In this article we will learn about nullity and rank of a linear transformation.
Definition
Let V and W be vector spaces over the field F and let T be a linear transformation from V to W. The nullity of T is the dimension of the subspace ker T of V and the rank of T is the dimension of the subspace Im(T) of W.
If V be a finite dimensional vector space then both ker T and Im(T) are finite dimensional.
Theorem |
Let V and W be vector spaces over the field F and V is finite dimensional. If T: V→W is a linear transformation then the nullity of T + rank of T = dim V.
Proof:
Let dim V = n.
Let {α1, α2, …, αk} be a basis of ker T. There are vectors αk+1, αk+2, …, αn in V such that {α1, α2, …, αn} is a basis for V. We shall now prove that {T(αk+1), T(αk+2), …, T(αn)} is a basis for Im(T). We have noted that the vectors T(α1), T(α2), …, T(αn) certainly span Im(T), and since α1, α2, …, αk ∈ ker T, we have T(α1) = T(α2) = … = T(αk) = 0’. Hence it follows that the vectors T(αk+1), T(αk+2), …, T(αn) span Im(T). We shall show that T(αk+1), T(αk+2), …, T(αn) are linearly independent.
Suppose that there are ck+1, ck+2, …, cn ∈ F such that
ck+1 T(αk+1) + ck+2 T(αk+2) + … + cn T(αn) = 0’
Since T is a linear transformation, it follows that
T(ck+1 αk+1 + ck+2 αk+2 + … + cn αn) = 0’
Hence α = ck+1 αk+1 + ck+2 αk+2 + … + cn αn ∈ ker T.
Again, by assumption, {α1, α2, …, αk} be a basis of ker T so
α = b1 α1 + b2 α2 + … + bk αk , b1, b2, …, bk ∈ F.
Thus, α = b1 α1 + b2 α2 + … + bk αk = ck+1 αk+1 + ck+2 αk+2 + … + cn αn
or, b1 α1 + b2 α2 + … + bk αk – ck+1 αk+1 – ck+2 αk+2 – … – cn αn = 0
But {α1, α2, …, αn} is a basis of V, so we must have,
b1 = b2 = … = bk = ck+1 = ck+2 = … = cn = 0
Hence T(αk+1), T(αk+2), …, T(αn) are linearly independent and as they span Im(T), {T(αk+1), T(αk+2), …, T(αn)} is a basis of Im(T).
Clearly, dim (Im(T)) = n – k = rank of T.
Also dim(ker T) = k = nullity of T
Hence rank of T + nullity of T = n = dim V.
Theorem |
Let V and W are both finite dimensional vector spaces of the same dimension and T is a linear transformation from V to W, then T is one-one if and only if T is onto.
Proof:
Let T be one-one then clearly ker T = {0} and so dim ker T = 0.
Now, ker T is a subspace of V and Im(T) is a subspace of W and also dim ker T + dim Im(T) = dim V.
Since dim ker T = 0, we have dim Im(T) = dim V = dim W (given)
Therefore, dim Im(T) = dim W ⇒ Im(T) = W
Hence the transformation T is onto.
Conversely, suppose that T is onto.
Then Im(T) = W
Therefore dim Im(T) = dim W = dim V.
Since dim ker T + dim Im(T) = dim V, we have dim ker T = 0
Hence ker T = {0} and T is one-one.
Theorem |
If V and W are finite dimensional vector spaces over a field F and if there exists a linear transformation T: V → W which is both one-one and onto, then dim V = dim W.
Proof:
Since by assumption T is one-one, we have ker T = {0}, therefore dim ker T = 0.
Again, since T is onto, we have Im(T) = W, therefore dim Im(T) = dim W.
Now, dim ker T + dim Im(T) = dim V
⇒ 0 + dim W = dim V
i,e., dim V = dim W.
Example 01 |
Let V be the vector space of n-square matrices over the field F. Let M be an arbitrary matrix in V. Let T: V → W be defined by T(A) = AM + MA where A ∈ V. Show that T is linear.
Solution:
For any A, B ∈ V and c ∈ F, we have
T(A + B) = (A + B)M + M(A + B)
=AM + BM + MA + MB
=(AM + MA) + (BM + MB)
=T(A) + T(B)
And T(cA) = (cA)M + M(cA)
=c(AM) + c(MA) = c(AM +MA) = cT(A)
Accordingly, T is linear.
Example 02 |
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