0
$\begingroup$

I have the set:

$$S = \left\{\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}, \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0\\1 & 1\end{bmatrix}, \begin{bmatrix}0 & 1\\1 & 2\end{bmatrix}\right\}$$ And I want to verify if it spans $M_2(\mathbb{R})$ (the space of all 2x2 matrices).

1) If I verify this set is linearly independent, and knowing dim $M_2(\mathbb{R}) = 4$, is it okay to say that $S$ is a basis for $M_2(\mathbb{R})$ because it is linearly independent and has 4 vectors? I can verify if my set is L.I. and has $n$ vectors, then it spans the space of dimension $n$ formed by that vectors?

2) I think i've read somewhere that if I check that the set is L.I. and then write each vector of $S$ as a linear combination of the canonical basis for $M_2(\mathbb{R})$, then S is a set that spans $M_2(\mathbb{R})$. Why this is true? Can somebody give me an explanation of why this works? Because, for me, any vector in any set could be written as a linear combination of the canonical basis...

$\endgroup$
  • $\begingroup$ 1) yes and yes again 2) I think it's presumably the other way around: if you can write any element of the canonical basis via your set of vectors then you know that these vectors must span the whole space. Clearly if their number is equal to the dimension of the space, then these must be L.I, i.e. a basis. $\endgroup$ – b00n heT Feb 1 '15 at 18:07
  • $\begingroup$ It is possible to simplify (2) if we know the dimension of the vector space, for which a basis is sought. Here you probably know, from the mention of a "standard basis", that $M_2(\mathbb{R})$ is dimension four (over the reals). So it would suffice to show the linear independence OR spanning property of a set of four vectors, to prove that a basis. $\endgroup$ – hardmath Feb 1 '15 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.