The trace of a square n×n matrix a=(aij) is the sum a11+a22+⋯+ann of the entries on its main diagonal. let v be the vector space of all 2×2 matrices with real entries. let h be the set of all 2×2 matrices with real entries that have trace 0. is h a subspace of the vector space v?

Let and be two vectors in .

is a subspace of if (1) and (2) for any scalar , we have .

(1) True;

so .

(2) Also true, since

Therefore is a subspace of .

Yes, H is a subspace of V

Step-by-step explanation:

We know that V is the space of all the 2x2 matrices with real entries.

H is the set of all 2x2 matrices with real entries that have trace equal to 0.

Obviusly the matrices that are in the space H also belong in the space V (because in H you have some selected matrices and in V you have all of them). The thing we need to prove is if H is an actual subspace.

Suppose we have two matrices that belong to H, A and B.

We must see that:

1) if A and B ∈ H, then (A + B)∈H

2) for a scalar number k, k*A ∈ H

lets write this as:

where a1 + a4 = 0 = b1 + b4

then:

the trace is:

a1 + b1 - (a4 + b4) = (a1 - a4) + (b1 - b4) = 0

then the trace is nule, and (A + B) ∈ H

and:

the trace is:

k*a1 - k*a4 = k(a1 - a4) = 0

so kA ∈ H

then H is a subspace of V

120 seems like the most sensible answer.

step-by-step explanation:

if there is only one topping on all caramel apples and there's 120 of them. then all apples should/will be covered in sprinkles. make sense?

you ignore the remainder, therefore the answer is y = x+2

step-by-step explanation:

you have to divide the denominator by the numerator. you can use synthetic division or long division.

[-5] .     1 .     7 .     11

x + 2 + 1/x+5