- #1
Kernul
- 211
- 7
Homework Statement
I have this exercise that tells me to determine a base and the dimension of the subspaces of ##\mathbb {R}^4##, ##U \cap Ker(f)## and ##U + Ker(f)##, knowing that:
##U = <\begin{pmatrix}
-10 \\
11 \\
2 \\
9
\end{pmatrix}
\begin{pmatrix}
1 \\
1 \\
1 \\
3
\end{pmatrix}
\begin{pmatrix}
14 \\
-7 \\
2 \\
21
\end{pmatrix}
\begin{pmatrix}
11 \\
-10 \\
-1 \\
12
\end{pmatrix}>##
##B_{Ker(f)} = \left[\begin{pmatrix}
1/2 \\
0 \\
0 \\
1
\end{pmatrix}
\begin{pmatrix}
0 \\
0 \\
1 \\
0
\end{pmatrix}
\begin{pmatrix}
-3/2 \\
1 \\
0 \\
0
\end{pmatrix} \right]## e ##dim(Ker(f)) = 3##
Homework Equations
The Attempt at a Solution
Now, I see that the dimension of ##U## is ##4##.
I proceed now by finding the vector ##\vec v \in Ker(f)##, so I will have
##\vec v = \alpha \begin{pmatrix}
1/2 \\
0 \\
0 \\
1
\end{pmatrix} + \beta \begin{pmatrix}
0 \\
0 \\
1 \\
0
\end{pmatrix} + \gamma \begin{pmatrix}
-3/2 \\
1 \\
0 \\
0
\end{pmatrix} = \begin{pmatrix}
1/2 \alpha - 3/2 \gamma \\
\gamma \\
\beta \\
\alpha
\end{pmatrix}##
How should I proceed now? Do I have to take one of the vector of ##U## and match it with the one just found for ##Ker(f)##?
What I mean is:
I take the first vector of ##U## and do ##-10(1/2 \alpha - 3/2 \gamma) + 11(\gamma) + 2(\beta) + 9(\alpha) = 0## and then I continue by finding that ##\beta = -2\alpha - 13\gamma## and I substitute it in the vector found before with ##Ker(f)##, finding then that the dimension of the intersection is ##2##.
It's just that after, when I use Sarruss to find the dimension of the sum, I found myself with ##5## as a dimension, and so ##U + Ker(f) \in \mathbb {R}^5##. Is it normal that the sum is this big so that we have ##\mathbb {R}^5## even though we started with ##\mathbb {R}^4##? It seems weird to me.