Linear Algebra (Transform a Basis Set)

[DanielFaraday]

Homework Statement

If $$T:U\rightarrow V$$ is any linear transformation from U to V and $$B=\left\{u_1,u_2,\text{...},u_n\right\}$$ is a basis for U, then the set $$T(B)=\left\{T\left(u_1\right),T\left(u_2\right),\text{...} T\left(u_n\right)\right\}$$

a. spans V.
b. spans U.
c. is a basis for V.
d. is linearly independent.
e. spans the range of T.

Homework Equations

None

The Attempt at a Solution

It seems to me like all of these things are true (which is wrong of course). But in the example below, the result of the transformation does in fact meet all of the above criteria. What am I missing here?

$$u_1=\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 0<br /> \end{array}<br /> \right)$$
$$u_2=\left(<br /> \begin{array}{c}<br /> 0 \\<br /> 1 \\<br /> 0<br /> \end{array}<br /> \right)$$
$$u_3=\left(<br /> \begin{array}{c}<br /> 0 \\<br /> 0 \\<br /> 1<br /> \end{array}<br /> \right)$$
$$T(x,y,z)=(x,2y,3z)$$
$$T\left(u_1\right)=\left(<br /> \begin{array}{c}<br /> 1 \\<br /> 0 \\<br /> 0<br /> \end{array}<br /> \right)$$
$$T\left(u_2\right)=\left(<br /> \begin{array}{c}<br /> 0 \\<br /> 2 \\<br /> 0<br /> \end{array}<br /> \right)$$
$$T\left(u_3\right)=\left(<br /> \begin{array}{c}<br /> 0 \\<br /> 0 \\<br /> 3<br /> \end{array}<br /> \right)$$

 

[Mark44]

T(B) couldn't very well span U if U and V are different sized vector spaces, so that let's out b. And T(B) might or might not span V, depending on the dimension of the kernel of T. For example, if T(u) = 0, T(B) couldn't possibly span V, and the vectors in T(B) couldn't possibly be linearly independent, and therefore couldn't be a basis for V.

 

[DanielFaraday]

T(B) couldn't very well span U if U and V are different sized vector spaces, so that let's out b. And T(B) might or might not span V, depending on the dimension of the kernel of T. For example, if T(u) = 0, T(B) couldn't possibly span V, and the vectors in T(B) couldn't possibly be linearly independent, and therefore couldn't be a basis for V.

Great point. Thinking of T(u) = 0 is very helpful. That basically eliminates everything except e doesn't it.

Okay, I think I got it. The reason why e is the correct answer is because we know that the transformation is linear, correct?

 

[Mark44]

I think there's more to it than just that the transformation is linear. Seems like I remember a theorem that says that if T:U --> V is a linear transformation, and B = {u₁, u₂, ..., u_n} is a basis for U, then B = {T(u₁), T(u₂), ..., T(u_n)} spans the range of T.

 

[DanielFaraday]

I think there's more to it than just that the transformation is linear. Seems like I remember a theorem that says that if T:U --> V is a linear transformation, and B = {u₁, u₂, ..., u_n} is a basis for U, then B = {T(u₁), T(u₂), ..., T(u_n)} spans the range of T.

Okay. It makes sense, too.

Thanks! I am posting one more question as well (a True/False this time). Maybe you could take a look?

 

[HallsofIvy]

Since $\{u_1, u_2, \cdot\cdot\cdot, u_n\}$ form a basis for U, any vector in U can be written as a linear combination $u= a_1u_1+ a_2u_2+ \cdot\cdot\cdot a_nu_n$. And that means that any vector in the range of T is of the form $T(u)= a_1T(u_1)+ a_2T(u_2)+ \cdot\cdot\cdot+ a_nT(u_n)$. That says that $\{T(u_1), T(u_2), \cdot\cdot\cdot, T(u_n)\}$ spans the range of T.