$Rank(A+B)\leq Rank(A)+Rank(B)$

Question

Let there be $A,B$ matrices.
Let $C=A+B$
$Span(Col(C))\subseteq Span(Col(A))$ because C is a linear combination of A . $Span(Col(C))\subseteq Span(Col(B))$ because C is a linear combination of B .
Therefore $Span(Col(C))\subseteq Span(Col(A))+Span(Col(B))$ and $Rank(A+B)\leq Rank(A)+Rank(B)$

The first and second inclusions are not trues. – Hamou Aug 16 '14 at 19:43 — Hamou, Aug 16 '14 at 19:43
@Memming I did not understand the hint there – gbox Aug 16 '14 at 20:13 — gbox, Aug 16 '14 at 20:13

score 3 · Accepted Answer · answered Aug 16 '14 at 19:50

3

$Span(Col(C))\subset Span(Col(A))+Span(Col(B)) $ then $Rank(C)\leq \dim(Span(Col(A))+Span(Col(B)))$, by using the inequality $\dim(F+G)\leq \dim F+\dim G$ we get $Rank(C)\leq \dim(Span(Col(A)))+\dim(Span(Col(B)))=Rank( A)+Rank(B)$

answered Aug 16 '14 at 19:50

Hamou

6,915

score 1 · Answer 2 · answered Aug 16 '14 at 19:59

1

Define $Lx = (Ax, Bx)$. Clearly $L$ is linear and ${\cal R} L \subset {\cal R} A \times {\cal R} B$ (this inclusion could be strict).

Hence $\dim {\cal R} L \le \dim ( {\cal R} A \times {\cal R} B ) = \dim {\cal R} A + \dim {\cal R} B$.

Now let $\phi(y) = y_1+y_2$, where $y = (y_1,y_2)$, again note that $\phi$ is linear. If $S$ is a subspace, it is easy to see that $\dim \phi(S) \le \dim S$ (this is true for any linear operator, not just $\phi$).

Letting $S = {\cal R} L$, and noting that $A+B = \phi \circ L$ gives the desired result.

answered Aug 16 '14 at 19:59

copper.hat

178,207

Thanks, I am trying to prove using Spans – gbox Aug 16 '14 at 20:04
1

Replace ${\cal R} A$ by $\operatorname{sp} (\operatorname{col} A)$. It is entirely equivalent. – copper.hat Aug 16 '14 at 20:05

$Rank(A+B)\leq Rank(A)+Rank(B)$

2 Answers2