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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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1  
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 $\mathbb R^{2 \times 2}$. –  Théophile Mar 17 '12 at 20:04
    
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 $2 < 4$.) –  Théophile Mar 17 '12 at 20:06
1  
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? –  TMM Mar 17 '12 at 20:17
    
As the above comment says you need $4$ matrices (provided they are linearly independent). –  Daniel Montealegre Mar 17 '12 at 20:18
    
@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? –  Chris Mar 17 '12 at 20:32

3 Answers 3

up vote 2 down vote accepted

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 :S –  Chris Mar 17 '12 at 20:36
    
Right. ${\sf M}_2(\mathbb{R})$ is the space of all $2\times 2$ matrices with entries from $\mathbb{R}.$ –  user2468 Mar 17 '12 at 20:37
    
For lack of better link, here is a pointer on how ${\sf M}_2(\mathbb{R})$ has dimension $n^2$ –  user2468 Mar 17 '12 at 20:39
    
Thank you, J.D.! –  Chris Mar 17 '12 at 20:47

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! –  Chris Mar 17 '12 at 22:01

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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