MATH 33AH Lecture Notes - Lecture 9: Elementary Matrix, Surjective Function, Linear MapExam
Course CodeMATH 33AH
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Math 33A Discussion Notes
October 21, 2017
Week 3 – Incomplete! Will update soon.
- A function T:Rk−→ Rnis called injective, or one-to-one, if each input gets a unique
output. The rigorous deﬁnition is: for all ~x, ~y ∈Rk, we have that [T(~x) = T(~y)] implies
[~x =~y]. In other words, the only time the outputs are the same is when in fact the inputs
are the same as well.
- A function T:Rk−→ Rnis called surjective, or onto, if everything in the target space gets
hit by T. The rigorous deﬁnition is: for all ~
b∈Rn, there exists ~x ∈Rksuch that T(~x) = ~
In other words, given anything in the co-domain/target space, we can ﬁnd a vector in the
domain that T sends to that target vector.
- If T:Rk−→ Rnis linear, with matrix representation, say, A, then Tis 1-1 (short-
hand for “one-to-one”) iﬀ rref(A) has a pivot in every column. If rref(A) has a column
without a pivot, there will be a free variable in the linear system A~x =~
0, so T(~x) = ~
have more than one solution, so Twon’t be 1-1. Conversely, if rref(A) has a pivot in every
column, then T(~x) = T(~y) means A~x =A~y, so A(~x −~y) = ~
0, and since rref(A) has a pivot
in each column, the only solution to A~u =~
0 is ~u =~
0, which forces ~x −~y =~
0, and thus
~x =~y, showing that Tis injective.
- For Tas above, Tis onto iﬀ rref(A) has a pivot in every row. If rref(A) had a row
without a pivot, then it’d have at least one row of zeros on the bottom. By setting the
right hand side of constants of ~en, and undoing each elementary row op taken to get A
into rref(A), we can get a system A~x =~
bthat’s inconsistent; in other words, T(~x) = ~
has no solutions ~x for our specially chosen ~
b, so Tis not onto. Conversely, if rref(A) has
a pivot in every row, then we will always be able to solve A~x =~
bfor any ~
b∈Rn, so Tis onto.
- We identify a matrix with its associated linear transform when we say that some ma-
trix is onto or 1-1
- some examples: ( 100
012) is onto but not 1-1 since its rref (which is conveniently itself...)
has a pivot in every row, but not in every column. 1 0
0 0 is 1-1 but not onto since it has a
pivot in every column, but not in every row. ( 110
000) is nethier injective nor surjective, and
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- If n×k matrix A is 1-1, then it must be square or tall (n≥k), since each pivot gets
a unique row. If A is onto, then it must be square or wide (n≤k), since each pivot gets a
- We say T:Rk−→ Rnis bijective if Tis both 1-1 and onto. In this case, by the above
characterizations of onto and 1-1, if Tis bijective, then its matrix representation must be
square (combine the two inequalities n≤kand n≥kto get this). In fact, Tis bijective
iﬀ its matrix representation A satisﬁes rref(A) = In, since Inis the only n×n matrix in
rref with a pivot in every row and column. We also say that a matrix A is bijective if its
associated linear transform (T(~x) = A~x) is bijective.
- If matrix A is n×n square and surjective, then its rref has n pivots (one for each row),
which means also that rref(A)’s columns all have pivots as well, so A is injective, so A is
bijective. Similarly, if A is n×n square and injective, then its rref has n pivots (one for each
column), which means also that rref(A)’s rows all have pivots as well, so A is surjective, so
A is bijective.
- Bijective functions are special because they have inverses.
- We say T:Rk−→ Rnhas an inverse, which we denote by T−1, if T−1:Rn−→ Rk,
for all ~x ∈Rkwe have T−1(T(~x)) = ~x, and for all ~y ∈Rnwe have T(T−1(~y)) = ~y. This
means that inverse functions switch inputs and outputs.
- The ﬁrst condition requires Tto be injective and T−1to be surjective; the second condi-
tion requires Tto be surjective and T−1to be injective. So every invertible function is also
bijective, and so is its inverse.
- Conversely, every bijective function has an inverse – namely, the function that takes el-
ements of the target space back to the unique elements of the domain that the original
function sends to those targets. For such a function to be deﬁned, the original function must
be onto (or else the inverse has some elements it can’t send anywhere) and 1-1 (or else the
inverse has to send elements to multiple places).
- Long story short, a function is bijective iﬀ it is invertible.
- Inverses are unique since where they send each element is determined entirely by the
original function, and the inverse of the inverse is the original function. Both of these facts
are consequences of the deﬁnition.
- If Tis bijective, then we know its matrix representation is square. So it must be the
case that n=k, so Thas the same domain and target space.
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