Are direct sums necessarily invertible?

Question

In my linear algebra text, the example used to introduce direct sums (shown below) results in an invertible vector space. I am struggling to understand whether this happened to be the case in this instance, or if this is necessarily true for all direct sums.

Answer 1

The question boils down to this:

What possible dimension can the sum $U_1 + \dots + U_m$ have?

It is bounded by the sum of the dimensions of the $U_i$, i.e., $\dim (U_1 + \dots + U_m) \leq \sum_{i = 1}^m \dim U_i$. This is essentially true by definition of the (internal) sum: $U_1 + \dots + U_m$ is the subspace generated by the $U_i$, hence the dimension cannot be greater than the sum of the individual dimensions. If you prefer, the internal sum is by definition the image of the map given in your source (another way to see that the inequality holds, see next part).

On the other hand, $\dim (U_1 \times \dots \times U_m) = \sum_i \dim U_i$.

Thus, if the map is injective, the dimension of the sum $U_1 + \dots + U_m$ is at least $\sum_i \dim U_i$. This enforces equality in the inequality.

Then by the usual theorem that for a linear map between vector spaces of equal dimension, being injective and surjective and bijective are equivalent, you get your desired result (the converse of course is trivial).

Small remark, the direct sum itself is not invertible (which makes not a lot of sense in this context).