Determine whether each given sequence...

Question

1. spans $M_{2\times 2}$
2. is a basis for $M_{2\times 2}$

$$ \mbox{(a)} \qquad \left( \begin{pmatrix} -4 & 1 \\ 0 & 5 \end{pmatrix}, \begin{pmatrix} -3 & 0 \\ -1 & 4 \end{pmatrix}, \begin{pmatrix} 0 & 3 \\ 4 & 6 \end{pmatrix} \right)$$
$$ \mbox{(b)} \qquad \left( \begin{pmatrix} 8 & 6 \\ -9 & 5 \end{pmatrix}, \begin{pmatrix} -3 & -2 \\ 4 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 5 & 7 \end{pmatrix}, \begin{pmatrix} -7 & -4 \\ 11 & 0 \end{pmatrix}, \begin{pmatrix} 2 & 4 \\ -7 & 10 \end{pmatrix} \right)$$

I have tried this problem but I just want to make sure that I'm going about it the correct way. For (a) I know that because only 3 vectors are present, it can't span $\mathbb{R}^4$ making it not span $M_{2,2}$ & because of that it also cannot be a basis for $M_{2,2}$.
In reality I only really need help making my reasoning clearer.

Answer 1

For (a), you are correct. $\dim(M_{2,2})=\dim(\mathbb{R}^4)=4$ so it cannot be spanned by less than 4 vectors. And, yes, since it does not span, it cannot be a basis.

For (b), it is easy to see that this is not a basis because (like (a)) there are the wrong number of vectors present. In this case, it is too big, so it cannot be a linearly independent set.

This just leaves question 1. for part (b): Does it span? To answer this, you need to calculate.

To do the calculation turn each $2 \times 2$ matrix into a column vector (i.e. a coordinate vector). Then slap them together into a matrix and row reduce...

$$ \begin{bmatrix} 8 & 6 \\ -9 & 5 \end{bmatrix} \qquad \Longrightarrow \qquad \begin{bmatrix} 8 \\ -9 \\ 6 \\ 5 \end{bmatrix}$$

You'll get...

$$ \begin{bmatrix} 8 & -3 & 0 & -7 & 2 \\ -9 & 4 & 5 & 11 & -7 \\ 6 & -2 & 2 & -4 & 4 \\ 5 & -1 & 7 & 0 & 10 \end{bmatrix} $$

If this matrix has rank 4, then you've got a spanning set (rank = dimension of the span). So if you row reduce and don't end up with a row a zeroes, you've got a spanning set.

Answer 2

You are right for part (a). As a vector space over $\Bbb{R}$, $M_{2\times 2}$ has dimension 4, so it can't be spanned by only 3 vectors. Any spanning set for $M_{2 \times 2}$ must have at least 4 vectors, and any basis must have exactly 4 vectors. Be careful to remember, though, that not every collection of 4 vectors is a basis, and not every collection of 4 or more vectors spans the space!

Answer 3

Reasoning (Without calculating) which is possible to take:

Let $V$ a vector space and $\emptyset \neq W \subseteq V$ ($W$ finite).

Prove (Or quote combinations of theorems) the following, and use them or their contrapositives:

1. $\dim({spanW}) \leq |W|$.

2. $ \dim{V} > |W| \implies V \neq spanW$.

3. $\dim{V} < |W| \implies W $ is linearly dependent $\implies W$ is not a basis of $V$.

((1.) Is pretty simple, and then (2.) follows naturally.)