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Hints only

Let $P = \text{Span } \{v_1, v_2\}$ be a plane in $\mathbb{R}^3$ with normal vector $n$. Show that $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. Hints only

Equation for $P$: $P = c_1v_1 + c_2v_2$. For real $c_1, c_2$. We have by definition, $n = v_1 \times v_2$.

To make sure $\{ v_1, v_2, n\}$ is a basis for $\mathbb{R}^3$. we must have

  1. Span $\{ v_1, v_2, n\}$ = $\mathbb{R}^3$
  2. $\{ v_1, v_2, n\}$ Linearly independent.

I am having major trouble showing (1).

Can I get a SMALL hint?

I am not allowed to use dimension.

Showing set is LI

By definition of a plane, $\{v_1, v_2 \}$ is linearly independent thus $\overrightarrow{0} \not \in \{v_1, v_2 \}$

So $c_1v_1 + c_2v_2 + 0 \overrightarrow{n} = 0$ Thus $\{v_1, v_2, n \}$ is LI.

The span part is confusing

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    Hint: If you can show 2, why do you still need 1? – 2017-01-21
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    @HagenvonEitzen, that is the condition for basis? – 2017-01-21
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    What is the dimension of $\text{Span}\left\{v_1,v_2,n\right\}$ given that you showed $2.$? – 2017-01-21
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    I'm not sure the OP knows how to show 2 either... he's stuck just showing 1? – 2017-01-21
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    "that is the condition for basis?" Not necessarily. A basis for a vector space has **three** properties: (1) the vectors span the space, (2) the vectors are independent, (3) the number of vectors is the same as the dimension of the space. Further, any two imply the third. Here you know that the vector space has dimension 3 so any set if three vectors that span the space **must** be independent and a basis for the space. – 2017-01-21
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    @user247327: No, a basis is defined by the first two properties. How would you define the dimension of a vector space without first defining a basis? – 2017-01-22

3 Answers 3

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You want to show that $\{ v_1, v_2, n\}$ is a basis, meaning it is a linearly-independent set generating all of $\mathbb{R}^3$.

Linear independency means that you need to show that the only way to get the zero vector is by the null linear combination.

Now what is left is to show the generating part meaning: span$\{ v_1, v_2, n\}=\mathbb{R}^3$.

Obviously, span$\{ v_1, v_2, n\}\subseteq \mathbb{R}^3$ by definition of span and definition of a vector space.

All you need to show is that $\mathbb{R}^3\subseteq$ span$\{ v_1, v_2, n\}$. That would mean that an arbitrary vector in $\mathbb{R}^3$ can be written as a linear combination of $v_1$, $v_2$ and $n$.

This is the long way to prove it (recommended to understand). Once you are familiar with this, the fast way is to use the fact that a span is a vector subspace and a lemma that states if a vector subspace $W\subseteq V$ has the same dimension as $V$ then $W=V$.

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    How can I show a vector in R^3 can be written as a combination though? – 2017-01-22
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    @Amad27 You could take any $\begin{pmatrix} x\\ y\\ z \end{pmatrix}=x\mathbf{e_{1}}+y\mathbf{e_{2}}+z\mathbf{e_{3}}$ where the $\mathbf{e_{i}}$ vectors make up the standard basis and argue that you can write each of the $\mathbf{e_{i}}$ vectors as a linear combination of $v_1$, $v_2$ and $n$ so you end up with your original vector written in the your new basis. – 2017-01-22
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    @Amad27 Another way would be to argue by case, suppose first your random vector $\mathbb{R}^3$ lies in the plane $P$ then it is written with just $v_1$ and $v_2$ and its coefficient with respect to $n$ is $0$. If it does not lie in the plane you take the projection of the vector onto the plane and so on. I don't know what you have seen and what you are allowed to use for a proof. – 2017-01-22
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Hints

$1$.If the set {$v_1,v_2,...v_r$} spans a vector space of dimension $r$ then the set is linearly independent.

$2$.Orthogonal pair of vectors are linearly independent.

$3$. For a subspace $W$ of vector space $V$, $dim(W)=dim(V)\implies W=V$.

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Have a look at the Gram-determinant of the three vectors, see here: https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. It's clearly not zero.

Edit: You may alternatively show for any vector $w$ that $$w-\frac{\langle n,w\rangle}{\|n\|^2}n\in\mathrm{Span}\{v_1,v_2\}.$$