The equation of a plane through the origin is x-y=0. The basis for the subspace

Question

The equation of a plane through the origin is x-y=0. The basis for the subspace of R^3 for the solution set of the given plane. if M22 denotes the set of all 2 X 2 matrices, then the set of all 2 X 2 invertible matrices with matrix standard addition and scalar multiplication is - THIS SECTION IS TO GIVE YOU THE BACKGROUND QS
A vector v= (v1, v2, .....,vn) is a specific single non null vector. The vector will always be linearly - PLEASE ANS THIS PART ONLY
i. dependant
ii independant
iii related
iv Unknown

Explanation / Answer

A family of vectors u1, u2, ..., un is linearly independent if and only if setting a linear combination of the vectors equal to the zero vector, as such:

c1u1 + c2u2 + .... + cnun = 0 (with c1, c2, ..., cn scalars)

has only the trivial solution, i.e., c1 = c2 = ... = cn = 0.

We will be able to find a solution with c1, c2, ..., cn not all zero if and only if the family of vectors is linearly dependent.

(Often the above conditions are used as the definitions of linear independence/dependence. Sometimes linear dependence is instead defined as the ability to write one vector as a linear combination of the others, with the vectors being linearly independent if that is impossible. These are really the same definition: after all, if the above equation has a solutoin other than the trivial one, there are some c's for which the c does not equal zero. Moving one of those terms to the other side of the equation will easily yield an expression for one vector being written as a linear combination of others in the set. And vice versa: if we can write one vector as a linear combination of others, we can subtract it from that linear combination to equal the zero vector in a case where the c's are not all 0. Therefore, these two ways of thinking about linear dependence/independence are equivalent.)

So, for this problem: our family of vectors consists of only one vector, v. By the above definition, v is linearly independent iff

cv = 0

has only the trivial solution c=0.

We know v = (v1, v2, ..., vn) to be non-null, so there must be at least one vk such that vk does not equal zero. Since cv = (cv1, cv2, ..., cvn), in order for this to equal the zero vector we must have cv1=0, cv2=0, ..., cvn=0. But since we have some vk such that vk does not equal 0, in order for cvk to equal 0, we must have c=0, giving this equation only the trivial solution. Therefore v is linearly independent, and the answer is (ii).

(Incidentally, this proves that any single non-null vector is linearly independent. What if we have a single null vector? Then cv = 0 is the same as c(0, 0, 0, ..., 0) = 0, and c can be anything. Since in that case we would have many solutions other than the trivial one for cv = 0, any lone null vector would be linearly dependent.)

Since v is (ii) linearly independent, (i) and (iv) are not true, and "linearly related" doesn't make sense in this context (you generally say two quantities x and y are "linearly related" if they are directly proportional, i.e., there is some k such that y=kx). So the only correct answer is (ii).

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