Let T : Rn-->Rn be a linear transformation, whose standard matrix is invertible. Show that if B = {v1, . . . , vn} is a basis for
Rn, then {T(v1), . . . ,T(vn)} is also a basis for Rn.
To show that {T(v1), . . . , T(vn)} is a basis for Rn, we need to prove two main things:
1. Span: We need to show that {T(v1), . . . , T(vn)} spans Rn, meaning that for any vector w in Rn, we can find scalars c1, . . . , cn such that w = c1T(v1) + . . . + cnT(vn).
2. Linear Independence: We need to show that {T(v1), . . . , T(vn)} is linearly independent, meaning that the only solution to c1T(v1) + . . . + cnT(vn) = 0 is c1 = c2 = . . . = cn = 0.
Let's prove each of these properties.
1. To show that {T(v1), . . . , T(vn)} spans Rn:
Since T is a linear transformation from Rn to Rn, we know that T maps any vector in Rn to another vector in Rn. Therefore, for any vector w in Rn, there exists a vector v in Rn such that T(v) = w.
Since B = {v1, . . . , vn} is a basis for Rn, we can express any vector v in Rn as a linear combination of the basis vectors v1, . . . , vn. So, v = c1v1 + . . . + cnvn for some scalars c1, . . . , cn.
Now, applying the linear transformation T to both sides of this equation, we get:
T(v) = T(c1v1 + . . . + cnvn)
Since T is a linear transformation, we have:
T(v) = c1T(v1) + . . . + cnT(vn)
But we know that T(v) = w, so substituting this in the equation above, we get:
w = c1T(v1) + . . . + cnT(vn)
This shows that for any vector w in Rn, we can find scalars c1, . . . , cn such that w = c1T(v1) + . . . + cnT(vn). Therefore, {T(v1), . . . , T(vn)} spans Rn.
2. To show that {T(v1), . . . , T(vn)} is linearly independent:
Assume that the equation c1T(v1) + . . . + cnT(vn) = 0 has a nontrivial solution, where not all scalars c1, . . . , cn are zero.
Since T is a linear transformation, it preserves linear combinations. So the equation can be rewritten as:
T(c1v1 + . . . + cnvn) = 0.
But we know that the only vector in Rn that T maps to 0 is the zero vector itself. So, c1v1 + . . . + cnvn = 0.
Since B = {v1, . . . , vn} is a basis for Rn, it is linearly independent. This means that the only solution to c1v1 + . . . + cnvn = 0 is c1 = c2 = . . . = cn = 0.
Therefore, the assumption that the equation has a nontrivial solution is false. This shows that {T(v1), . . . , T(vn)} is linearly independent.
Since we have shown that {T(v1), . . . , T(vn)} both spans Rn and is linearly independent, it is a basis for Rn.