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Show that $v_1,v_2,v_3$ and $w_1,w_2,w_3$ are two bases for the same $3$ dimensional subspace $V\subset \mathbb{R}^4$

I'm given the entries in each vector but I'm simply looking for direction in solving this.

My first thought is that I need to prove that $v_1,v_2,v_3$ and $w_1,w_2,w_3$ span the same subspace. So I set up something like: $$c_1v_1+c_2v_2+c_3v_3=d_1w_1+d_2w_2+d_3w_3$$ I then set up a matrix $$A= \begin{pmatrix} | & | & | & | & | & |\\ v_1 & v_2 & v_3 & w_1 & w_2 & w_3\\ | & | & | & | & | & |\\ \end{pmatrix} = \mathbf0 $$ And then I rref this matrix and get: $$ \begin{pmatrix} 1 & 0 & 0 & 1 & 1 & 1\\ 0 & 1 & 0 & 0 & 1 & 1\\ 0 & 0 & 1 & 1 & 2 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{pmatrix} = \mathbf0 $$

Did this prove $v_1,v_2,v_3$ and $w_1,w_2,w_3$ are bases of the same subspace?

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  • $\begingroup$ If you prove that they span the same subspace, then by default they are both bases as they are the same in number. A set of $k$ vectors that spans a $k$-dimensional space is always a basis by definition. $\endgroup$
    – whoisit
    Oct 14, 2022 at 4:58
  • $\begingroup$ @KB So if I prove the $v$'s are linearly independent and the $w$'s are linearly independent, this necessarily means they span $\mathbb{R}^3=V\in\mathbb{R}^4$ then I'm done? $\endgroup$ Oct 14, 2022 at 5:03
  • $\begingroup$ @Seeker "I'm given the entries in each vector" $\endgroup$ Oct 14, 2022 at 6:08
  • $\begingroup$ The two "=0" is to be suppressed. $\endgroup$
    – Jean Marie
    Oct 14, 2022 at 8:02
  • $\begingroup$ @JeanMarie I couldn't figure out how to make it augmented $\endgroup$ Oct 15, 2022 at 0:03

2 Answers 2

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One of the basic method to prove that two different sets $\{v_1,v_2,v_3\}$ and $\{w_1,w_2,w_3\}$ span the same space is to prove that they span each other i.e. for every $i\in \{1,2,3\}$, $$v_i = a_iw_1+b_iw_2+c_iw_3\qquad(Eq\;1)$$ where $a_i, b_i, \text{ and } c_i$ are scalars, and $$w_i = d_iv_1+e_iv_2+f_iw_3\qquad (Eq\;2)$$ where $d_i, e_i, \text{ and } f_i$ are scalars.

Each of $(Eq\;1)$ and $(Eq\;2)$ can be written as a system of $3$ equations with $3$ unknowns and transformed into matrices to help with finding the scalars. We need more information about the vectors to help more.

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  • $\begingroup$ I have completely re-written my answer. $\endgroup$
    – Jean Marie
    Oct 15, 2022 at 18:25
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Yes, performing a RREF is a good way.

Moreover one can give an interesting interpretation to the colored entries below :

$$R:=\begin{pmatrix} 1 & 0 & 0 & \color{red}{1} & \color{orange}{1} & \color{blue}{1}\\ 0 & 1 & 0 & \color{red}{0} & \color{orange}{1} & \color{blue}{1}\\ 0 & 0 & 1 & \color{red}{1} & \color{orange}{2} & \color{blue}{1}\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}=\begin{pmatrix}I_3&R'\\0&0\end{pmatrix},\tag{1}$$

Indeed, we have an unexpected fact : these 3 last columns are exactly the coefficients of the linear combinations :

$$\begin{cases}w_1&=&\color{red}{1}v_1+\color{red}{0}v_2+\color{red}{1}v_3 \\ w_2&=&\color{orange}{1}v_1+\color{orange}{1}v_2+\color{orange}{2}v_3 \\ w_2&=&\color{blue}{1}v_1+\color{blue}{1}v_2+\color{blue}{1}v_3 \end{cases}$$

Having this row-reduced matrix $R$ written in the block-form (1) with a last line filled by zeros is a necessary condition.

You must also check that $3 \times 3$ block $R'$ is invertible (this is the case in the example: $\det(R') \ne 0$).

Let us explain this "unexpected fact".

If the $w_i$ are dependent upon the $v_i$ under the form (as used in the answer of @Sam) :

$$\begin{cases}w_1&=a_1v_1+a_2v_2+a_3v_3 \\ w_2&=b_1v_1+b_2v_2+b_3v_3 \\ w_3&=c_1v_1+c_2v_2+c_3v_3 \end{cases}\tag{2}$$

one can write (2) under the matrix equivalent form:

$$ \begin{pmatrix} | & | & |\\ w_1 & w_2 & w_3\\ | & | & |\\ \end{pmatrix} = \begin{pmatrix} | & | & |\\ v_1 & v_2 & v_3\\ | & | & |\\ \end{pmatrix} \begin{pmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3 \end{pmatrix} \tag{3}$$

Using (3):

$$ \begin{pmatrix} | & | & | & | & | & |\\ v_1 & v_2 & v_3 & w_1 & w_2 & w_3\\ | & | & | & | & | & |\\ \end{pmatrix}=\left(\begin{pmatrix} | & | & |\\ v_1 & v_2 & v_3\\ | & | & | \end{pmatrix}I_3 \left|\begin{pmatrix} | & | & |\\ v_1 & v_2 & v_3\\ | & | & | \end{pmatrix} \begin{pmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3 \end{pmatrix}\right. \right) \tag{4}$$

(4) allows to write, by left factorization:

$$A=\begin{pmatrix} | & | & |\\ v_1 & v_2 & v_3\\ | & | & | \end{pmatrix}\begin{pmatrix} 1&0&0&a_1&b_1&c_1\\0&1&0&a_2&b_2&c_2\\0&0&1&a_3&b_3&c_3 \end{pmatrix}\tag{5}$$

where the last matrix is nothing else than matrix $(I_3\ | \ R')$ as given in (1). In fact, (5) expresses the fact that the row reduction has been "driven" by the first set of vectors $v_1,v_2,v_3$.

Remark: Another way to explain this "phenomena" is by using the fact that row reduction is equivalent to left multiplication by a succession of elementary matrices (see explanations here), but I think it is less clear.

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