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How can I prove that:

$\dim(M_1\vec{\oplus}M_2\vec{\oplus}\ldots\vec{\oplus}M_n) = \dim(M_1) + \dim(M_2) + \ldots + \dim(M_m)$

Where $M_i\lt L,\forall i\in\{1,\ldots,n\}$, and $\vec{\oplus}$ is the direct sum.

Matcha Latte
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Ramon
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1 Answers1

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Let $U_1$ and $U_2$ be subspaces of vector space $V$. We show first that $$ \dim(U_1 \oplus U_2) = \dim(U_1) + \dim(U_2) \; . $$ Let $(v_1, \ldots v_n)$ a basis of $U_1$ and $(w_1, \ldots, w_m)$ a basis of $U_2$. We can show that $(v_1, \ldots, v_n, w_1, \ldots, w_m)$ is a basis of $U_1 \oplus U_2$.

You have to check two things:

  1. Let $u \in U_1 \oplus U_2$. Then there are scalars $\alpha_1, \ldots,\alpha_n, \beta_1, \ldots, \beta_m$, such that $$ u = \sum_{i=1}^n \alpha_i v_i + \sum_{j=1}^m \beta_j w_j \; . $$
  2. The vectors $\{v_1, \ldots, v_n, w_1, \ldots, w_m\}$ are linearly independent, i.e. $$ \sum_{i=1}^n \alpha_i v_i + \sum_{j=1}^m \beta_j w_j = 0 \quad \implies \quad \alpha_1 = \ldots = \alpha_n = \beta_1 = \ldots = \beta_m = 0 $$

Finally, you can use induction to show the general case.

aexl
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