I know what is expected, yet I have troubles to calculation or proof it as soon as a rather large matrix is offered with numbers and letters.

You are watching: How to tell if a matrix is onto

This is the function I have:

How deserve to I examine if it"s one come one? perform I require to examine for everything that is in my function T, therefore let"s speak 3x-z, and also take a random number for x and z, and also check whether the worth that rolls out, is unique? i did this, however I don"t think this is exactly to be honest.

What is the method to technique this? I"d really evaluate some assist and tips...

linear-algebra matrices features
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edited Mar 17 "17 in ~ 11:29
Siyah
request Mar 16 "17 at 16:59

SiyahSiyah
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T is one-to-one duty if the kernel is trivial, so friend you neeed to check implication \$T(x,y,z)=0implies x=y=z=0\$.

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answer Mar 16 "17 at 17:04

alansalans
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Find the REF that the traditional matrix (it"s not necessary to acquire to RREF). Then, look in ~ the pivots (the top 1"s the the rows).

If we have actually a pivot in every column, then the nullspace of the procession (and hence the kernel the \$T\$) is zero-dimensional. So, \$T\$ is one-to-one if and also only if the REF has actually pivot in every column.

If we have a pivot in every row, then \$T(x,y,z) = wcchristchurchappeal.combf b\$ has a solution for any kind of \$ wcchristchurchappeal.combf b\$. The is, the measurement of the photo is same to the dimension of the space. So, \$T\$ is ~ above if and also only if the REF has a pivot in every row.

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reply Mar 16 "17 at 17:15

Ben GrossmannBen Grossmann
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alans has provided you the technique for b)

One that the countless ways to fix for c) :

Find the selection of the linear transformation. I.e. \$R(T) = spanT(1,0,0),T(0,1,0),T(0,0,1) = span(3,1,1,0),(0,2,-1,1),(-1,0,1,2)\$

Note: You have the right to use any other communication you like. Also, i would certainly strongly introduce you to prove the the over statement is true in general.

Suppose top top the contrary that \$R(T)\$ is equal to \$ wcchristchurchappeal.combbR^4\$.

Then over there exist \$lambda_1,lambda_2,lambda_3in wcchristchurchappeal.combbR\$ such the \$(0,0,0,1) = lambda_1cdot (3,1,1,0) + lambda_2cdot (0,2,-1,1) + lambda_3cdot (-1,0,1,2)\$

Now, this mechanism of straight equation gives you no solution, i m sorry is a contradiction.

Note: mean you are given one more example i beg your pardon \$T\$ is without doubt onto, climate you will likewise find \$R(T)\$ and also show that it is equal to the vector room that \$T\$ is mapped to.

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edited Mar 17 "17 in ~ 18:03
reply Mar 17 "17 in ~ 17:47

tiny RookieLittle Rookie
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