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Let $\{v_1,\ldots,v_n\}$ be a basis of $\mathbb R^n$. Which of the following must be true? (Check all that apply).

A. For all $i=1,\ldots,n, \|v_i\|=0$.

B. For any real numbers $c_1,\ldots,c_n, c_1v_1+\cdots+c_nv_n=0$, $\Longleftrightarrow$ $c_1=\cdots=c_n=0$.

C. Any vector in $v\in\mathbb R^n$ may be written $v=c_1v_1+\cdots+c_nv_n$ for some real numbers $c_1,\ldots,c_n$.

D. For all $i=1,\ldots,n, \|v_i\|=1$.

E. Removing any $v_i$ from the basis also yields a basis of $\mathbb R^n$.

F. For any real numbers $c_1,\ldots,c_n, c_1v_1+\cdots+c_nv_n=0$, $\Longleftrightarrow$ $c_1=\cdots=c_n\neq0$.

G. For all $i=1,\ldots,n, \|v_i\|\neq0$.

H. The dot product of any two vectors $v_i,v_j$ is zero unless $i=j$.

I. For any real numbers $c_1,\ldots,c_n, c_1v_1+\cdots+c_nv_n\neq0$.

J. Removing any $v_i$ from the basis also yields a basis of $\mathbb R^{n-1}$.

K. For any real numbers $c_1,\ldots,c_n, v=c_1v_1+\cdots+c_nv_n$, is a vector in $\mathbb R^n$.

Any help with this would be appreciated. I have scoured notes and textbooks to try and figure out the problem, but I can't seem to get it right.
I'm having the most difficulty with the options that seem similar, such as C and K. C and K confuse me because if the set of vectors is linearly independent, then doesn't that mean that any one vector can't be written as a combination of the other vectors? If it can, then that would make it dependent on the other vectors, and wouldn't be a basis of $\mathbb R^n$.

I do know that the dot product of linearly independent vectors do not necessarily have to be zero. That is, the vectors don't have to be orthogonal. However, since the basis is a set of linearly independent vectors, then B must be true.

Additionally, I'm not sure about the magnitude of any vector $v$ in the basis. Does the magnitude have to be one? If so, then does that mean the basis is made up of all linearly independent unit vectors?

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closed as off-topic by Shailesh, Siong Thye Goh, Leucippus, Claude Leibovici, Cyclohexanol. Aug 19 '17 at 2:45

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  • $\begingroup$ Any thoughts? Which ones are you having the most trouble with? $\endgroup$ – John Griffin Aug 19 '17 at 0:42
  • $\begingroup$ That's a hard question, since it seems rather disorganized. I'm having the most difficulty with the options that seem similar, such as C and K. $\endgroup$ – torikolpacoff Aug 19 '17 at 0:44
  • $\begingroup$ I do know that the dot product of linearly independent vectors do not necessarily have to be zero. That is, the vectors don't have to be orthogonal. However, since the basis is a set of linearly independent vectors, then B must be true. $\endgroup$ – torikolpacoff Aug 19 '17 at 0:46
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    $\begingroup$ C and K aren’t really similar, though. The first says that the set of vectors $\{v_1\,\dots,v_n\}$ spans all of $\mathbb R^n$. The other basically just says that $\mathbb R^n$ is a vector space. $\endgroup$ – amd Aug 19 '17 at 0:59
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    $\begingroup$ K says that every linear combination of the $v$’s is an element of $\mathbb R^n$, which is just closure under addition and scalar multiplication, part of what it means to be a vector space in the first place. $\endgroup$ – amd Aug 19 '17 at 7:28
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I'll denote by $e_1:=(1,0)$ and $e_2:=(0,1)$ the standard orthonormal basis on $\mathbb{R}^2$ and by $V$ the set $\{v_1,\ldots,v_n\}$.

A. False. Consider the basis $e_1,e_2$ on $\mathbb{R}^2$.

B. True. Any basis must be linearly independent, and this is the definition of linear independence for $V$.

C. True. Any basis must span $\mathbb{R}^n$, and this is the definition that $V$ spans $\mathbb{R}^n$.

D. False. While this is true for the standard basis, the vectors $2e_1,2e_2$ form a basis on $\mathbb{R}^2$.

E. False. If $V\setminus\{v_i\}$ spans $\mathbb{R}^n$, then $V$ is not linearly independent.

F. False. This is false for any set $V$ because $c_1=\cdots=c_n=0$ implies $c_1v_1+\cdots+c_nv_n=0$.

G. True (unless $n=0$ is allowed). If $n\geq 1$ and $0\in V$, then $V$ is not linearly independent.

H. False. Consider the basis $e_1+e_2,e_2$ on $\mathbb{R}^2$.

I. False. Take $c_1=\cdots=c_n=0$.

J. False. While removing a vector from $V$ will yield a basis for an $n-1$ dimensional subspace of $\mathbb{R}^n$, this subspace is only isomorphic to $\mathbb{R}^{n-1}$.

K. True. This is true for any set of vectors $V$.

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