Problem :

Suppose $A$ is a $3$ by $4$ and $B$ is $4$ by $5$ and $AN=0$. So $N(A)$ contains $C(B)$. Prove from the dimensions of $N(A)$ and $C(B)$ that $\text{rank}(A)+\text{rank}(B)\le4$

This is my solution:

Because matrix A is $3\times4$, then $\max(\text{rank}(A)) = 3$ by definition.

We know that $N(A)$ contains $C(B)$. We can consider $AB = 0$ as $Ax = 0$, where $x = B$. As we know, the dimension of Null space is $n - r$, where $n$ is number of columns and $r$ is a rank of a matrix. So, if matrix $A$ has maximum $\text {rank}(3)$, than dimension of $N(A) = 4-3 = 1$. When matrix $A$ has minimum $\text{rank}(1)$, then dimension of $N(A) = 4-1 = 3$. As we know, $C(B)$ lies in the $N(A)$ (because of $AB = 0$).

Be definition, dimension of a space = number of vectors in every basis. It means, that matrix$(B) = N(A)$ has rank number of columns of $A$ - rank of $A$.

So rank of $A$ varies from $1$ to $3 \implies$ rank of matrix B varies from $3$ to $1$.

As a result $\text{rank}(A) + \text{rank}(B) \le4$

Is my logic(proof) correct? Can you check it please.


This follows from the rank nullity theorem:

$\operatorname{rk}A + \dim \ker A = 4$.

Since ${\cal R}B \subset \ker A$, we have $\operatorname{rk} B \le \dim \ker A$, hence $\operatorname{rk}A + \operatorname{rk}B \le 4$.

  • $\begingroup$ Thank you for the answer! But can you check my logic, if it is possible to proof by my approach? $\endgroup$ Sep 26 '16 at 15:06
  • 1
    $\begingroup$ @DanielYefimov: It looks correct, but is a long winded way of writing the above. $\endgroup$
    – copper.hat
    Sep 26 '16 at 15:13
  • $\begingroup$ What is $\mathcal R$ here? I've never seen this notation before in linear algebra. $\endgroup$ Sep 26 '16 at 15:24
  • $\begingroup$ @ParclyTaxel: It means the range space of a linear operator. It is fairly standard notation. $\endgroup$
    – copper.hat
    Sep 26 '16 at 15:35

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