How to find representation of polynomial w.r.t different basis
$begingroup$
Let $B$ be the basis of the vector space of polynomials of degree less than or equal to 2. $B = {1, t-1,(t-1)^2}$.
Let $u = 2t^2-5t+6$. How do you find $u_b$, the coordinate vector of $u$ relative to $B$? In particular, I'm looking how would you do this using matrices.
polynomials vector-spaces linear-transformations
asked Apr 15 '15 at 23:56
larrylarry
36611130
$endgroup$
add a comment |
$begingroup$
Let $B$ be the basis of the vector space of polynomials of degree less than or equal to 2. $B = {1, t-1,(t-1)^2}$.
Let $u = 2t^2-5t+6$. How do you find $u_b$, the coordinate vector of $u$ relative to $B$? In particular, I'm looking how would you do this using matrices.
polynomials vector-spaces linear-transformations
asked Apr 15 '15 at 23:56
larrylarry
36611130
$endgroup$
add a comment |
$begingroup$
Let $B$ be the basis of the vector space of polynomials of degree less than or equal to 2. $B = {1, t-1,(t-1)^2}$.
Let $u = 2t^2-5t+6$. How do you find $u_b$, the coordinate vector of $u$ relative to $B$? In particular, I'm looking how would you do this using matrices.
polynomials vector-spaces linear-transformations
asked Apr 15 '15 at 23:56
larrylarry
36611130
$endgroup$
Let $B$ be the basis of the vector space of polynomials of degree less than or equal to 2. $B = {1, t-1,(t-1)^2}$.
Let $u = 2t^2-5t+6$. How do you find $u_b$, the coordinate vector of $u$ relative to $B$? In particular, I'm looking how would you do this using matrices.
polynomials vector-spaces linear-transformations
polynomials vector-spaces linear-transformations
asked Apr 15 '15 at 23:56
larrylarry
36611130
asked Apr 15 '15 at 23:56
larrylarry
36611130
asked Apr 15 '15 at 23:56
larrylarry
36611130
asked Apr 15 '15 at 23:56
larrylarry
36611130
asked Apr 15 '15 at 23:56
larrylarry
36611130
36611130
add a comment |
add a comment |
1 Answer
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$begingroup$
$$mathcal{B}= left{{ begin{bmatrix}0t^2 \ 0t \ 1 end{bmatrix}, begin{bmatrix}0t^2 \ 1t \ -1 end{bmatrix}, begin{bmatrix}1t^2 \ -2t \ 1 end{bmatrix}}right}$$
This gives us the change of basis matrix:
$$mathbf{B} = begin{bmatrix}0 & 0 & 1\ 0 & 1 & -2\ 1 & -1 & 1end{bmatrix} $$
which is invertible, because $mathcal{B}$ is a basis and $mathbf{B}$ is square:
$$mathbf{B}^{-1} = begin{bmatrix}1 & 1 & 1\ 2 & 1 & 0\ 1 & 0 & 0end{bmatrix} $$
To convert a vector to your desired coordinate system, multiply the vector on the left by this matrix:
$$left[{u}right]_{mathcal{B}} = mathbf{B}^{-1} begin{bmatrix}2 \ -5 \ 6 end{bmatrix} = color{blue}{begin{bmatrix}3 \ -1 \ 2 end{bmatrix}} $$
Which is to say:
$$2t^2 -5t + 6 = color{blue}{3} color{blue}{-1}(t-1) + color{blue}{2}(t-1)^2$$
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
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1 Answer
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$begingroup$
$$mathcal{B}= left{{ begin{bmatrix}0t^2 \ 0t \ 1 end{bmatrix}, begin{bmatrix}0t^2 \ 1t \ -1 end{bmatrix}, begin{bmatrix}1t^2 \ -2t \ 1 end{bmatrix}}right}$$
This gives us the change of basis matrix:
$$mathbf{B} = begin{bmatrix}0 & 0 & 1\ 0 & 1 & -2\ 1 & -1 & 1end{bmatrix} $$
which is invertible, because $mathcal{B}$ is a basis and $mathbf{B}$ is square:
$$mathbf{B}^{-1} = begin{bmatrix}1 & 1 & 1\ 2 & 1 & 0\ 1 & 0 & 0end{bmatrix} $$
To convert a vector to your desired coordinate system, multiply the vector on the left by this matrix:
$$left[{u}right]_{mathcal{B}} = mathbf{B}^{-1} begin{bmatrix}2 \ -5 \ 6 end{bmatrix} = color{blue}{begin{bmatrix}3 \ -1 \ 2 end{bmatrix}} $$
Which is to say:
$$2t^2 -5t + 6 = color{blue}{3} color{blue}{-1}(t-1) + color{blue}{2}(t-1)^2$$
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
$endgroup$
add a comment |
$begingroup$
$$mathcal{B}= left{{ begin{bmatrix}0t^2 \ 0t \ 1 end{bmatrix}, begin{bmatrix}0t^2 \ 1t \ -1 end{bmatrix}, begin{bmatrix}1t^2 \ -2t \ 1 end{bmatrix}}right}$$
This gives us the change of basis matrix:
$$mathbf{B} = begin{bmatrix}0 & 0 & 1\ 0 & 1 & -2\ 1 & -1 & 1end{bmatrix} $$
which is invertible, because $mathcal{B}$ is a basis and $mathbf{B}$ is square:
$$mathbf{B}^{-1} = begin{bmatrix}1 & 1 & 1\ 2 & 1 & 0\ 1 & 0 & 0end{bmatrix} $$
To convert a vector to your desired coordinate system, multiply the vector on the left by this matrix:
$$left[{u}right]_{mathcal{B}} = mathbf{B}^{-1} begin{bmatrix}2 \ -5 \ 6 end{bmatrix} = color{blue}{begin{bmatrix}3 \ -1 \ 2 end{bmatrix}} $$
Which is to say:
$$2t^2 -5t + 6 = color{blue}{3} color{blue}{-1}(t-1) + color{blue}{2}(t-1)^2$$
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
$endgroup$
add a comment |
$begingroup$
$$mathcal{B}= left{{ begin{bmatrix}0t^2 \ 0t \ 1 end{bmatrix}, begin{bmatrix}0t^2 \ 1t \ -1 end{bmatrix}, begin{bmatrix}1t^2 \ -2t \ 1 end{bmatrix}}right}$$
This gives us the change of basis matrix:
$$mathbf{B} = begin{bmatrix}0 & 0 & 1\ 0 & 1 & -2\ 1 & -1 & 1end{bmatrix} $$
which is invertible, because $mathcal{B}$ is a basis and $mathbf{B}$ is square:
$$mathbf{B}^{-1} = begin{bmatrix}1 & 1 & 1\ 2 & 1 & 0\ 1 & 0 & 0end{bmatrix} $$
To convert a vector to your desired coordinate system, multiply the vector on the left by this matrix:
$$left[{u}right]_{mathcal{B}} = mathbf{B}^{-1} begin{bmatrix}2 \ -5 \ 6 end{bmatrix} = color{blue}{begin{bmatrix}3 \ -1 \ 2 end{bmatrix}} $$
Which is to say:
$$2t^2 -5t + 6 = color{blue}{3} color{blue}{-1}(t-1) + color{blue}{2}(t-1)^2$$
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
$endgroup$
$$mathcal{B}= left{{ begin{bmatrix}0t^2 \ 0t \ 1 end{bmatrix}, begin{bmatrix}0t^2 \ 1t \ -1 end{bmatrix}, begin{bmatrix}1t^2 \ -2t \ 1 end{bmatrix}}right}$$
This gives us the change of basis matrix:
$$mathbf{B} = begin{bmatrix}0 & 0 & 1\ 0 & 1 & -2\ 1 & -1 & 1end{bmatrix} $$
which is invertible, because $mathcal{B}$ is a basis and $mathbf{B}$ is square:
$$mathbf{B}^{-1} = begin{bmatrix}1 & 1 & 1\ 2 & 1 & 0\ 1 & 0 & 0end{bmatrix} $$
To convert a vector to your desired coordinate system, multiply the vector on the left by this matrix:
$$left[{u}right]_{mathcal{B}} = mathbf{B}^{-1} begin{bmatrix}2 \ -5 \ 6 end{bmatrix} = color{blue}{begin{bmatrix}3 \ -1 \ 2 end{bmatrix}} $$
Which is to say:
$$2t^2 -5t + 6 = color{blue}{3} color{blue}{-1}(t-1) + color{blue}{2}(t-1)^2$$
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
edited Apr 16 '15 at 0:24
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
answered Apr 16 '15 at 0:16
GFauxPasGFauxPas
4,16311229
4,16311229
add a comment |
add a comment |
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