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Found in: Page 146

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Consider two n x m matrices A and B. What can you say about the relationship among the quantities rank(A), rank(B), rank(A+B).

If we have two n x m matrices A and B then

$\mathrm{rank}\left(\mathrm{A}+\mathrm{B}\right)\le \mathrm{rank}\left(\mathrm{A}\right)+\mathrm{rank}\left(\mathrm{B}\right).$

See the step by step solution

## Step 1:   Mentioning concept

Since we know that

Dim (Im(A)) = rank(A) and dim (Im(B)) = rank(B)

So, dim (Im(A+B)) = rank(A+B)

## Step 2:   Finding the rank of (A+B)

Let $A=\left[{\stackrel{\to }{v}}_{1}{\stackrel{\to }{v}}_{2}...{\stackrel{\to }{v}}_{m}\right]\mathrm{and}B=\left[w{⃗}_{1}w{⃗}_{2}...w{⃗}_{m}\right]$

Where $v{⃗}_{i}andw{⃗}_{i}$ are column vectors of A and B respectively.

Then,

role="math" localid="1659355348377" $\mathrm{rank}\left(\mathrm{A}\right)=\mathrm{dim}\left(\mathrm{span}\left(\mathrm{v}{⃗}_{1},\mathrm{v}{⃗}_{2},...,\mathrm{v}{⃗}_{\mathrm{m}}\right)\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{rank}\left(\mathrm{B}\right)=\mathrm{dim}\left(\mathrm{span}\left(\mathrm{w}{⃗}_{1},\mathrm{w}{⃗}_{2},...\mathrm{w}{⃗}_{\mathrm{m}}\right)\right)$

$⇒rank\left(A+B\right)=dim\left(span\left(v{⃗}_{1}+w{⃗}_{1},v{⃗}_{2}+w{⃗}_{2},...v{⃗}_{m}+w{⃗}_{m}\right)\right)$

## Step 3: Finding the dimension of image of (A+B)

Let us claim,

$span\left(v{⃗}_{1}+w{⃗}_{1},\stackrel{\to }{v}{⃗}_{2}+w{⃗}_{2},...,v{⃗}_{m}+w{⃗}_{m}\right)\subset span\left(v{⃗}_{1},v{⃗}_{2},...,v{⃗}_{m}\right)+span\left(w{⃗}_{1},w{⃗}_{2},...,w{⃗}_{m}\right)$

Then any vector can be written as

$x⃗={a}_{1}\left(v{⃗}_{1}+w{⃗}_{1}\right)+{a}_{2}\left(v{⃗}_{2}+w{⃗}_{2}\right)+...+{a}_{m}\left(v{⃗}_{m}+w{⃗}_{m}\right)\phantom{\rule{0ex}{0ex}}⇒x⃗=\left({a}_{1}v{⃗}_{1}+{a}_{2}v{⃗}_{2}+...+{a}_{m}v{⃗}_{m}\right)+\left({a}_{1}w{⃗}_{1}+{a}_{2}w{⃗}_{2}+...+{a}_{m}w{⃗}_{m}\right)\phantom{\rule{0ex}{0ex}}⇒x⃗\in span\left(v{⃗}_{1},v{⃗}_{2},...,v{⃗}_{m}\right)+span\left(w{⃗}_{1},w{⃗}_{2},...,w{⃗}_{m}\right)$

for some scalars $a{⃗}_{1},a{⃗}_{2},...,{a}_{m}$.

$⇒span\left(v{⃗}_{1}+w{⃗}_{1},v{⃗}_{2}+w{⃗}_{2},...,v{⃗}_{m}+w{⃗}_{m}\right)\subseteq span\left(v{⃗}_{1},v{⃗}_{2},...,v{⃗}_{m}\right)+span\left(w{⃗}_{1},w{⃗}_{2},...,w{⃗}_{m}\right)\phantom{\rule{0ex}{0ex}}⇒Im\left(A+B\right)\subseteq Im\left(A\right)+Im\left(B\right)\phantom{\rule{0ex}{0ex}}⇒dim\left(Im\left(A+B\right)\right)\le dim\left(Im\left(A\right)\right)+dim\left(Im\left(B\right)\right)\phantom{\rule{0ex}{0ex}}⇒rank\left(A+B\right)\le rank\left(A\right)+rank\left(B\right)$

$⇒Im\left(A+B\right)\subseteq Im\left(A\right)+Im\left(B\right)\phantom{\rule{0ex}{0ex}}⇒dim\left(Im\left(A+B\right)\right)\le dim\left(Im\left(A\right)\right)+dim\left(Im\left(B\right)\right)\phantom{\rule{0ex}{0ex}}⇒rank\left(A+B\right)\le rank\left(A\right)+rank\left(B\right)$