- #1

- 211

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## 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.