# Is this a basis for the set S?

Consider the set: $\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}$

Would the basis be found by doing: a$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ + $b\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$

So those two 2x2 matrices would form the basis? Is this the proof that they form a basis? How do I prove they're linearly independent? Thanks!

Put

a $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ + $b\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$ $=0$, the zero matrix

If you show that $a=b=0$, then the matrices are linearly independent.

Also, the fact that they span the set is obvious.

• I don't get how to show a=b=0. That's the part I was confused on, because if I multiply them back out, I just get my set. – Math StackExchange Nov 29 '14 at 14:17
• Why don't you multiply the scalars to the corresponding matrices and add the two up. – Swapnil Tripathi Nov 29 '14 at 14:18
• When I do that, I just get $\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}$ – Math StackExchange Nov 29 '14 at 14:19
• i mean, if a=b=0 for this, then it would be the zero matrix....? – Math StackExchange Nov 29 '14 at 14:20
• Yes!! You got it. – Swapnil Tripathi Nov 29 '14 at 14:20

Yes those two matrices form a basis for the vector space $$V = \left\{\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} | \ a, b \in K \right\}$$ as evidently every member of $V$ can be written as such a linear combination.

The show linear independence, just explain why

$$a \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} + b\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix} = 0 \ \ \ \ \ \ \text{ if and only if } \ \ \ \ \ a = b = 0$$