
– Chris Custer
Jan 3 at 4:41
$begingroup$
Nice proof, Jack.
$endgroup$
– Chris Custer
Jan 3 at 4:41
add a comment |
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3
$begingroup$
Yep that works.
$endgroup$
– John Doe
Jan 3 at 4:33
1
$begingroup$
Nice proof, Jack.
$endgroup$
– Chris Custer
Jan 3 at 4:41