I sort of recognize one to one, but I cannot, for the life of me, recognize anything about if a matrix is onto. Every definition I look at is really difficult to understand.

You are watching: How to tell if a matrix is onto The summary onto (or surjective) go not use to matrices only, therefore I'll simply lay out the full definition.

Def. A function f indigenous A come B is dubbed onto if for every bB over there is aA such that f(a) = b.

In other words, f is onto if there is no point in B that have the right to not be reached as f(a). For example, if we take into consideration functions RR (that map from the real line to the genuine line), we have:

f(x) = x2 is not onto, because no issue which actual number you put right into f, you can never acquire a an unfavorable real number out. In other words, there space some number in the codomain the cannot it is in reached. (Remember the for a duty f:AB, A is referred to as the domain and B the codomain.)

f(x) = x3 is onto, since for any number x (negative, optimistic or zero) the cube root is fine defined and also real, so you deserve to reach any real number through inputting one more real number right into f.

f(x) = 1 is not onto, due to the fact that no matter what you put in, you deserve to only with the number 1. No other output is possible.

f(x) = sin(x) is not onto, due to the fact that you can only acquire numbers in <-1, 1> together output. No other outputs room possible.

f(x) = ax+b (assuming a is nonzero) is onto, due to the fact that if you want to reach the calculation number y, simply let x=(y-b)/a.

So anyway, ago to matrices. The reason we have the right to use words onto because that matrices is the a procession M is a representation of a direct map ~ you settle a basis. A procession M is then taken into consideration onto if the straight map it represents is onto. If linear map and basis don't sound familiar, here's what this means for matrices. A m×n matrix M is favor a duty that maps vectors native Rn to Rm by way of multiplication. So if v is in Rn climate Mv is in Rm. The matrix M is then onto if every element in Rm deserve to be reached from some element in Rn by multiplying v M.

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A simple means to test if a m×n procession M is ~ above is to minimize it to heat echelon form and examine for a heat that consists of just zeroes. If over there is together a row, the matrix is not onto. However, if every rows room nonzero, the procession is onto. You deserve to think the this as having to carry out with levels of freedom. If only m-1 rows space nonzero (so there's one zero row), then you have actually only m-1 levels of flexibility in specifying what value Mv must take if you gain to select v freely. In various other words, the collection of reachable vectors is m-1 dimensional. In other words, it's not Rm (because Rm is m dimensional). The very same intuition holds if over there is more than one nonzero row.