Decide if the following statements are true or false. Provide arguments or a couterexample to support your answers:

1) The set of matrices $A$ with $\det(A)=1$ is a subspace in the vector space $\mathcal{M}_{2 \times 2}(\mathbb{R})$ of $2 \times 2$ matrices.

2) $\dim(\operatorname{Null}(A))=\dim(\operatorname{Null}(S_{A})).$

3) If $S=\operatorname{span}(u_{1},u_{2}, \ldots, u_{n})$ then $\dim(S)=n$.

4) The intersection of two vector subspaces of a vector space $V$ cannot be empty.

5) In the vector space $\mathcal{M}_{2 \times 2}(\mathbb{R})$ consider $M$, the set of matrices with positive elements. The subspace spanned by the matrices from $M$ is $\mathcal{M}_{2 \times 2}(\mathbb{R})$ itself.

I think the affirmative answers are for the questions 1), 3). But about the remaining questions I can't tell anything. I am not sure, but I think 4) is affirmative, also--but I'm not sure.

Thanks :)

  • 1
    $\begingroup$ about 4 - what element must be included in all sub-spaces? $\endgroup$ – mousomer Dec 4 '12 at 10:53
  • 2
    $\begingroup$ What is $S_A$ in 2)? $\endgroup$ – Rudy the Reindeer Dec 4 '12 at 10:59
  • $\begingroup$ @mousomer the null vector :) am I right ? $\endgroup$ – Iuli Dec 4 '12 at 11:07
  • $\begingroup$ @MattN. I don't know. Can be a subspace ? $\endgroup$ – Iuli Dec 4 '12 at 11:13
  • $\begingroup$ Null? Are you referring to the zero vector? As for $S_A$ - that's probably the transform described by the matrix A. $\endgroup$ – mousomer Dec 4 '12 at 12:02

1) The matrices having determinant equal to 1 form a group called special linear group. They do not form a vector subspace of $M_{2,2}$, since zero matrix does not belong to $\operatorname{SL}(2,F)$.

3) What if $u_1=\dots=u_n$? This is true only if the vectors are linearly independent.

5) Notice that this subspace contains matrices $A_1=\begin{pmatrix}2&1\\1&1\end{pmatrix}$, $A_2=\begin{pmatrix}1&2\\1&1\end{pmatrix}$, $A_3=\begin{pmatrix}1&1\\2&1\end{pmatrix}$, $A_4=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ and $A_5=\begin{pmatrix}1&1\\1&1\end{pmatrix}$. Can you obtain matrices from the standard basis as their linear combinations?

The part 4 was solved in comments. Without knowing what $S_A$ means, I can't really say anything about the part 2.

  • $\begingroup$ What is $A_5$ for? Over real numbers, we perfectly can divide by 5. $\endgroup$ – Incnis Mrsi Aug 29 '14 at 17:17
  • $\begingroup$ I have simply suggested set of matrices which belong to $M$ and it is easy to generate standard basis for them. Choosing any 4 linearly independent matrices would suffice (for example $A_1$, $A_2$, $A_3$, $A_4$), but I wanted to choose some elements for which it is immediately obvious that they generate the whole space. $\endgroup$ – Martin Sleziak Aug 29 '14 at 17:44
  • $\begingroup$ In other words: You are right in saying that $A_5=\frac{A_1+\dots+A_4}5$, so it belongs to span(M). But it is easier to argue that this matrix belongs to span(M) simply because it belongs to M. $\endgroup$ – Martin Sleziak Aug 29 '14 at 17:44

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