Subspace, Basis

**millerst** · February 11th 2010, 05:17 PM

I posted a question, which I have attempted all except for c. Could someone tell me if I did the question correctly? And how do I begin (c). Thanks, there is a PDF file attached.

**NonCommAlg** · February 11th 2010, 07:07 PM

Originally Posted by millerst

I posted a question, which I have attempted all except for c. Could someone tell me if I did the question correctly? And how do I begin (c). Thanks, there is a PDF file attached.

in part (a) you have a couple of mistakes: in proving that W is closed under addition that $-a+b$ should be changed to $-(a+b).$ also $\sigma$ is not in $W.$ it's in your base field.

finally in order to show that a non-empty set is a subspace you do not need to prove that it contains $0.$ unfortunately some instructors give this wrong idea to some students.

for part (b), the basis you found is wrong! you should at least see that none of the elements of the set that you think is a basis basically belongs to $W.$ anyway, a basis of $W$

has only one element. for example $\left \{ \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} \right \}.$

for part (c) first see that an $n \times n,$ matrix $A=[a_{ij}],$ with real (or complex) entries, is antisymmetric iff $a_{ii}=0$ for all $1 \leq i \leq n$ and $a_{ij}=-a_{ji}$ for all $1 \leq i \neq j \leq n.$ can you

see the general form of $A$ ? if not, try to do it for n = 3 first. now it should be easy to show that a basis for $n \times n$ antisymmetric matrices has $\frac{n(n-1)}{2}$ elements. what are they?

**millerst** · February 11th 2010, 07:12 PM

So would it be correct to say that a = 0, therefore the set contains {0}. ?

**NonCommAlg** · February 11th 2010, 07:14 PM

Originally Posted by millerst

So would it be correct to say that a = 0, therefore the set contains {0}. ?

sure, you can put a = 0 to get the zero matrix. but, as i said, this part is not necessary.

**millerst** · February 11th 2010, 07:17 PM

Then how exactly do you prove it contains a zero vector?

**NonCommAlg** · February 11th 2010, 07:21 PM

Originally Posted by millerst

Then how exactly do you prove it contains a zero vector?

well, $W$ is not empty and it's closed under addition and scalar multiplication. so if we choose $w \in W,$ then we must have $0=w + (-1)w \in W.$

**HallsofIvy** · February 12th 2010, 04:31 AM

The reason why we often do prove that the 0 vector is in the set is simply to prove it is non-empty- and 0 is typically easiest to work with. Here, you are asked to show that the set of anti-symmetric 2 by 2 matrices is a subspace and I would disagree with millerst- you are NOT given that it is non-empty, you need to show that. Of course, you could do that as well by showing that $\begin{bmatrix}0 & 1 // -1 & 0\end{bmatrix}$ is in the set as by showing that $\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ is in the set.

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