Question

How do you determine that a 2 x 2 matrix is a subspace of R(2x2)

Answer 1

For example,
\[\left[\begin{matrix}a & b \\ b-a & c\end{matrix}\right]\]
is a subspace of \[\mathbb{R}^(2\times 2)\]

Answer 2

Can you tell me what properties have to be satisfied for something to be a subspace?

Answer 3

Closed under addition and subtraction as well as in scalar multiplication?

Answer 4

Right, so take two arbitrary matrices that take on that form and add them together. If the sum has the same form, the set is closed under vector addition. (Closure under subtraction is the same as closure under addition.)

Do the same to establish closure under multiplication. Take an arbitrary constant and multiply it to an arbitrary matrix from the set, then check if the product is in the set as well. If it is, the set is closed under scalar multiplication.

Also, check to see if the set can have a zero element, in this case \(\begin{pmatrix}0&0\\0&0\end{pmatrix}\).

Answer 5

Would the following matrices be valid?
\[\left[\begin{matrix}a & b \\ b & c\end{matrix}\right] - \left[\begin{matrix}0 & 0 \\ a & 0\end{matrix}\right] = \left[\begin{matrix}a & b \\ b-a & c\end{matrix}\right]\]

Answer 6

And for the scalar multiple,
\[k\left[\begin{matrix}a & b \\ b & c\end{matrix}\right] -k \left[\begin{matrix}0 & 0 \\ b & 0\end{matrix}\right] =\left[\begin{matrix}ka & kb \\ k(b-a) & kc\end{matrix}\right]=k \left[\begin{matrix}a & b \\ b-a & c\end{matrix}\right]\]
?

Answer 7

"b" in the second matrix was suppose to be an "a"

Answer 8

No, to prove the subspace is non-empty you need to find values of a, b and c which give the zero element (pretty trivial) which @SithsAndGiggles mentioned.

Then, to prove it is closed under addition you'll need to show something like this:
\[\left[\begin{matrix}a _{1} & b _{1} \\ b_{1} - a _{1} & c_{1}\end{matrix}\right] + \left[\begin{matrix}a _{2} & b _{2} \\ b_{2} - a _{2} & c_{2}\end{matrix}\right] = \left[\begin{matrix}(a _{1} + a_{2}) & (b _{1}+b_{2}) \\ (b_{1} + b_{2}) - (a _{1} + a_{2}) & (c_{1} + c_{2})\end{matrix}\right]\]And for the scalar multiplication proof, something like this:
\[r\left[\begin{matrix}a & b \\ b-a & c\end{matrix}\right] = \left[\begin{matrix}ra & rb \\ rb-ra & rc\end{matrix}\right]\]
As the three properties hold, this is a subspace :)

Answer 9

Thank you very much I see what you mean and so this is the method I would use for all 2x2 invertible matrices

Answer 10

No problem, happy to help! I learnt about subspaces last year and I really struggled so it's nice to be able to pass on my knowledge to someone.