Find dimension and a basis of a subspaces $U+V$, $U \cap V$ in terms of the parameter $\alpha$ Let $U=span((1,1,1,2),(1,2,2,\alpha))$ and $V=span((1,3,4,\alpha+2),(1,4,\alpha,\alpha+1))$ are the subspaces of $\mathbb{R^4}$. ($\alpha\in\mathbb{R}$).
Find dimension and one basis of $U+V$, $U \cap V$ in terms of $\alpha$.
For values $\alpha$, which satisfies that $\dim(U+V)$ is the smallest, check if $v=(1,2,3,4)$ is an element of $U+V$.
Attempt:
From the definition of vector space span $\Rightarrow$
$U=\{(x+y,x+2y,x+2y,2x+\alpha y):x,y\in\mathbb{R}\}$
$V=\{(t+p,3t+4p,4t+\alpha p,(\alpha+2)t+(\alpha+1)p):t,p\in\mathbb{R}\}$
$U\cap V\Rightarrow$
$$x+y=t+p$$
$$x+2y=3t+4p$$
$$x+2y=4t+\alpha p$$
$$2x+\alpha y=(\alpha+2)t+(\alpha+1)p$$
We can use Kronecker-Capelli's theorem to find for which $\alpha$ the system has a unique solution.
When will $\dim(U+V)$ be the smallest?
$\dim(U+V)=\dim(U)+\dim(V)-\dim(U\cap V)$ 
 A: First, note that $\DeclareMathOperator{rref}{rref}$
\begin{align*}
\rref
\begin{bmatrix}
1&1&1&2\\
1&2&2&\alpha
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&4-\alpha\\
0&1&1&\alpha-2
\end{bmatrix}
&
\rref
\begin{bmatrix}
1&3&4&2+\alpha\\
1&4&\alpha&1+\alpha
\end{bmatrix}
&=
\begin{bmatrix}
1&0&16-3\,\alpha&\alpha+5\\
0&1&\alpha-4&-1
\end{bmatrix}
\end{align*}
This proves that $\dim(U)=2$ and that $\dim(V)=2$.
Next, note that
$$
\det
\begin{bmatrix}
1&1&1&2\\
1&2&2&\alpha\\
1&3&4&\alpha+2\\
1&4&\alpha&\alpha+1
\end{bmatrix}
=(\alpha-7)(\alpha-3)
$$
This proves that $\dim(U+V)=4$ provided that $\alpha\neq7$ and $\alpha\neq 3$. Consequently
$$
\dim(U\cap V)=\dim(U)+\dim(V)-\dim(U+V)=2+2-4=0
$$
provided that $\alpha\neq7$ and $\alpha\neq 3$. 
Now, note that
\begin{align*}
\rref
\begin{bmatrix}
1&1&1&2\\
1&2&2&7\\
1&3&4&7+2\\
1&4&7&7+1
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&-3\\
0&1&0&8\\
0&0&1&-3\\
0&0&0&0
\end{bmatrix}
&
\rref
\begin{bmatrix}
1&1&1&2 \\
1&2&2&3 \\
1&3&4&3+2\\
1&4&3&3+1
\end{bmatrix}
&=
\begin{bmatrix}
1&0&0&1\\
0&1&0&0\\
0&0&1&1\\
0&0&0&0
\end{bmatrix}
\end{align*}
This proves that $\dim(U+V)=3$ provided that $\alpha=7$ or $\alpha=3$. Consequently
$$
\dim(U\cap V)=\dim(U)+\dim(V)-\dim(U+V)=2+2-3=1
$$
provided that $\alpha=7$ or $\alpha=3$.
This covers the questions about dimension. Can you use the above to answer the question about bases?
