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Expert-verified Found in: Page 146 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider a 4 x 2 matrix A and 2 x 5 matrix B.a. What are the possible dimensions of the kernel of AB?b. What are the possible dimensions of the image of AB?

a. Possible dimensions of the kernel of AB = {5,4,3,2,1}

b. Possible dimensions of the image of AB ={0,1,2,3,4}.

See the step by step solution

## Step 1:   Mentioning the concept

For any matrix A of order n x m

Dim (Im(A)) = rank(A).

And

Dim (ker(A))= number of free variables

=total number of variables −number of leading variables

= m − rank(A).

## Step 2:   Given quantities

We have given two matrices A of order 4 x 2, and B of order 2 x5

Then the matrix AB is of order 4 x 5.

Now we have to find the possible dimensions of ker(AB) and Im(AB).

## Step 3:   (b) Finding the possible dimensions of Im(AB).

Now, for a n x m matrix A, rank (A) ≤ min (m,n).

Then for a matrix AB of order 4 x 5, rank (AB) ≤ min (4,5)

$⇒\mathrm{rank}\left(\mathrm{AB}\right)\le 4\phantom{\rule{0ex}{0ex}}⇒\mathrm{ran}\left(\mathrm{AB}\right)=0\mathrm{or}1\mathrm{or}2\mathrm{or}3\mathrm{or}4$

Thus, possible value of rank(AB)={0,1,2,3,4}.

Since, dim(Im(AB))=rank(AB)

Then possible dimensions of image of AB = {0,1,2,3,4}.

## Step 4:  (a) Finding the possible dimensions of ker(AB).

Since the number of unknown variables in an n x m matrix = m then the number of unknown variables in a 4 x 5 matrix = 5.

Now, the dim(ker(A)) = m – rank(AB) = 5 - {0,1,2,3,4}.

$⇒$ Possible values of dim(ker(AB)) = {5,4,3,2,1}

If we have a 4 x 2 matrix A and a 2 x 5 matrix B then .

Then,

c. Possible dimensions of the kernel of AB = {5,4,3,2,1}

d. Possible dimensions of the image of AB ={0,1,2,3,4}. ### Want to see more solutions like these? 