Question

Let U and V be vector spaces over a field F and dim U = n. Let
T:U→V
be a linear operator, then rank (T) + nullity (T) = ……
0
1
n-1
n

Answer 1

Let $U$ and $V$ be vector spaces over a field $F$ and $\dim U = n$. Let $T: U \to V$ be a linear operator, then $\operatorname{rank}(T) + \operatorname{nullity}(T) = \dots$

a. 0

b. 1

c. $n - 1$

d. $n$

Solution.

Use the rank-nullity theorem to prove it. The theorem states that the rank and the nullity of a matrix add up to the number of columns of the matrix. Specifically, if $A$ is an m-by-n matrix (with m rows and n columns) over some field, then

$$\operatorname{rank} A + \operatorname{nullity} A = n.$$

So we have two vector spaces $U$ and $V$ over a field $F$ and $T: U \to V$. Then the rank of $T$ is the dimension of the image of $T$ and the nullity of $T$ is the dimension of the kernel of $T$, so we have

$$\dim(\operatorname{im} T) + \dim(\ker T) = \dim U$$

or, equivalently,

$$\operatorname{rank} T + \operatorname{nullity} T = \dim U$$

Such that $\dim U = n$ we have

$$\operatorname{rank} T + \operatorname{nullity} T = n$$

Answer:

$$d. \operatorname{rank} T + \operatorname{nullity} T = n$$