Question

Show that the following forms a basis for M_22 (given in comments)

Answer 1

\[B = \left\{ \left[\begin{matrix}0 & 1 \\ 1 & 1\end{matrix}\right],\left[\begin{matrix}1 & 0 \\ 1 & 1\end{matrix}\right] , \left[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right], \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\right\}\]

Answer 2

Any help would be appreciated :) This whole linear algebra thing is not going so well for me!

Answer 3

what i would do is change them into 4 dimensional vectors, and prove the vectors are linearly independent.

if the matrix is:

a b
c d

then change it into the vector (a, b, c, d).

so basically you want to show the vectors:
(0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 1), and (1, 1, 1, 0)
are linearly independent, and you can do that by putting them in a matrix and row reducing it to the identity matrix.

Answer 4

I was having trouble reducing the matrix \[\left[\begin{matrix}0 & 1 &1 &1 \\ 1 & 0&1&1\\1&1&0&1\\1&1&1&0\end{matrix}\right]\] so saught the help of an "automatic row operations" applet... And it seems like a very long process to get the identity matrix (though it does, in the end, work). Is this the only way of doing this question? It seems too many steps for what is worth only a few marks :/ Thanks so much for the help!

Answer 5

Never mind! Trusting technology too much :P Got it OK in the end.

Do all linearly independent systems like this one reduce to the identity matrix?? Or was it just this example??

Answer 6

all linearly independent systems will reduce to some form of the identity matrix. If its a square matrix, it will be exactly the identity, if its a rectangular matrix, it might look like:

1 0 0
0 1 0
0 0 1
0 0 0

or something else. But that lets you know the columns are independent.

Answer 7

Great, cheers :)

Answer 8

Would it be too much to ask for you to take a quick look at the most recent question I posted? I'm chasing someone to check my working out, for a question regarding a coordinate vector for the same system as I posted above :)