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I have a linear transformation $F: \mathbb R^3 \to \mathbb R[t]_2$, where $\mathbb R[t]_2$ is the vector space of polynomials up to the second degree.

I also have three different bases: $$B = \{(1,1,0),(0,1,0),(0,0,1)\}$$ $$C = \{1, 1 + t, 1 + t^2\}$$ $$D = \{1, t, t^2\}$$

The linear transformation matrix $[F]_{B,C}$ has also been given:

$$\begin{pmatrix}1&2&-1\\1&0&-1\\0&1&0\end{pmatrix}.$$

The problem asks to write the linear transformation matrix of $[F]_{B,D}$.

What I've tried so far:

$$F(1,1,0) = something = 1(1,0,0) + 1(1,1,0) + 0(1,0,1)$$ $$F(0,1,0) = something = 2(1,0,0) + 0(1,1,0) + 1(1,0,1)$$ $$F(0,0,1) = something = (-1)(1,0,0) + (-1)(1,1,0) + 0(1,0,1)$$

So this is basically my attempt to find the linear transformation formula so I can write the matrix. But I'm not sure what to put where I wrote "something" as I'm dealing with a polynomial.

I've also thought that there may be a way to find the $[F]_{B,D}$ matrix using matrices of changing basis, but I could not think how.

Can anyone give me any hint?

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  • $\begingroup$ Hint: you are told $F(1,1,0)=1+(1+t)$, $F(0,1,0)=2\cdot1+0\cdot(1+t)+1\cdot(1+t^2)$, .... turning polynomials into 3-tuples doesn't help at all. $\endgroup$ Nov 26, 2020 at 17:02
  • $\begingroup$ Right, so if I write: F(1,1,0) = 2 + t F(0,1,0) = 3 + tˆ2 F(0,0,1) = -2 - t Where do I go from here? If I sum all of that I get the formula? $\endgroup$ Nov 26, 2020 at 17:07

2 Answers 2

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If you denote the identity by $I:R[t]_2\rightarrow R[t]_2$, the corresponding representation from basis $C$ to basis $D$ is (just reading from the given polynomials)

$$[I]_{C,D} =\begin{pmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$

So, the matrix looked for is

$$[F]_{B,D} = [I]_{C,D}[F]_{B,C} = \begin{pmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1&2&-1\\1&0&-1\\0&1&0\end{pmatrix}=\begin{pmatrix}2&3&-2\\1&0&-1\\0&1&0\end{pmatrix}$$

Tudo claro agora?

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From your work so far, it seems as though you have misinterpreted the matrix of the transformation. In particular, we have $$ [F(0,1,0)]_{C} = (2,0,1) \implies\\ F(0,1,0) = 2 \cdot 1 + 0 \cdot (1 + t) + 1\cdot (1 + t^2)\\ = 3 + t^2 = 3\cdot 1 + 0 \cdot t + 1\cdot t^2. $$ With that, we see that the coordinate vector of $F(0,1,0)$ with respect to $D$ is given by $$ [F(0,1,0)]_D = (3,0,1). $$ So, the second column of $[F]_{B,D}$ is given by $(3,0,1)$. That is, we have $$ [F]_{B,D} = \pmatrix{? & 3 & ?\\? & 0 & ?\\ ? & 1 & ?}. $$ The first column is equal to $[F(1,1,0)]_D$, and the third column is equal to $[F(0,0,1)]_D$.

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  • $\begingroup$ I'm trying to understand what I got wrong. Can you please clarify why you wrote [F(0,1,0)_C = (2,0,1)? It's my understanding, that if we apply F to vector (0,1,0), which belongs to base B, then we get another vector written as linear combination of base C and the coefficients of this linear combination form the column of the linear transformation matrix. Is not that true? $\endgroup$ Nov 26, 2020 at 18:24
  • $\begingroup$ @Brasilian_student I don't understand your second sentence, so I can't tell you whether it's correct. $(0,1,0)$ is the second element of the basis $B$, which means that $F(0,1,0)$ corresponds to the second column of the matrix of $[F]_{B,C}$. In particular, the vector $(2,0,1)$ is the coordinate vector of $F(0,1,0)$ relative to the basis $C$, which is to say that $$ F(0,1,0) = 2 \cdot 1 + 0 \cdot (1 + t) + 1\cdot (1 + t^2). $$ $\endgroup$ Nov 26, 2020 at 22:45

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