if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that ? why or why not? the second question is identical just switch m<n. I dont understand this question at all. any help?
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Originally Posted by p00ndawg if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that ? why or why not? If is such a matrix so that is invertible then it can be shown satisfies . For example, let and follow the computations above to convince yourself. There is a similar formula for the second part as well.
Originally Posted by ThePerfectHacker If is such a matrix so that is invertible then it can be shown satisfies . For example, let and follow the computations above to convince yourself. There is a similar formula for the second part as well. can you explain the bolded part. What does the raised to the -1 do the transpose and A? woops it didnt bold, look at the [b]'s placed. sorry.
Originally Posted by p00ndawg can you explain the bolded part. What does the raised to the -1 do the transpose and A? woops it didnt bold, look at the [b]'s placed. sorry. oh nvm, I got it, now im just having trouble with the m<n 2nd one. Should I just use the same assumption but pick a different matrix in which m<n? and solve it similarly, to see what I geT?
Originally Posted by p00ndawg if A is any mxn matrix such that m > n, is it possible to find an nxm matrix B such that ? why or why not? the second question is identical just switch m<n. The rank of B cannot be greater than m, nor can the rank of BA. So if m<n then BA cannot be equal to I_n.
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