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Chapter 3: Q41E (page 164)

There exists a 2 x 2 matrix A such that Im (A) = ker (A).

Short Answer

Expert verified

The above statement is true.

There exists a 2 x 2 matrix A such that Im (A) = ker (A).

Step by step solution

01

Considering a 2 x 2 matrix 

Let A be a 2 x 2 matrix such that

A=0100鈬�x2=0andx1isafreevariable

02

Finding the ker (A) and Im (A)

To find ker (A), let us suppose Ax = 0.

0100x1x2=00

鈬�x2=0andx1is a free variable.

Letx2=k.

Then,ker鈦�(A)=0k=span01

AndIm鈦�(A)=10

dim (ker(A)) = 1 = dim (Im(A))

03

Final Answer

There exists a 2 x 2 matrix A=0100such that Im (A) = ker (A).

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Most popular questions from this chapter

Consider the matrices

A=[102040013050000160000001]andB=[102040013050000170000001]

Show that the kernels of the matrices A and B are different

Find a basis of the subspace of 鈩�4defined by the equation

2x1-x2+2x3+4x4=0.

Find a basis of the kernel of the matrix

[1203500146]

Justify your answer carefully; that is, explain how you know that the vectors you found are linearly independent and span the kernel.

In Problem 46 through 55, Find all the cubics through the given points. You may use the results from Exercises 44 and 45 throughout. If there is a unique cubic, make a rough sketch of it. If there are infinitely many cubics, sketch two of them.

(0,0),(1,0),(2,0),(3,0),(4,0),(0,1),(0,2),(0,3),(1,1)

In Exercises 21 through 25, find the reduced row-echelon form of the given matrix A. Then find a basis of the image of A and a basis of the kernel of A.

23.[102401-3-134-680-131]
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