What Is the Dimensio of the Range of T if T Is One to One Mapping

The Kernel and the Range of a Linear Transformation

One to Ane Linear Transformations

Recall that a function is 1-1 if

f(x)  =  f(y)

implies that

        ten  =  y

Since a linear transformation is divers as a office, the definition of 1-1 carries over to linear transformations.  That is

Definition

        A linear transformation Fifty is ane-1 if for all vectors u and v ,

L(u)  =  L(v)

        implies that

u  = v

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Example

Permit L exist the linear transformation from Rⁱⁱ to P_ii defined by

       L((10,y))  =  xt² + yt

We can verify that L is indeed a linear transformation.  We now check that 50 is i-1.  Let

Fifty(x_i,y_ane)  =  Fifty(10₂,y₂)

and so

x_it² + y_anet  =  ten_twotⁱⁱ + y₂t

If two polynomials are equal to each other, then their coefficients are all equal.  In particular,

  ten₁  =  10_ii       and y₁  =  y₂

We can conclude that L is a i-1 linear transformation.

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Case

Let L be the linear transformation from P_i to R¹ divers by

  L(f(t))  =  f(0)

So L is non a ane-1 linear transformation since

        L(0)  =  Fifty(t)

and

0 1

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The Kernel

Related to one-1 linear transformations is the thought of the kernel of a linear transformation.

Definition

The kernel of a linear transformation L is the gear up of all vectors v such that

L(five)  = 0

Instance

Let Fifty exist the linear transformation from M^2x2 to P^ane divers past

Then to find the kernel of 50, we fix

  (a + d) + (b + c)t  =  0

     d  =  -a        c  =  -b

so that the kernel of Fifty is the set of all matrices of the form

Find that this set is a subspace of M^2x2 .

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Theorem

The kernel of a linear transformation from a vector space V to a vector space W is a subspace of V .

Proof

Suppose that u and v are vectors in the kernel of L.  Then

50(u)  =  L(v)  =  0

We have

        L(u + v)  =  50(u) + (v)  = 0 + 0  = 0

and

    L(cu)  =  cL(u)  =  c 0  = 0

Hence u + v and cu are in the kernel of L.  We tin can conclude that the kernel of L is a subspace of V.

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In light of the above theorem, it makes sense to ask for a basis for the kernel of a linear transformation.  In the previous case, a basis for the kernel is given by

Adjacent we show the relationship between i-1 linear transformations and the kernel.

Theorem

A linear transformation L is ane-1 if and only if Ker(L)  = 0 .

Proof

Let L be ane-one and allow v be in Ker(Fifty).  We need to testify that 5 is the zero vector.  We have both

   L(v)  = 0       and     50(0)  = 0

Since L is 1-one,

v  =  0

Now let Ker(L)  =  0.  And so

L(u) =  L(v)

implies that

0  =  L(v) - 50(u)  =  Fifty(v - u)

Hence v - u is in Ker(L), so that

v - u  = 0      or u  = v

and L is ane-1.

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Range

We have seen that a linear transformation from Five to West defines a special subspace of V chosen the kernel of L.  Now we plough to a special subspace of W.

Definition

Let Fifty be a linear transformation from a vector space 5 to a vector space W .  Then the range of L is the set of all vectors w in Due west such that there is a v in V with

   L(v)  = west

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Theorem

The range of a linear transformation L from V to W is a subspace of W .

Proof

Let w_i and w₂ vectors in the range of Westward.  And so there are vectors 5_ane and five₂ with

L(5_one)  =  westward_i    and 50(v₂ )  = w₂

We must show closure nether addition and scalar multiplication.  We have

    L(v₁ + 5₂ )  =  L(5_ane ) + 50(v₂ )  = westward_ane + west₂

and

  Fifty(c5₁ )  =  cL(v₁ )  =  cw_one

hence w₁ + w₂ and c w₁ are in the range of L.  Hence the range of L is a subspace of West.

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Nosotros say that a linear transformation is onto W if the range of L is equal to W.

Example

Let Fifty be the linear transformation from R² to R^three defined by

Fifty(v)  =  A5

with

A.  Observe a ground for Ker(L).

B.  Decide of L is one-1.

C.  Find a footing for the range of L.

D.  Determine if L is onto.

Solution

The Ker(L) is the same as the null infinite of the matrix A.  We accept

Hence a basis for Ker(50) is

{(3,-ane)}

Fifty is not 1-1 since the Ker(L) is non the aught subspace.

Now for the range.  If we allow {east_i } be the standard basis for R² , then

{L(e₁ ), L(e_ii )}

volition span the range of 50.  These two vectors are merely the columns of A.  In general

The columns of A span the range of 50 .

A basis for the cavalcade space is L is given by the starting time cavalcade of A (the only corner of rref(A)).  That is a basis is

{(1,2,iii)}

Since the dimension of the range of A is ane and the dimension of R³ is 3, 50 is not onto.

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A Few Theorems

In the last example the dimension of R² is 2, which is the sum of the dimensions of Ker(L) and the range of L.  This will be true in full general.

Theorem

Let L be a linear transformation from Five to Due west .  And so

        dim(Ker(50))  + dim(range(L))  =  dim(V)

Proof

Permit

     S  =  {v₁ , ..., v_k }

be a basis for Ker(50). Then extend this basis to a total basis for 5.

        T  =  {v₁ , ..., v_k , v_k+1 , ..., v_n }

We need to prove that

       U  =  {50( v _thou+1 k+1), ..., L(v_n )}

is a basis for range L.  If w is in the range of L so at that place is a v in 5 with L(5)  = w .  Since T spans V, we can write

v  =  c₁ v₁ + ... + c_k five₁₀₀₀ + c_k+1 v_yard+1 + ... +  c_north v_n

and then that

     west  =  L(v)  =  50(c₁ five₁ + ... + c_k 5_grand + c_thou+ane 5_k+one + ... +  c_n v_northward )

     = c_one50( 5₁ ) + ... + c_k L( v₁₀₀₀ ) + c_m+1 Fifty( v_k+1 ) + ... +  c_n Fifty( v_northward )

     = c_i 0 + ... + c_k 0 + c_{one thousand+ane} L( 5_k+1 ) + ... +  c_{due north} L( v_north )

     =  c_k+ane L( five_k+1 ) + ... +  c_n 50( v_{due north} )

hence U spans the range of L.  Now nosotros demand to show that U is a linearly independent set of vectors.  If

c_chiliad+1 L( v_k+1 ) + ... +  c_n L( five_n )  =  0

then

     0  =  L(c_{one thousand+i} v_k+1 + ... +  c_n v_n )

hence

5  =  c_{one thousand+1} 5_chiliad+one + ... +  c_northward five_{due north}

is in Ker(50).  But then five can be written equally a linear combination of vectors in S.  That is

c_k+1 v_chiliad+1 + ... +  c_n v_north   = c_ane v_i + ... + c_k v₁₀₀₀

Which ways that all of the constants are zero since these are linearly contained.

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We call the dimension of Ker(L) the nullity of 50 and the dimension of the rang of Fifty the rank of L.

Nosotros terminate this discussion with a corollary that follows immediately from the in a higher place theorem.

Theorem

Let L be a linear transformation from a vector infinite V to a vector infinite W with dim V   =  dim Due west , and so the post-obit are equivalent:

        1. L is one-1 .

        2. Fifty is onto.

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Source: http://ltcconline.net/greenl/courses/203/matrixonvectors/kernelRange.htm