Raising the bar

Rank of a matrix and the rank-nullity theorem

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Rank and its properties

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

Exercise 1. For any matrix $latex A$

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

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

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

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

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

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

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

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

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

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

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

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