Question

Question: If the subset of $\left\{u_1,\_\_\_\_\_,u_i\right\}$is linearly independent over and $\mathit{w}\mathbf{\in }\mathit{V}$is not a linear combination of the . Prove that$\left\{u_1,\_\_\_\_\_,u_i,w\right\}$ is linearly independent.

Short answer

Answer

Yes, $\left\{u_1,\dots ,u_i,w\right\}$is linearly independent

Step-by-step solution

Explanation

Let$\left\{u_1,\dots ,u_i,w\right\}$ is a linearly independent set.

Then we shall have to prove that the linear equation of the form,

$a_1{u}_1+\dots +a_i{u}_i+aw=0$________(1)

Where $a_1,a_2,\dots ,a_i$, a are scalar

This should imply that all must be equal to zero.

Concept

From equation (1)

$au=-\left(a_1{u}_1+\dots +a_i{u}_i\right)$

If,$a\ne 0$ we get $a^{-1}$by multiplication.

( is the inverse of a )

$au=-\left(a_1{u}_1+\dots +a_i{u}_i\right)$

This means that u is a linear combination of the vectors,

$u_1,\dots ,u_i$

That is $v\in \left[u_1,\dots ,u_i\right]$.

It is given that u does not belong to$\left[u_1,\dots ,u_i\right]$ and hence, we must have .

If we substitute a = 0 in (i), we get

$a_1{u}_1+\_\_\_+a_i{u}_i=0$

Solution

The vectors$u_1,\dots ,u_i$ are given to be linearly independent and hence,

.$u_1=\dots =u_i=0$

Thus we show that the equation (1),

Can hold only when all the coefficients are zero.

This proves that the vectors $u_1,\dots ,u_i,w$are linearly independent.