# I don't understand why the rank = n - Rank-nullity theorem - nullity

• s3a

#### s3a

I don't understand why the rank = n -- Rank-nullity theorem -- nullity

## Homework Statement

I'm working on #1 (the solutions are also included in that pdf) here ( http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/exam-1/MIT18_06SCF11_ex1s.pdf ).

## Homework Equations

Ax = b
Ax = 0
rank A+ nullity A = n

## The Attempt at a Solution

For #1(a), I don't understand why the rank = n. I've been told to look at the Rank-nullity theorem which states that rank A+ nullity A = n but, I don't understand what nullity means exactly. In fact, I'm not very solid on the meaning of rank either (but, I kind of get it).

Any input would be greatly appreciated!

A linear transformation, A:U->V, maps vectors in a vector space, U, to a vector space V. It is fairly straight forward to show that the "range" of A, {v| v= Au for some u in U} is a subspace of V. The 'rank' of A is the dimension of that subspace. It is similarly straight forward to show that the null space of A, {u| AU= 0}. The 'nullity' of A is the dimension of the null space. One way of approaching the proof of the "rank-nullity" theorem is to choose a basis for the null space, $\{u_1, u_2, ..., u_m\}$ and expand it to a basis for U by adding vectors $\{u_{m+1}, ..., u_n\}$. Show that every vector in the range of A can be written as a linear combination of $\{Au_{m+1}, ..., Au_n\}$ and so has dimension n- m.

Two hints:

The first statement,
$$Ax = \left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] \textrm{ has no solutions}$$
means that ##A## is not ...? (Injective, surjective?)

The second statement,
$$Ax = \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] \textrm{ has exactly one solution}$$
means that the kernel (null space) of ##A## is ...?