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 I_{m}(T) of W.

If V be a finite dimensional vector space then both ker T and I_{m}(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 {α_{1}, α_{2}, …, α_{k}} be a basis of ker T. There are vectors α_{k+1}, α_{k+2}, …, α_{n} in V such that {α_{1}, α_{2}, …, α_{n}} is a basis for V. We shall now prove that {T(α_{k+1}), T(α_{k+2}), …, T(α_{n})} is a basis for I_{m}(T). We have noted that the vectors T(α_{1}), T(α_{2}), …, T(α_{n}) certainly span I_{m}(T), and since α_{1}, α_{2}, …, α_{k} ∈ ker T, we have T(α_{1}) = T(α_{2}) = … = T(α_{k}) = 0’. Hence it follows that the vectors T(α_{k+1}), T(α_{k+2}), …, T(α_{n}) span I_{m}(T). We shall show that T(α_{k+1}), T(α_{k+2}), …, T(α_{n}) are linearly independent.

Suppose that there are c_{k+1}, c_{k+2}, …, c_{n} ∈ F such that

c_{k+1} T(α_{k+1}) + c_{k+2} T(α_{k+2}) + … + c_{n} T(α_{n}) = 0’

Since T is a linear transformation, it follows that

T(c_{k+1} α_{k+1} + c_{k+2} α_{k+2} + … + c_{n} α_{n}) = 0’

Hence α = c_{k+1} α_{k+1} + c_{k+2} α_{k+2} + … + c_{n} α_{n} ∈ ker T.

Again, by assumption, {α_{1}, α_{2}, …, α_{k}} be a basis of ker T so

α = b_{1} α_{1} + b_{2} α_{2} + … + b_{k} α_{k} , b_{1}, b_{2}, …, b_{k} ∈ F.

Thus, α = b_{1} α_{1} + b_{2} α_{2} + … + b_{k} α_{k} = c_{k+1} α_{k+1} + c_{k+2} α_{k+2} + … + c_{n} α_{n}

or, b_{1} α_{1} + b_{2} α_{2} + … + b_{k} α_{k} – c_{k+1} α_{k+1} – c_{k+2} α_{k+2} – … – c_{n} α_{n} = 0

But {α_{1}, α_{2}, …, α_{n}} is a basis of V, so we must have,

b_{1} = b_{2} = … = b_{k} = c_{k+1} = c_{k+2} = … = c_{n} = 0

Hence T(α_{k+1}), T(α_{k+2}), …, T(α_{n}) are linearly independent and as they span I_{m}(T), {T(α_{k+1}), T(α_{k+2}), …, T(α_{n})} is a basis of I_{m}(T).

Clearly, dim (I_{m}(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 I_{m}(T) is a subspace of W and also dim ker T + dim I_{m}(T) = dim V.

Since dim ker T = 0, we have dim I_{m}(T) = dim V = dim W (given)

Therefore, dim I_{m}(T) = dim W ⇒ I_{m}(T) = W

Hence the transformation T is onto.

Conversely, suppose that T is onto.

Then I_{m}(T) = W

Therefore dim I_{m}(T) = dim W = dim V.

Since dim ker T + dim I_{m}(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 I_{m}(T) = W, therefore dim I_{m}(T) = dim W.

Now, dim ker T + dim I_{m}(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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