If $\{0_v\} \neq T \leq S_1$ and $S_1 \oplus S_2 = S_1 \oplus S_3$, then $T+S_2 = T+S_3$?

If $$\{0_v\} \neq T \leq S_1$$ and $$S_1 \oplus S_2 = S_1 \oplus S_3$$, then $$T+S_2 = T+S_3$$?

I tried this way:

Let $$S_1=span\{(1,0,0),(0,1,0)\}$$ and $$T$$, it's subspace, $$=span\{(1,0,0)\}$$.

Also lets have $$S_2=span\{(0,0,1)\}$$ and $$S_3=span\{(1,1,1)\}$$

Therefore, $$S_1 \cap S_2 = \{0_v\}$$ and $$S_1 \cap S_3 = \{0_v\}$$ so we have the direct sum.

Thus, $$T+S_2 = span\{(1,0,0)\} +span\{(0,0,1)\} \neq T+S_3 = span\{(1,0,0)\} +span\{(1,1,1)\}$$

Since for example, the vector $$(1,1,1)$$ is not contained in $$T+S_2$$

How does it look?

• Looks good to me! – zipirovich Nov 4 '18 at 20:17
• Your counterexample looks good! There are even counterexamples with the $S_i$ subspaces of $\Bbb{R}^2$. – Servaes Nov 4 '18 at 20:57
• By $$S_1 \oplus S_2 = S_2 \oplus S_3$$ did you mean $$S_1 \oplus S_2 = S_1 \oplus S_3\ ?$$ – bof Nov 4 '18 at 23:42
• @bof yes I meant that thanks for figurint it out..! It's still correct ? – iggykimi Nov 5 '18 at 19:00