2
$\begingroup$

I came across this problem:

Given the set of linearly independent vectors $\{u,v,w\} \subseteq \mathbb{R}^n$, determine whether or not the following set of vectors is linearly dependent: $\{u-v-w, 2u+w, 3u+v+3w\}$

There's a theorem which says that if at least one of the vectors in a set is a linear combination of other vectors, then the set is linearly dependent. In this particular question, I had to go through all of the three vectors $\{u-v-w, 2u+w, 3u+v+3w\}$ and check whether any of them is a linear combination of others (and I think the answer is that the set is linearly independent).

But is there more efficient way of doing this? I tried checking linear dependence using matrix, that is, checking for a non-trivial solution of the following system:

$$\begin{bmatrix} (u_1-v_1-w_1) & (2 u_1 + w_1) & (3 u_1 + v_1 + 3 w_1) & 0\\ (u_2-v_2-w_2) & (2 u_2 + w_2) & (3 u_2 + v_2 + 3 w_2) & 0\\ & \vdots \\ (u_n-v_n-w_n) & (2 u_n + w_n) & (3 u_n + v_n + 3 w_n) & 0 \end{bmatrix}$$

But I have no idea how to do this because I don't know the components of the vectors $u, v, w$ and I don't know $n$.

$\endgroup$
3
$\begingroup$

Suppose $$ \DeclareMathOperator{Null}{Null} \DeclareMathOperator{Span}{Span} \lambda_1\cdot(u-v-w)+\lambda_2\cdot(2u+w)+\lambda_3\cdot(3u+v+3w)=\mathbf 0\tag{1} $$ Then $$ (\lambda_1+2\lambda_2+3\lambda_3)\cdot u + (-\lambda_1+\lambda_3)\cdot v + (-\lambda_1+\lambda_2+3\lambda_3)\cdot w = \mathbf0\tag{2} $$ Since $\{u,v,w\}$ is linearly independent (2) implies \begin{align*} \lambda_1+2\lambda_2+3\lambda_3 &= 0 \\ -\lambda_1+\lambda_3&= 0 \\ -\lambda_1+\lambda_2+3\lambda_3 &= 0 \end{align*} which is equivalent to the equation $A\vec\lambda=\mathbf 0$ where $$ A= \begin{bmatrix} 1 & 2 & 3 \\ -1 & 0 & 1 \\ -1 & 1 & 3 \end{bmatrix} $$ But $\DeclareMathOperator{Rank}{Rank}\Rank(A)=2$ and we see that $$ \Null(A)= \Span \left\{ \begin{bmatrix} 1\\-2\\ 1 \end{bmatrix} \right\} $$ Hence the equation (1) is solved by $\lambda_1=1$, $\lambda_2=-2$, and $\lambda_3=1$ and we see that $\{u-v-w, 2u+w, 3u+v+3w\}$ is linearly dependent.

$\endgroup$
  • $\begingroup$ Brilliant. Thank you, sir! How I didn't think of that. $\endgroup$ – user175788 Oct 22 '14 at 7:54
0
$\begingroup$

The information that $u$, $v$ and $w$ belong to $\mathbb{R}^n$ is irrelevant; it's even better thinking to those vector in an arbitrary $n$-dimensional space $V$. You can complete $\{u,v,w\}$ to a basis of $V$, say to the set $\{u,v,w,z_4,\dots,z_n\}$ and there is a unique linear map $f\colon V\to\mathbb{R}^n$ such that \begin{align} f(u)&=e_1\\ f(v)&=e_2\\ f(w)&=e_3\\ f(z_k)&=e_k &&(4\le k\le n) \end{align} and this map is bijective, because it sends a basis to a basis. Thus a set of vectors $\{x_1,x_2,\dots,x_m\}$ is linearly independent if and only if $\{f(x_1),\dots,f(x_n)\}$ is linearly independent in $\mathbb{R}^n$.

This shows that it's not restrictive to assume $u=e_1$, $v=e_2$ and $w=e_3$, so you just need to check whether the set $$ \left\{ \begin{bmatrix}1\\-1\\-1\\0\\\vdots\\0\end{bmatrix}, \begin{bmatrix}2\\0\\1\\0\\\vdots\\0\end{bmatrix}, \begin{bmatrix}3\\1\\3\\0\\\vdots\\0\end{bmatrix} \right\} $$ is linearly independent which is clearly equivalent to showing that the matrix $$ \begin{bmatrix} 1&2&3\\ -1&0&1\\ -1&1&3 \end{bmatrix} $$ has rank $3$. The rank is easily computed with Gaussian elimination: $$ \begin{bmatrix} 1&2&3\\ -1&0&1\\ -1&1&3 \end{bmatrix} \to \begin{bmatrix} 1&2&3\\ 0&2&4\\ 0&3&6 \end{bmatrix} \to \begin{bmatrix} 1&2&3\\ 0&1&2\\ 0&3&6 \end{bmatrix} \to \begin{bmatrix} 1&2&3\\ 0&1&2\\ 0&0&0 \end{bmatrix} $$ so the rank is $2$ and the three given vectors are linearly dependent. If you go on with backwards elimination, you reach $$ \begin{bmatrix} 1&0&\color{red}{-1}\\ 0&1&\color{red}{2}\\ 0&0&0 \end{bmatrix} $$ so you conclude that $$ 3u+v+3w=\color{red}{-1}(u-v-w)+\color{red}{2}(2u+w) $$

$\endgroup$
0
$\begingroup$

Forget what space $u,v,w$ live in; the only important thing is that they are linearly independent. Therefore they form a basis of the subspace $V=\langle u,v,w\rangle$ spanned by themselves (they are a spanning set by definition of $V$, and independent by hypothesis). This means that the operation of formation of linear combinations $L:(a,b,c)\mapsto au+bv+cw$ is an isomorphism $\def\R{\Bbb R}\R^3\to V$ so it has an inverse isomorphism $L^{-1}:V\to\R^3$ which is expression of vectors of $V$ in coordinates on the basis $[u,v,w]$. Now directly form the definition $$ \begin{align} L^{-1}(u−v−w)&=(1,-1,-1), \\L^{-1}(2u+w)&=(2,0,1), \\ L^{-1}(3u+v+3w)&=(3,1,3). \end{align} $$ Since an isomorphism preserves linear dependence or independence, the question whether the vectors $u−v−w$, $2u+w$, $3u+v+3w$ of $V$ are linearly dependent is equivalent to the question whether their images $(1,-1,-1)$, $(2,0,1)$, and $(3,1,3)$ in $\R^3$ under the isomorphism $L^{-1}$ are linearly dependent.

Thus a somewhat abstract question in $\R^n$ is translated into a concrete question in$~\R^3$. To actually answer that question, one has to set up and find all solutions of a $3\times 3$ system of linear equations, which is exactly the system in the answer by Brain Fitzpatrick. Since there are actually non-zero solutions, the given vectors are linearly dependent. I won't repeat that (easy) computation; the main point I want to make is that the reduction to a problem in $\R^3$ results directly from the expression of relevant vectors in coordinates with respect to an appropriate basis (here $[u,v,w]$), which is why for computations the spaces of the form $\R^k$ are of such a central importance, and why the notion of basis (of a space or of a subspace) is so important.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.