What is the relationship between $S \oplus T$ and $T \oplus S$? Is the direct sum operation commutative? Formulate and prove a similar statement concerning associativity. Is there an "identitiy" for direct sum? What about "negatives"?
$S \oplus T = T \oplus S$ because elements of $S \oplus T$ are of the form $s+t$ where $s \in S$ and $t \in T$, and we know that $s + t = t + s$. Associativity is valid with the same argument. The identity is $\{0\}$. However, we cannot have negatives, because a negative subspace doesn't make sense, since if $s \in S$ then $-s \in S$.
Let V be a finite-dimensional vector space over an infinite field F. Prove that if $S_1, … ,S_k$ are subspaces of $V$ of equal dimension, then there is a subspace $T$ of $V$ for which $V = S_i \oplus T$ for all $I = 1, … ,k$. In other words, $T$ is a common complement of the subspaces $S_i$.
We know that every subspace has a complement. So let's say that the complement of $S_m$ where $1 \leq m \leq k$ is $T$. Then $$\dim(V) = \dim(S_m) + \dim(T).$$
For any $S_n$ where $1 \leq n \leq k$, we know that $$\dim(S_n) + \dim(T) = \dim(S_n + T) + \dim(S \cap T).$$
Since $\dim(S_n) = \dim(S_m)$,
$$(\dim(V) – \dim(T)) + \dim(T) = \dim(S_n + T) + \dim(S \cap T)$$
$$\implies \dim(V) = \dim(S_n + T) + \dim(S \cap T) = \dim(S_n) + \dim(T)$$
So $T$ must also be a complement of $S_n$ for all $1 \leq n \leq k$.
Do you think my answer is correct?
Thank you in advance
Best Answer
This conclusion is incorrect. For example, consider $V=M_2(\mathbb{R})$, two-by-two matrices with real coefficients. Set $$S_1=\left\{\left(\begin{smallmatrix}a&b\\c&0\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
$$S_2=\left\{\left(\begin{smallmatrix}a&b\\0&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
$$S_3=\left\{\left(\begin{smallmatrix}a&0\\b&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
$$S_4=\left\{\left(\begin{smallmatrix}0&a\\b&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$
Then $T=\left\{\left(\begin{smallmatrix}0&0\\0&d\end{smallmatrix}\right):d\in\mathbb{R}\right\}$ has $S_1\oplus T=V$, but $S_2\oplus T\neq V$.