How would I find the rank and nullity of the following matrix? (Given in comments)
\[\left[\begin{matrix}2 & 2 & 0 & 2 \\ 2&3&0&3 \\ 1&2&0&3 \\ 1& 0 &0&0\end{matrix}\right]\]
Am I correct in thinking that to find the nullity, I first find the dimension of ker(A) [assuming the above matrix is 'A'], by taking the augmented matrix, and finding all possible answers to Ax=0?
Then rank(A) + nullity(A) = m for an n x m matrix?
that is correct.
Would you possibly be able to help with augmenting the matrix etc? I'm getting a kind of strange answer :/
sure. We need to row reduce that matrix first. Im not pro enough to type the matrices on here >.< so i'll do it on paper and post. one sec.
Thanks so much :)
The matrix needs to be in row reduced echelon form (i think thats how you spell it.)
I'm getting an answer where the bottom row is [0 0 0 -6] ... Don't think thats correct :/
Sorry, -2, not -6 ... Not helpful, regardless...
alright, im double checking this on my phone, but this is what im getting for the RREF of A.
Yeah thats right. do you have any questions about anything in those three pictures?
Just one; In the second row of your final answer, how did [0 1 0 1] become [0 1 0 0]?
i used the 1 in row 3 to kill it. I did: \[-R_3+R_2\]
Alrighty! So the non-leading variable is only "x_4" (if the notation is similar to what you're familiar with) ... This is where I get really stuck - the Ax=0 thing :/
As far as your question is concerned, all the information we need is in that RREF form. The number of pivots is the rank of A, and the number of free variables is the Nullity of A.
Oh really?? So I dont have to let the non-leading variable be denoted some constant ('t' or something) and find the possible solutions to Ax=0?? :D
you could still get the answer by doing that, but what you are interested in is the dimension of the Null Space, not really the Null Space itself. If the question was, "Find a basis for the Null Space", or "Describe the Ker(A)", then i would do that. In this problem, just knowing its one-dimensional (since there is only one freee variable) is enough.
Great! Thanks so much!
Morning Zarkon!
morning
nice work
thanks :) oh, i found out a few days ago there might be a Complex Variables class in the spring (pending enough people register). i remember you talking about that class, i hope i get to take it lol >.<
nice...definitely take it if you can. very Interesting stuff
Haha Zarkon thanks for your help on almost exactly the same question earlier today :P Much obliged :)
np
well I hope you all enjoy your day...as for me I get to be in meetings all day :(
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