How do you establish measurement

Suppose $U$ and also $V$ are $2$ dimensional subspaces of $\mathbb{R}^4$. Just how do you establish the measurement of $U \cap V$? I recognize that

\begin{equation*} \text{dim}(U + V) = \text{dim}(U)+ \text{dim}(V) - \text{dim}(U \cap V). \end{equation*}

So it appears that $\text{dim}(U \cap V) = 0$.

2019-05-18 22:48:46
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Answers: 2

If you have B and also B', basis of U and also V specifically, you can examine if the vectors of B are linearly independent with those of B'. If so, after that $\dim(U\cap V) = 0$. Otherwise, you might have the ability to write a new basis of $U+V$ removing the linearly reliant vectors. $\dim(U\cap V)$ is the quantity of vectors you have actually removed from $U+V$ is basis.

Additionally, $\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$.

2019-05-21 09:08:25

First: it is wrong, as a whole, to also write $\dim(U\cup V)$, due to the fact that $U\cup V$ is virtually never ever a subspace: it is a subspace if and also just if $U\subseteq V$ or $V\subseteq U$. Not being a subspace, it does not also make good sense to speak about its measurement.

Instead, what you possibly suggested is the proper formula $$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$ (though I favor to place the junction on the left hand side, due to the fact that because kind it stands also in the boundless dimensional instance). Below, $U+V$ is the tiniest subspace which contains $U$ and also $V$, and also it takes place to amount to the set of all vectors of the kind $u+v$ with $u\in U$ and also $v\in V$.

The formula does not offer you the complete solution, due to the fact that there are numerous scenarios that can take place: you can have $U$ and also $V$ converge trivially: this is what takes place when $U+V=\mathbb{R}^4$. For a specific instance, you can have $U=\{(a,b,0,0):a,b\in\mathbb{R}\}$, and also $V=\{(0,0,c,d) : c,d\in\mathbb{R}\}$. Below, $\dim(U\cap V)=0$.

Or you can have that $U$ and also $V$ converge in a one - dimensional subspace (as an example, take $U$ as above, yet take $V=\{(0,b,c,0) : b,c\in\mathbb{R}\}$).

Or you can have $U=V$, in which instance the junction has measurement $2$.

What you can claim is: if the rooms stand out, after that the junction will certainly either have measurement 1 or measurement 0. Yet that is all you can claim with the offered details.

2019-05-21 09:06:59