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Linear Algebra With Applications
Found in: Page 146
Linear Algebra With Applications

Linear Algebra With Applications

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

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Short Answer

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

rank(A+B)rank(A)+rank(B).

See the step by step solution

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=v1 v2 ...vmandB=[w1w2...wm]

Where vi and wi are column vectors of A and B respectively.

Then,

role="math" localid="1659355348377" rank(A)=dim(span(v1,v2,...,vm))andrank(B)=dim(span(w1,w2,...wm))

rank(A+B)=dim(span(v1+w1,v2+w2,...vm+wm))

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

Let us claim,

span(v1+w1,v2+w2,...,vm+wm)span(v1,v2,...,vm)+span(w1,w2,...,wm)

Then any vector can be written as

x=a1(v1+w1)+a2(v2+w2)+...+am(vm+wm)x=(a1v1+a2v2+...+amvm)+(a1w1+a2w2+...+amwm)xspan(v1,v2,...,vm)+span(w1,w2,...,wm)

for some scalars a1,a2,...,am.

span(v1+w1,v2+w2,...,vm+wm)span(v1,v2,...,vm)+span(w1,w2,...,wm)Im(A+B)Im(A)+Im(B)dim(Im(A+B))dim(Im(A))+dim(Im(B))rank(A+B)rank(A)+rank(B)

Step 4:   Final Answer

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

Im(A+B)Im(A)+Im(B)dim(Im(A+B))dim(Im(A))+dim(Im(B))rank(A+B)rank(A)+rank(B)

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