True or false: $V$ is a real vector space and $\left\{a,b,c \right \}$ is linearly independent in $V$. Then for any $x,y,z \in \mathbb{R}$ , $\left\{xa+yb+zc,b,c \right\}$ is also linearly independent.

This is a task from a test-exam but sadly we didnt't get the solutions. I'm not sure how to solve that correctly that's why I ask here.

What is very confusing is $zc,b,c$.. do you know what's meant by that?

Anyway, I think this is true because it's said that $\left\{a,b,c \right \}$ is linearly independent in $V$, so multiplying them with real numbers will keep them linearly independent as well but I think the importance is this $zc,b,c$ which might make it wront the problem is I don't understand that notation.

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    $\begingroup$ It's asking you to consider the linear in/dependence of the set containing the vectors $x\cdot a+y\cdot b+z\cdot c,$ $b,$ and $c$ (dots added to emphasize scalar multiplication of a vector). That is, $a$ is replaced with a linear combination of $a,b,$ and $c$ with coefficients $x,y,$ and $z.$ $\endgroup$ – RideTheWavelet Feb 28 '17 at 22:33
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    $\begingroup$ False, unless you add the condition that $x\neq 0$. $\endgroup$ – Thomas Andrews Feb 28 '17 at 22:34
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    $\begingroup$ if $x=0$, then $xa+yb+zc = yb+zc$ and is a linear combination of $b$ and $c$. $\endgroup$ – Guangliang Feb 28 '17 at 22:35
  • $\begingroup$ Can some of you please make a short explaining answer of it? :) $\endgroup$ – cnmesr Feb 28 '17 at 22:36

The span of $\{xa+yb+zc,b,c\}$ is contained in the span of $\{a,b,c\}$; call $V$ this space and compute the coordinate vectors with respect to the basis $\{a,b,c\}$: you get the three columns $$ \begin{bmatrix}x\\y\\z\end{bmatrix} \quad \begin{bmatrix}0\\1\\0\end{bmatrix} \quad \begin{bmatrix}0\\0\\1\end{bmatrix} $$ and the problem reduces to computing the rank of $$ \begin{bmatrix} x & 0 & 0 \\ y & 1 & 0 \\ z & 0 & 1 \end{bmatrix} $$ The rank is clearly $2$ if $x=0$. If $x\ne0$, we can easily compute the RREF of the matrix as $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ which has rank $3$. So the set $\{xa+yb+zc\}$ is linearly independent if and only if $x\ne0$.


If $\{a,b,c\}$ are linearly independent, you cannot build $a$ from summing scalar multiples of $b,c$. The same applies for any permutation of these three. Using the same notation as lappen68,

$$a \ne \lambda_2b + \lambda_3c$$

$$b \ne \lambda_1a + \lambda_3c$$

$$c \ne \lambda_1a + \lambda_2b$$

This is equivalent to the statement that $\lambda_1a+\lambda_2b+\lambda_3c = 0$ can only be satisfied trivially. None of these vectors are linear combinations of the others.

In the new set $\left\{xa+yb+zc,b,c \right\}$, you can construct the first vector as a linear combination of the latter two

$$xa+yb+zc = \lambda_2b+\lambda_3c$$

if $x=0, y=\lambda_2, z=\lambda_3$.


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