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

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?

• 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. Oct 14, 2022 at 4:58
• @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? Oct 14, 2022 at 5:03
• @Seeker "I'm given the entries in each vector" Oct 14, 2022 at 6:08
• The two "=0" is to be suppressed. Oct 14, 2022 at 8:02
• @JeanMarie I couldn't figure out how to make it augmented Oct 15, 2022 at 0:03

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.

• I have completely re-written my answer. Oct 15, 2022 at 18:25

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.

• Thanks, I appreciate it Oct 15, 2022 at 22:32