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My example: Proof that the subspace of strictly upper triangular matrices in $M_{n}(\mathbb R)$ and the subspace of strictly lower triangular matrices in $M_{n} (\mathbb R)$ have a null intersection. Is their direct sum whole vector space $M_{n} (\mathbb R)$? $M_{n} (\mathbb R)$ is the set of $n \times n$ matrices.

If I write down strictly upper ($U$)and strictly lower ($L$) triangular matrices, is obvious that $L\cap U = \{0\}$. But do I have to prove it more formally?

Those subspaces do not form whole $M_{n} (\mathbb R)$, because:

$$\dim L + \dim U = \frac {n(n-1)}2 + \frac {n(n-1)}2 = n^2 -n$$

and that doesn't equal to $n^2$.

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    Note: `$\mathbb R$` produces $\Bbb R$ as opposed to $R$. If you prefer, there's also `$\mathbf R$` which is $\mathbf R$. – 2017-02-07
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    Also, `$n \times n$` produces $n \times n$ as opposed to $nxn$. Also, use `$\dim$` rather than `$dim$`. – 2017-02-07

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Here's one formal approach:

$U$ consists of the matrices $A$ such that $A_{ij} = 0$ whenever $i\geq j$. On the other hand, $L$ consists of matrices such that $A_{ij} = 0$ whenever $i \leq g$. Thus, if $A \in U \cap L$, then $A_{ij} = 0$ for all $i,j$, which is to say that $A = 0$.

Your proof for the second part is fine. Alternatively, we can observe that all elements $A + B$ with $A \in U$ and $B \in L$ have zeros along the main diagonal. So, for example, the identity matrix is not in $U + V$.

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    @Leif sure, that's fine. Or to be even lazier: $\dim (L \vee U) \leq \dim L + \dim U$ (I assume $L \vee U$ denotes the subspace generated by $L$ and $U$). – 2017-02-07
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    Can I prove it by this? $dimL + dimU = dim(L \cap U) + dim(L \lor U)$, so $dim(L \cap U) = dimL + dimU - dim(L \lor U) = {{n(n-1)}\over 2} + {{n(n-1)}\over 2} - (n^2 - n) = 0$, because if we sum those two types of matrices its dimension would be the dimension of $M_n(\mathbb R)$ minus $n$, because $n$ is the dimension of diagonal matrix and in our case that diagonal is made just of zeros. – 2017-02-07
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    And thank you for helping me. – 2017-02-07
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    That also is fine. I don't think it's any easier, though – 2017-02-07