Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\mathbb{F} = \mathbb{C}$, $V = \mathbb{C}^{3\times 3}$, the set of all complex $3\times 3$ matrices, and S is all the set of all matrices of the form
$$ \left( \begin{array}{ccc} a & a & a \\ 0 & 0 & a \\ a & a & a \end{array} \right)$$

where a is an arbitrary complex number.

determine if $S \subset \mathbb{C}^{3\times 3}$ is a subspace.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Yes, this is indeed a subspace.

To see that it is a subspace we need to check that it is closed under scalar multiplication and addition.

Choose any scalar $z\in \mathbb{C}$ then $$ z \left( \begin{array}{ccc} a & a & a \\ 0 & 0 & a \\ a & a & a \end{array} \right) = \left( \begin{array}{ccc} za & za & za \\ 0 & 0 & za \\ za & za & za \end{array} \right) \in S$$

and for any $a, b\in \mathbb{C}$ we have, $$ \left( \begin{array}{ccc} a & a & a \\ 0 & 0 & a \\ a & a & a \end{array}\right) + \left( \begin{array}{ccc} b & b & b \\ 0 & 0 & b \\ b & b & b \end{array} \right) = \left( \begin{array}{ccc} a+b & a+b & a+b \\ 0 & 0 & a+b \\ a+b & a+b & a+b \end{array} \right) \in S$$

Thus we see that $S$ is closed under scalar multiplication and addition of vectors, so it is a subspace.

share|improve this answer
    
thank you so much –  chanson Sep 26 '12 at 6:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.