Let $U$ be a subspace of $V$ , $\dim U=k$ also Let $B=\{u_1,\ldots,u_k\}$ be a basis of $U$ and given vector $w\in V$ but $w \notin U$ we will define $A=\{u_1-w,\ldots,u_k-w\}$ and $W=\operatorname{Sp}(A)$… find
$\dim W$ , $\dim U+W$ and $\dim U \cap W$.
I don't know if I approached it right but here is my try:
Given $W = \operatorname{Sp}\{u_1-w,…,u_k-w\}$
we will check linear independence
$\alpha_1\cdot(u_1-w)+\cdots+ \alpha_k\cdot(u_k-w)$ $\iff$ $\alpha_1 u_1-\alpha_1w +\cdots+ \alpha_k u_k- \alpha_kw=0$ $\iff$ $(u_1+\cdots+u_k)\cdot(\alpha_1+\cdots+\alpha_k)-w\cdot(\alpha_1+\cdots+\alpha_k)=0$
we know that $B=\{u_1,\ldots,u_k\}$ is a basis therefore it is linearly independent so $\alpha_1\cdot(u_1)+\cdots+ \alpha_k\cdot(u_k)=0$ and $\alpha_1=\alpha_2=\cdots=\alpha_k=0$ but $w$ can be any value and multiplied by zero so it is linearly dependent(?) and from here we can remove the vector $w$ (?) so $\dim W=k$
from here I could not continue , the textbook has final answers: $\dim W=k$ , $\dim(U+W)=k+1$ and $\dim(U \cap W)=k-1$ and that $C=\{u_1,\ldots,u_k,w\}$ is linearly independent but where does this come from ?
Appreciate any tips and hints so I can try to solve it
Edit (solving according to Átila Correia):
first we will prove that $A$ is LI.
$\alpha_1(u_1-w)+…+\alpha_k(u_k-w)=0$ $\iff$ $\alpha_1 u_1 +… + \alpha_k u_k-(\alpha_1+…+\alpha_k)w=0$
according to given information about $B$ being a basis we get that $\alpha=…\alpha_k=0$ therefore $A$ is LI since all the scalars are equal to zero. from here the dimension of $W=Sp\{A\}$ is $\dim W=k$
now $dim(U+W)$ :
consider $x \in U+W$ then
$x=(\alpha_1 u_1 +…+ \alpha_k u_k) + (\beta_1 (u_1-w) +…+\beta_k (u_k-w)$
$x=(\alpha_1 +\beta_1)u_1 +…+ (\alpha_k +\beta_k)u_k -w(\alpha_1 +…+ \alpha_k)$
we know that $\{u_1,…,u_k\}$ is LI so $\alpha_1+\beta_1 =0$ $…$ $\alpha_k + \beta_k=0$ and we know that $\alpha_1=…=\alpha_k=0$ therefore $\beta_1=…=\beta_k=0$ and so $\{u_1,…,u_k,w\}$ is LI and $dim(U+W)=k+1$
and lastly $dim(U \cap W)$
if $x \in U \cap W$ then $x \in U$ and $x \in W$
$x=(\alpha_1 u_1 +…+ \alpha_k u_k) = \beta_1 (u_1-w)+…+ \beta_k (u_k-w) $
$(\alpha_1 u_1 +…+ \alpha_k u_k) = \beta_1 (u_1-w)+…+ \beta_k (u_k-w) $
$(\alpha_1 u_1 +…+ \alpha_k u_k) – (\beta_1 (u_1-w)+…+ \beta_k (u_k-w))=0 $
$(\alpha_1 -\beta_1)u_1 +…+ (\alpha_k -\beta_k)u_k – w(\alpha_1+…+\alpha_k)=0$
Best Answer
HINT
To begin with, let us prove that $A$ is LI.
In order to conclude so, let us consider the linear combination: \begin{align*} \alpha_{1}(u_{1} - w) + \ldots + \alpha_{n}(u_{n} - w) = 0 & \Longleftrightarrow \alpha_{1}u_{1} + \ldots + \alpha_{n}u_{n} - (\alpha_{1} + \ldots + \alpha_{n})w = 0 \end{align*}
from there, we can conclude that $\alpha_{1} + \ldots + \alpha_{n} = 0$.
Otherwise $w$ would belong to $U$, which is a contradiction.
Hence it can be deduced that $\alpha_{1} = \ldots = \alpha_{n} = 0$, and $A$ is LI.
Based on such result, we conclude that $\dim W = k$.
Now let us consider that $x\in U + W$.
Then there are vectors $a\in U$ and $b\in W$ such that $x = a + b$.
In other words, we have \begin{align*} x = a + b & = (a_{1}u_{1} + \ldots + a_{n}u_{n}) + (b_{1}(u_{1} - w) + \ldots + b_{n}(u_{n} - w))\\\\ & = (a_{1} + b_{1})u_{1} + \ldots (a_{n} + b_{n})u_{n} - (b_{1} + b_{2} + \ldots + b_{n})w \end{align*}
Hence $\dim(U + W) = k + 1$ because $\{u_{1},\ldots,u_{n},w\}$ is LI (why?).
Finally, $x\in U\cap W$ iff $x\in U$ and $x\in W$. In other words,
\begin{align*} x = a_{1}u_{1} + \ldots + a_{n}u_{n} = b_{1}(u_{1} - w) + \ldots + b_{n}(u_{n} - w) \end{align*} whence we conclude that \begin{align*} (a_{1} - b_{1})u_{1} + \ldots + (a_{n} - b_{n})u_{n} + (b_{1} + \ldots + b_{n})w = 0 \end{align*}
based on the same previous argument, we deduce that $a_{k} = b_{k}$ and $a_{1} + a_{2} + \ldots + a_{n} = 0$.
Consequently, $\dim(U\cap W) = k - 1$. That is because the set \begin{align*} \{u_{2} - u_{1}, u_{3} - u_{1},\ldots, u_{n} - u_{1}\} \end{align*}
is linear independent (why?).
and we are done.
Hopefully this helps !
EDIT
Since $a_{1} + a_{2} + \ldots + a_{n} = 0$, $x$ is expressed as
\begin{align*} x & = a_{1}u_{1} + a_{2}u_{2} + \ldots + a_{n}u_{n}\\\\ & = -(a_{2} + a_{3} + \ldots + a_{n})u_{1} + a_{2}u_{2} + \ldots + a_{n}u_{n}\\\\ & = a_{2}(u_{2} - u_{1}) + \ldots + a_{n}(u_{n} - u_{1}) \end{align*}
To prove that $\{u_{2} - u_{1}, u_{3} - u_{1},\ldots, u_{n} - u_{1}\}$ is LI, let us consider the linear combination:
\begin{align*} \alpha_{1}(u_{2} - u_{1}) + \ldots + \alpha_{n-1}(u_{n} - u_{1}) = 0 \Longleftrightarrow -(\alpha_{1} + \ldots + \alpha_{n-1})u_{1} + \alpha_{1}u_{2} + \alpha_{2}u_{3} + \ldots \alpha_{n-1}u_{n} = 0 \end{align*}
Since the set $\{u_{1},u_{2},\ldots,u_{n}\}$ is LI, we conclude that \begin{align*} \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n-1} = \alpha_{1} + \alpha_{2} + \ldots + \alpha_{n - 1} = 0 \end{align*}
whence it results the proposed claim is true.