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Chapter 4: Q33E (page 200)

State true or false, every two-dimensional subspace of 鈩�2脳2contains at least one invertible matrix.

Short Answer

Expert verified

The given stamen is false.

Step by step solution

01

Definition of the invertible matrix

A matrix A of dimension n to n is called invertible if and only if there exists another matrix B of the same dimension, such that AB = BA = I, where I is the identity matrix of the same order.

Matrix B is known as the inverse of matrix A. Inverse of matrix A is symbolically represented by A-1.

For example

A=1325

02

Non-invertible matrix

Two-dimensional subspace spanned with 0000and role="math" localid="1659428405618" a0b0contains all matrices of form a0b0.

The determinant of every matrix of this type is 0. So, all of them are singular or non-invertible.

Hence, the statement is false.

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Most popular questions from this chapter

T denotes the space of infinity sequence of real numbers, T(f(t))=[f7f11]fromP2toR2.

Find the image, kernel, rank, and nullity of the transformationT inT(f(t))=f(7) from P2to R.

Find the basis of each of the spaceV=R3脳2,and determine its dimension.

T denotes the space of infinity sequence of real numbers,T(f(t))=[f(5)f(7)f(11)]fromP2toR3. .

Let Vbe the space of all infinite sequences of real numbers. See Example 5. Which of the subsets of Vgiven in Exercises 12 through 15 are subspaces of V? The sequences (x0,x1,...)that converge to zerorole="math" localid="1659412709897" (i.e.,limn鈫�鈭�xn=0)
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