Linear Transformations and matrix representation

[henry3369]

Assume the mapping T: P2 -> P2 defined by:
T(a₀ + a₁t+a₂t²) = 3a₀ + (5a₀ - 2a₁)t + (4a₁ + a₂)t²
is linear.Find the matrix representation of T relative to the basis B = {1,t,t²}

My book says to first compute the images of the basis vector. This is the point where I'm stuck at because I'm not sure how the books arrives at the images:
T(b1) = T(1) = 3+5t
T(b2) = T(t) = -2t+4t²
T(b3) = T(t²) = t²

Where are these results coming from?
I don't understand where 1 is supposed to go to solve for T(1). I guess its the notation that is throwing me off. Usually when solving for a transformation, it has something such as T(x) = x^2, and you solve the transformation by substituting the value of the input for x. But now my input is 1 for an entire expression (a₀ + a₁t+a₂t²)

 

[Svein]

Where are these results coming from?

Start with what you know. If you want T(1), look at your definition. In order to get T(1), put a₀=1, a₁=0 and a₂=0. Then the definition says T(1)= 3⋅1 + (5⋅1 - 2⋅0)t + (4⋅0 + 0)t². In the same way, to find T(t) , put a₀=0, a₁=1 and a₂=0. The rest is left as an exercise...

 

[Fredrik]

I'm not a fan of calling the functions 1, t and t² (these are notations for numbers, not functions). I would define functions $e_0, e_1, e_2$ by
\begin{align*}
&e_0(x)=1\\
&e_1(x)=x\\
&e_2(x)=x^2
\end{align*} for all real numbers x. Then T is defined by $T(a_0e_0+a_1e_1+a_2e_2)=3a_0e_0+(5a_0-2a_1)e_1+(4a_1+a_2)e_2$ for all real numbers $a_1,a_2,a_3$. Now let's do what Svein did, in my notation:
$$T(e_0)=T(1e_0+0e_1+0e_2)=3\cdot 1 e_0+(5\cdot 1-2\cdot 0)e_1+(4\cdot 0+0)e_2=3e_0+5e_1.$$

 

[henry3369]

Start with what you know. If you want T(1), look at your definition. In order to get T(1), put a₀=1, a₁=0 and a₂=0. Then the definition says T(1)= 3⋅1 + (5⋅1 - 2⋅0)t + (4⋅0 + 0)t². In the same way, to find T(t) , put a₀=0, a₁=1 and a₂=0. The rest is left as an exercise...

How do you know which a_x corresponds to the input? For T(1), you set a₀ =1 and for T(t) you set a₁ = 1. I don't think it could be the order in which it appears in B, because B is just a set of vectors and the order shouldn't matter.

 

[henry3369]

I'm not a fan of calling the functions 1, t and t² (these are notations for numbers, not functions). I would define functions $e_0, e_1, e_2$ by
\begin{align*}
&e_0(x)=1\\
&e_1(x)=x\\
&e_2(x)=x^2
\end{align*} for all real numbers x. Then T is defined by $T(a_0e_0+a_1e_1+a_2e_2)=3a_0e_0+(5a_0-2a_1)e_1+(4a_1+a_2)e_2$ for all real numbers $a_1,a_2,a_3$. Now let's do what Svein did, in my notation:
$$T(e_0)=T(1e_0+0e_1+0e_2)=3\cdot 1 e_0+(5\cdot 1-2\cdot 0)e_1+(4\cdot 0+0)e_2=3e_0+5e_1.$$

So what makes
e₀(x)=1 and not e₀(x)=x? If the order of the vectors in the Basis changed, how would I know that e₀(x)=1? . Also, why are the others always zero?

 

[Fredrik]

How do you know which a_x corresponds to the input? For T(1), you set a₀ =1 and for T(t) you set a₁ = 1. I don't think it could be the order in which it appears in B, because B is just a set of vectors and the order shouldn't matter.

Yes, strictly speaking, it's ambiguous to talk about the components of a vector in a specific basis. We should always be talking about the components of a vector with respect to an ordered basis like the triple $(b_1,b_2,b_3)$ rather than the components of a vector with respect to the basis $\{b_1,b_2,b_3\}$. Unfortunately people are sloppy with the language. But they're at least being sloppy in a consistent way. When they talk about the components of a vector with respect to $\{b_1,b_2,b_3\}$, they always mean with respect to $(b_1,b_2,b_3)$, and never with respect to e.g. $(b_3,b_1,b_2)$.

So what makes
e₀(x)=1 and not e₀(x)=x? If the order of the vectors in the Basis changed, how would I know that e₀(x)=1?

The way I did it is just a convention. You could number the functions differently if you want to.

Now you're probably thinking "wait a minute, the formula for the number on row i, column j of the matrix depends on the order of the basis vectors, so each choice of how to order them could give me a different matrix". This would be a correct observation. A linear operator and a basis don't uniquely determine a matrix. A linear operator and an ordered basis on the other hand...

In this problem, it's safe to assume that you should find the matrix of T with respect to the ordered basis $(b_1,b_2,b_3)$ (i.e. my $(e_0,e_1,e_2)$).

Also, why are the others always zero?

I'm not sure what others you're referring to.