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We have these two matrices:

$$K = \left(\begin{matrix} 2 & 1 \\ 8 & 7\end{matrix}\right), \quad L = \left(\begin{matrix} 2 & 1 \\ 2 & 7 \end{matrix} \right)$$

We have been asked if every matrix of $\mathbb{R}^{2 \times 2}$ can be written as a linear combination of $K$ and $L$ matrices. This means that the set $\{K,L\}$ is a base of $\mathbb{R}^{2 \times 2}$, right?

I've thought of this: For $K$ and $L$ matrices to be a base of $\mathbb{R}^{2 \times 2}$ they must be linearly independent, is that correct?

$a,b$ numbers of $\mathbb{R}$
$a \cdot K + b \cdot L = 0$, where $0$ is the $\left(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}\right)$ matrix.

So:

$$\begin{array}{cccc} 2a &+& 2b &=& 0 \\ a &+& b &=& 0 \\ 8a &+& 2b &=& 0 \\ 7a &+& 7b &=& 0 \end{array}$$

(the solution set of this system is empty set?)

How can I think of that?

Thank you!

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    First, notice that the solution set is not empty, but rather $(a,b)=(0,0)$, which proves that $K$ and $L$ are independent. Still, that does not prove that they form a basis for $\mathbb R^{2 \times 2}$.2012-03-17
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    Now, as for a basis, consider that every $2 \times 2$ matrix has 4 independent variables. Is it possible to cover all possibilities using only 2 matrices? (No, since $2 < 4$.)2012-03-17
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    What you are proving is that $K$ and $L$ are linearly independent, which they indeed are (so $a = b = 0$ is the only solution). However, for certain vectors to form a basis of a vector space, you need more than independence. Can any matrix be written as a linear combination of these two matrices?2012-03-17
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    As the above comment says you need $4$ matrices (provided they are linearly independent).2012-03-17
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    @Théophile: Yes that's right, (a,b)=(0,0)! Why the variables of a 2x2 matrix are independent? :S So we would need to have 4 matrices, right? Because with the given 2 we cannot create the other 2 variables, right?2012-03-17
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    @TMM: "Can any matrix be written as a linear combination of these two matrices?" -> This is the question I have to answer for my assignment and I am not sure what steps to follow to prove it..2012-03-17

4 Answers 4

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Hint: The dimension of ${\sf M}_2({\mathbb R})$ is $4.$ So the basis has to have how many matrices?!

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    You mean that dim$\mathbb{R}^{2 \times 2}$ = 4, right? It **must** have 4 matrices! Sorry, I got confused with subspaces :S2012-03-17
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    Right. ${\sf M}_2(\mathbb{R})$ is the space of all $2\times 2$ matrices with entries from $\mathbb{R}.$2012-03-17
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    For lack of better link, [here is a pointer on how ${\sf M}_2(\mathbb{R})$ has dimension $n^2$](http://math.stackexchange.com/questions/117226/dimension-of-gln-mathbbr#comment272363_117226)2012-03-17
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Instead of writing a $2\times 2$ matrix as
$$\begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ write it unconventionally as $(a,b,c,d)$. Now do you see that the vector space of $2\times 2$ matrices, with the usual addition, is $4$-dimensional?

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    Yes, thank you Andre! I will try to "see" it that way!2012-03-17
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What about noting that $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} $

forms a basis of $\mathbb{R}^{2\times 2}$ and so as all basis of a vector space have the same number of elements and as you only have 2 elements then these cannot span.

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Your matrices $K,L$ are indeed linearly independent, but $\{K,L\}$ is not a basis in $\mathbb R^{2\times 2}$, because $\dim \mathbb R^{2\times 2} = 4$. But I think it would be an interesting exercise for you to find other two matrices $M,N\in \mathbb R^{2\times 2}$ such that $\{K,L,M,N\}$ is a basis.

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    Maybe in time you will learn that this doesn't qualify as an answer, but instead can post it as a comment.2018-01-16