Can SOMEBODY PLS HELP ME
2 answers:
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Answer:
Step-by-step explanation:
You need to remember the mulplication of signs:
Knowing this, you can distributive the negative sign. Then you get:
Now you need to add the like terms. Then:
Finally, you can order the polynomial obtained in descending order. Therefore, the answer is:
The first two rows of coefficients are identical, so by inspection, the determinant is 0.
Answer:
Step-by-step explanation:
To prove that form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w
Recall that the set is linearly independent if and only if the equation
implies that
.
Recall that for i=1,...,n. Consider
to be the inverse transformation of T. Consider the equation
If we apply to this equation, then, we get
Since T is linear, its inverse is also linear, hence
which is equivalent to the equation
Since are linearly independt, this implies that
, so the set
is linearly independent.
Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist such that
Since T is surjective, there exists a vector v in V such that T(v) = w. Since is a basis of v, there exist
, such that
Then, applying T on both sides, we have that
which proves that generate the whole space W. Hence, the set
is a basis of W.
Consider the linear transformation , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of
given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of