Finding the Dimension of the Union of Subspaces

[AwesomeTrains]

Evening everyone, I have a problem with addition of subspaces.

Homework Statement

I have to find the dimension of U and dim(V), of the union dim(U+V) and of dim(U\capV)
U is spanned by
\begin{align}
\begin{pmatrix}
1 \\
-2 \\
0
\end{pmatrix},
\begin{pmatrix}
1 \\
1 \\
2
\end{pmatrix}
\end{align} and V is spanned by
\begin{align}
\begin{pmatrix}
3 \\
0 \\
4
\end{pmatrix},
\begin{pmatrix}
0 \\
3 \\
a
\end{pmatrix}
\end{align} a\in\textbf{R}

Homework Equations

dim(U)+dim(V)-dim(U\capV)=dim(U+V)

The Attempt at a Solution

Because the vectors spanning U and V are lin. independent:
dim(U) = dim(V) = 2

I find the intersection by equaling the two subspaces and then solving the linear system. But how do I find the sum of the two subspaces without calculating the intersection first?
Any hints are very appreciated :)

 

[HallsofIvy]

First, a minor point: you titled this "sum or union o subspaces". The union of two subspaces of a vector space is not, in general a subspace. I presume you accidently wrote "union" when you meant "intersection". It is true that the direct sum of two subspaces is spanned by the union of their bases (or any two sets that span the respective spaces) so perhaps that was what you meant.

We are told that U is spanned by the two vectors \begin{pmatrix}1 \\-2 \\ 0 \end{pmatrix} and \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}. Those vectors are independent (one is not a multiple of the other) so U is two dimensional.

We are told that V spanned by the two vectors \begin{pmatrix}3\\ 0 \\ 4 \end{pmatrix} and \begin{pmatrix}0 \\ 3 \\ a\end{pmatrix}. Those vectors are independent, for any a, (one is not a multiple of the other) so V is two dimensional.

The direct sum of these two subspaces is spanned by the union of the bases for these two subspaces so to determine the dimension (it could be 2, or 3- since these are all subspaces of R³, they can't all be independent), determine whether those four vectors are independent.
One way to determine that is from the definition of "linearly independent": the only values of \alpha, \beta, \gamma, and \delta, such that \alpha\begin{pmatrix}1 \\-2 \\ 0 \end{pmatrix}+ \beta\begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}+ \gamma\begin{pmatrix}3\\ 0 \\ 4 \end{pmatrix}+ \delta\begin{pmatrix}0 \\ 3 \\ a\end{pmatrix}= \begin{pmatrix}0<br /> \\ 0 \\ 0 \end{pmatrix} which gives the three equations \alpha+ \beta+ 3\gamma= 0, -2\alpha+ \beta+ 3\delta= 0, and 2b+ 4c+ ad= 0.

An equivalent way of doing this is to construct a matrix having those vectors as columns and "row reduce":
\begin{pmatrix}1 &amp; 1 &amp; 3 &amp; 0 \\ -2 &amp; -1 &amp; 0 &amp; 3 \\ 0 &amp; 2 &amp; 4 &amp; a\end{pmatrix}
Add twice the first two to the second to get
\begin{pmatrix}1 &amp; 1 &amp; 3 &amp; 0 \\ 0 &amp; 3 &amp; 6 &amp; 3 \\ 0 &amp; 2 &amp; 4 &amp; a\end{pmatrix}
Subtract 2/3 the second row from the first to get
\begin{pmatrix}1 &amp; 1 &amp; 3 &amp; 0 \\ 0 &amp; 3 &amp; 6 &amp; 3 \\ 0 &amp; 0 &amp; -5 &amp; a- 2\end{pmatrix}.

Since we still have three non-zero rows the dimension of U+ V is 3.

 

[AwesomeTrains]

First of all thanks for taking your time to write such a throughout explanation, it helped me understand it better.
Though I think you made a mistake at the row reduction. I think the last row vector in the last matrix should be
\begin{pmatrix}
0 & 0 & 0 & a-2\\
\end{pmatrix}
That would also explain why they put the a in there. When a=2 the dimension of the sum is 2 else 3, am I right?
Oh, by the way there's also a sign error in the first matrix, at 2,2.
And thank you for the fast response :)