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Chapter 3: Q64E (page 145)

Consider a subspace V of鈩�nwith dim(V)=n. Explain why V=鈩�n.

Short Answer

Expert verified

V=鈩�nbecause V is subspace of鈩�n .

Step by step solution

01

Describe the dimension of the image

For any matrix A,

dim(颈尘鈥�A)=rank(A)

02

Explain the reason why V=ℝn

Consider thatV in鈩�n and dim(V)=n.

This means that there are exactlyn linearly independent vectorsv鈫�1,...,v鈫�n that form a basis of V.

From the previous exercise, it is known that ifdimV=dimW andV is subspace ofW thenV=W .

Here, it is known thatV is subspace of 鈩�nand that dimV=dim鈩�n, therefore, V=鈩�n.

Hence,V=鈩�nbecause Vis subspace of鈩�n .

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Most popular questions from this chapter

(a) Consider a linear transformationT from 鈩�5to 鈩�3. What are the possible values of dim(kerT)? Explain.

(b) Consider a linear transformationT from 鈩�4to 鈩�7. What are the possible values of dim(imT)? Explain.

Suppose a matrix A in reduced row-echelon form can be obtained from a matrix M by a sequence of elementary row operations. Show thatA=rref(M). Hint: Both A and rref(M)are in reduced row-echelon form, and they have the same kernel. Exercise 88 is helpful.

Consider the matrices

A=[102040013050000160000001]andB=[102040013050000170000001]

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

Let V be the subspace of 鈩�4defined by the equation

x1-x2+2x3+4x4=0

Find a linear transformation T from 鈩�4to 鈩�4such that ker(T)={0鈫�}and im(T) = V. Describe T by its matrix A.

Describe the images and kernels of the transformations in Exercises 23through 25 geometrically.

24. Orthogonal projection onto the planex+2y+3z=0 in鈩�3.
See all solutions

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