prove or disprove direct sum question

**jayshizwiz** · November 7th 2010, 08:28 AM

Prove or disprove

$W, U_1, U_2$ are subspaces of $V$ .

If $V=U_1 \oplus U_2$

then $W = (U_1 \cap W) \oplus (U_2 \cap W)$

Attempt:

False

let's say $dimV = 8, dimU_1 = dimU_2 = 4 (dimV = dimU_1 + dimU_2)$
and $dimW = 5$

then $dim(U_1 \cap W)$ may equal 4
and $dim(U_2 \cap W)$ may equal 4

and then we get $5 \neq 4+4$

**jayshizwiz** · November 7th 2010, 12:32 PM

**Is the question understandable? Please let me know if there is something I need to explain better.

Thanks!

**tonio** · November 8th 2010, 02:16 AM

Originally Posted by jayshizwiz

Prove or disprove

$W, U_1, U_2$ are subspaces of $V$ .

If $V=U_1 \oplus U_2$

then $W = (U_1 \cap W) \oplus (U_2 \cap W)$

Attempt:

False

let's say $dimV = 8, dimU_1 = dimU_2 = 4 (dimV = dimU_1 + dimU_2)$
and $dimW = 5$

then $dim(U_1 \cap W)$ may equal 4
and $dim(U_2 \cap W)$ may equal 4

and then we get $5 \neq 4+4$

You haven't disproved anything: "may" it's not existence. You have to come up with a particular example that shows

clearly that the claim is false.

Hint: Take a look at $\mathbb{R}^2\,,\,\,U_1=\left\{\,\binom{x}{x}\in\ma thbb{R}^2\right\}\,,\,U_2=\left\{\,\binom{x}{-x}\in\mathbb{R}^2\right\}\,,\,\,W=\left\{\binom{x} {0}\in\mathbb{R}^2\right\}$

Tonio

**jayshizwiz** · November 10th 2010, 04:24 AM

Originally Posted by tonio

You haven't disproved anything: "may" it's not existence. You have to come up with a particular example that shows

clearly that the claim is false.

Hint: Take a look at $\mathbb{R}^2\,,\,\,U_1=\left\{\,\binom{x}{x}\in\ma thbb{R}^2\right\}\,,\,U_2=\left\{\,\binom{x}{-x}\in\mathbb{R}^2\right\}\,,\,\,W=\left\{\binom{x} {0}\in\mathbb{R}^2\right\}$

Tonio

I truly hope you aren't grading my exam (;

i sohld get at least partial credit

**Tinyboss** · November 11th 2010, 03:27 PM

The misconception that some students have, and that this problem is attempting to dispel, is that the direct sum is like a (set) union. But in fact, the direct sum of U1 and U2 contains a whole lot of elements that are not in U1 or U2. In case Tonio's excellent example wasn't clear, think about this and read it again.

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