Linear Transformations with a Change of Basis

Question

There is a linear transformation P1 -> P2

$T(a +bx) = (2a-3b) + (b-5a)x +(a+b)x^2$

Determine the matrix representation of the transformation relative to the given bases B and C.

$B = \{ 1,~ x+5 \}$ $C = \{1, 1 + x, 1 + x^2 \}$

Solution attempt: I first practiced this with the standard basis vectors

$B = \{ 1,~ x \}$

$C = \{1, x, x^2 \}$

And got the following:

$T(a +bx) = (2a-3b) + (b-5a)x +(a+b)x^2$

$a + b = b - 5a = a + b = 0$

Therefore, $ a = b = 0$

$a(2 - 5x + x^2) + b(-3 + x + x^2)$

So the span is calculated to be:

{$2-5x + x^2, ~ -3 + x + x^2$}

So the transformation matrix consists of

|$2~ -3 $|

|$-5~~ 1 $|

|$ 1~ ~1$|

I am unsure of how to begin when the bases change to the ones given above. Should I be taking a similar approach, just with the new bases? Any help would be appreciated - self studying a course over the summer. Thanks in advance!

Answer 1

You can either:

1. Set up the matrix representation with respect to the given bases directly: determine the images under the linear transformation of the basis vectors in $B$ and write these images with respect to the basis $C$; fill the columns of the transformation matrix with these coordinates.

2. Use the matrix representation with respect to the standard bases you already found and multiply (on "both sides") with the appropriate matrices for the change of bases between the standard bases and the given bases $B$ and $C$ respectively.

And since you're studying this material, ideally you do both and verify if you get the same result :-).

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Let me give it in a slightly more general framework first.

Say we have:

• a vector space $V$, with bases $B$ and $E$;
• a vector space $W$, with bases $C$ and $F$;
• a linear transformation $T : V \to W$.

Change of basis

If we write $x_B$ and $x_E$ for the coordinate representations of a vector $x \in V$ with respect to the bases $B$ and $E$ respectively, then the matrix for the change of basis between $B$ and $E$ is the matrix satisfying $x_B = M_{BE}x_E$ and it can be shown that $x_E = M_{EB}x_B$ with $M_{EB}=M_{BE}^{-1}$.

You construct this matrix $M_{BE}$ by filling the columns with the coordinate representations of the basis vectors of $E$ written with respect to $B$: $M_{BE} = \bigl( (e_1)_B \vert \ldots \vert (e_n)_B \bigr)$ and vice versa for $M_{EB}$.

Similarly, for $y = T(x) \in W$, you have $y_C = M_{CF}y_F$ and $y_F = M_{FC}y_C$ with $M_{FC}=M_{CF}^{-1}$.

Matrix of a linear transformation

The matrix of a linear transformation $T : V \to W$ depends on the choice of bases for $V$ and $W$. If we choose $B$ and $C$ and denote the associated matrix as $A_{CB}$, then you have: $$T(x)_C = \color{blue}{A_{CB}}x_B$$ You can construct this matrix as described in 1. in the beginning of my answer. For the other bases: $$T(x)_F = \color{red}{A_{FE}}x_E$$ Now you can use the change of basis matrices to get a relation between $A_{CB}$ and $A_{FE}$: $$\color{blue}{A_{CB}}= M_{CF}\color{red}{A_{FE}}M_{EB}$$

Your case

The bases $B$ and $C$ are given; take the standard bases for $E$ and $F$:

$$\begin{array}{ll} B = \{ 1,~ x+5 \} & C = \{1, 1 + x, 1 + x^2 \} \\ E = \{ 1,~ x \} & F = \{1, x, x^2 \} \end{array}$$

You can easily find with the notations from above: $$M_{EB} = \begin{pmatrix}1 & 5 \\ 0 & 1\end{pmatrix} \quad\mbox{and}\quad M_{FC} = \begin{pmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \implies M_{CF} = M_{FC}^{-1} = \begin{pmatrix}1 & -1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} $$

You already found: $$A_{FE} = \begin{pmatrix}2 & -3 \\ -5 & 1 \\ 1 & 1\end{pmatrix} $$

So then $A_{CB}$ follows as: $$A_{CB} = M_{CF}A_{FE}M_{EB} = \begin{pmatrix}1 & -1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}2 & -3 \\ -5 & 1 \\ 1 & 1\end{pmatrix} \begin{pmatrix}1 & 5 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}6 & 25 \\ -5 & -24 \\ 1 & 6\end{pmatrix}$$ You can verify this by finding $A_{CB}$ with the "direct method" (of 1.).