2
$\begingroup$
  1. The invertible $3 \times 3$ matrices
  2. The $3\times 3$ matrices whose entries are all integers
  3. The $3\times 3$ matrices with all zeros in the third row
  4. The non-invertible $3\times 3$ matrices
  5. The diagonal $3\times 3$ matrices
  6. The symmetric $3\times 3$ matrices

So the subspace must be closed under linear combinations and include $0$. For these reasons, I picked answers 2-4, but this is not correct. How can I determine which of these are subspaces?

$\endgroup$
3
  • 2
    $\begingroup$ You're right, subspaces are closed under linear combinations and include $0$. What was your reasoning for each of the entries, 1 through 6? $\endgroup$ Oct 21, 2014 at 7:43
  • 1
    $\begingroup$ None, those collections aren't subsets of $\mathbb{R}^3$. Are you sure you don't mean something like $\mathbb{R}^{3\times 3}$, which is usually used to denote the space of all $3\times 3$ matrices with entries in $\mathbb{R}$? $\endgroup$
    – Hayden
    Oct 21, 2014 at 7:57
  • 1
    $\begingroup$ @Hayden Yes, sorry, I did mean $\mathbb{R^{3\times 3}}$. I will update the title. $\endgroup$ Oct 21, 2014 at 11:17

2 Answers 2

1
$\begingroup$

If the entries are integers and the linear combination includes $\pi$ as a coefficient, then the entries in the resulting matrix will usually not be integers (unless some coincidental cancellations happen). Try it in some simple cases where there are just two terms, with coefficients $\pi$ and $0$.

As for (3), a sum of non-invertible matrices is often invertible. For example $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$.

$\#5$ and $\#6$ are subspaces.

It is redundant to say they include $0$ if they're closed under linear combinations, because there is a linear combination in which all coefficients are zero.

$\endgroup$
0
$\begingroup$
  1. No, 'cause $0\cdot A$ is not invertible

  2. No, 'cause they are not closed under scalar multiplication over $\mathbb{R}$

  3. I would say yes... check it by hands!

  4. No, the sum of two of them may be invertible

  5. Yes!! Easy to check

  6. I would say "YES", but again, try to check it by hands!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .