# Basis change matrix for polynomials.

$\mathcal{B}$ is the canonical basis of $\mathbb{R}_3 [X]$.

We have the polynomials:

$\begin{cases} P_0 = 1, \\ P_1 = 1 - X, \\ P_2 = X - X^2, \\ P_3 = X^2 - X^3. \\ \end{cases}$ and $\begin{cases} Q_0 = 1, \\ Q_1 = 1 + X, \\ Q_2 = 1 + X + X^2, \\ Q_3 = 1 + X + X^2 + X^3. \end{cases}$

1. Prove that $P = 3X - X^2 + 8X^3$ can be expressed as linear combination of the polynomials $P_0,P_1,P_2,P_3$ and determine the coefficients.

2. Prove that the $\mathcal{B_1} = (P_0, P_1, P_2, P_3)$ is a basis of $\mathbb{R}_3 [X]$ and determine the matrix $P_1$ of change of basis from $\mathcal{B}$ to $\mathcal{B}_1$.

3. Prove that $\mathcal{B}_2 = (Q_0, Q_1, Q_2, Q_3)$ is a basis of $\mathbb{R}_3[X]$ and determine the matrix $P_2$ of change of basis from $\mathcal{B}$ to $\mathcal{B}_2$.

4. Applying the formula of basis change on a vector, find the results of question $1$.

I have found:

1. $P = 10P_0 - 10P_1 - 7P_2 - 8P_3$

The coefficients are: $10, -10, - 7$ and $- 8$.

1. $\mathcal{B_1}$ is a basis of $\mathbb{R}_3[X]$.

$P_{\mathcal{B}, \mathcal{B_1}} = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}$

1. $\mathcal{B_2}$ is a basis of $\mathbb{R}_3[X]$.

$P_{\mathcal{B}, \mathcal{B_2}} = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$

Are basis change matrices correct?

1. The basis change formula is: $B = P^{-1}.A.P$. I don't know how to find the results of question 1 using.

## 2 Answers

Please bear with me if you know the initial part of this answer.

How do we write a matrix transformation? Given $(V,B_V)$ and $(W,B_W)$, where $V,W$ are vector spaces with bases $B_V,B_W$ respectively, and a linear transformation $T : V \to W$, how do we write the matrix for $T$, with respect to $B_V$ and $B_W$?

1. Arrange $B_V,B_W$ in some order.

2. Find where the elements of $B_V$ go under $T$.

3. Express the image vectors in terms of the basis $B_W$, and write the coefficients as columns one after the other to get a matrix.

For example, if $T : \mathbb R^2 \to \mathbb R^2$, both with standard bases, is given by $T(x) = -x$, then we find where $(1,0)$ goes : it goes to $(-1,0)$ in the standard basis, and then find where $(0,1)$ goes : it goes to $(0,-1)$ in the standard basis. Therefore, the matrix of $T$ will be : first column $(-1,0)$, second column $(0,-1)$. You can see that you will get minus the identity matrix if you put this together, for the matrix of $T$.

Now, what is a change of basis matrix? Here is a nice interpretation :

The change of basis matrix from basis $A$ to basis $B$ (in a vector space $V$), is the matrix of the identity linear transformation from $(V,A)$ to $(V,B)$.

That is, find where all elements of $A$ go, and express them in basis $B$.

What have you done? For the change of basis matrix from $B$ to $B_1$, this is the matrix of the identity map from $(V,B)$ to $(V,B_1)$. So, express the elements of $B$ in terms of the elements of $B_1$. You have done the opposite. You expressed $B_1$ in terms of $B$, when you wrote own your change of basis matrix.

The same with the third question.

Consequently, we see that the change of basis matrices are the inverses of what you have written, because base conversion needs to go the other way. The right answer for b and c, therefore, is the inverse of the matrices you have written.

Finally, for $d$, of course the given polynomial can be interpreted as an element of $(V,B)$. Now, to change the basis to $B_1$ we want to shift to $(V,B_1)$, so simply multiply the change of basis matrix you have with you, and the vector expressed in $B_1$.

That is, the inverse of the $P_{B,B_1}$ which you got, when multiplied with $[0,3,-1,8]^T$, gives you $[10,-10,-7,-8]$. You can check this is right.

• $P_{ \mathcal{B}, \mathcal{B_1} } = \begin{pmatrix} 1 & -1 & -1 & -1 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$ and $P_{ \mathcal{B}, \mathcal{B_2} } = \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$ Are the matrices correct? Thank you so much for explaining the concept to me. Commented Jul 2, 2018 at 12:11
• You are welcome, but did you find the matrix inverse of $P_{B,B_1}$? The first matrix is not correct. You can check the right answer here : wolframalpha.com/input/… The answer to the second is right! Nevertheless, good job. Commented Jul 2, 2018 at 12:19
• I have found $P_{\mathcal{B}, \mathcal{B}_1} = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & -1 & -1 & -1 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}$. When I computed $P_{\mathcal{B}, \mathcal{B}_1}. \begin{pmatrix} 0 \\ 3 \\ -1 \\ 8 \\ \end{pmatrix}$. I found: $v_{\mathcal{B}_2} = \begin{pmatrix} 10 \\ -10 \\ -7 \\ - 8 \\ \end{pmatrix}$, which is the result I found in question 1. Am I correct? Thank you. Commented Jul 2, 2018 at 17:02
• Yes that is correct. Commented Jul 3, 2018 at 3:09

The base change matrices are correct.
To prove the given polynomials form a basis, just note that the determinants of these matrices are nonzero.

For question 4, the formula $P^{-1}AP$ applies only to a matrix $A$.
Here we only have a vector $b$ with $[b]_\mathcal B =(0,3,-1,8)^T$.
To express $[b]_{\mathcal B_1}$, calculate $P^{-1}$ (e.g. by manually expressing the standard basis elements by $P_i$'s) and then apply/verify $$[b]_{\mathcal B_1}=P^{-1}[b]_\mathcal B$$ Note that, for a base change matrix $P:\mathcal B\leadsto\mathcal B_1$ and a square matrix $A$, the columns of $AP$ are just $A$ applied to the elements of $\mathcal B_1$, coordinated in $\mathcal B$, so $P^{-1}$ of this will contain the $\mathcal B_1$-coordinates of the same vectors.

• $P^{-1}_{\mathcal{B}, \mathcal{B_1}}$ ? Commented Jul 2, 2018 at 11:07