2
Aug 18

Rank of a matrix and the rank-nullity theorem




Rank of a matrix and the rank-nullity theorem

If you find proofs shorter than mine, shoot me an email.

Rank and its properties

Definition 1. The dimension of the image \text{Img}(A) is called a rank of the matrix A and denoted \text{rank}(A).

Exercise 1. For any matrix A

(1) \text{rank}(A)=\text{rank}(A^{T}).

Proof. Suppose A is of size n\times k. We know that

(2) R^k=N(A)\oplus\text{Img}(A^T).

By (2) any x\in R^k can be represented as

(3) x=y+z, with y\in N(A), z\in\text{Img}(A^T) and y\perp z.

From (3) Ax=Ay+Az=Az. By Definition 1, \text{Img}(A^T) is spanned by some linearly independent vectors z^{(1)},...,z^{(m)} with m=\dim\text{Img}(A^T). Hence z=\sum_{i=1}^ma_iz^{(i)} with some a_i depending on z and Ax=\sum_{i=1}^ma_iAz^{(i)}. This shows that \text{Img}(A) is spanned by Az^{(1)},...,Az^{(m)}. If we prove linear independence of these vectors, (1) will follow.

Suppose \sum_{i=1}^mb_iAz^{(i)}=0 with some b_{i}. Then A\left(\sum_{i=1}^mb_iz^{(i)}\right)=0. This tells us that \sum_{i=1}^mb_iz^{(i)}\in N(A). Since the vector \sum_{i=1}^mb_iz^{(i)} also belongs to \text{Img}(A^T), by Exercise 3 it is zero. By linear independence, \sum_{i=1}^mb_iz^{(i)}=0 is possible only when b_1=...=b_m=0. This proves the linear independence of Az^{(1)},...,Az^{(m)} and the statement.

Exercise 2 (rank-nullity theorem) If A is n\times k, then \dim N(A)+\dim \text{Img}(A)=k.

Proof. Exercise 4 and equation (2) imply k=\dim N(A)+\dim\text{Img}(A^T). But we know from Exercise 1 that \dim\text{Img}(A^T)=\dim\text{Img}(A).

Exercise 3. Let A be of size n\times k. Then \text{rank}(A)\leq\min\{n,k\}.

Proof. From Exercise 2 we see that \text{rank}(A)=\dim\text{Img}(A)\leq k. Applying this inequality to A^T we get \text{rank}(A)=\text{rank}(A^T)=\dim\text{Img}(A^T)\leq n.

Leave a Reply

You must be logged in to post a comment.