Given an associate matrix for a linear transformation, find the bases of two given vector space
$begingroup$
Linear transfomation $T: mathbb{R}[X]_{≤2} rightarrow mathbb{R}^3$ with $T(f)=(f(0),f'(1),f(2))$
Find the bases such that the associative matrix is:
(i)begin{bmatrix}0&0&0\0&0&0\0&0&0end{bmatrix} (ii)begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix} (iii)begin{bmatrix}1&0&0\0&1&0\0&0&0end{bmatrix} (iv)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}
Probably the (iv) is the easiest, but my real question is, if do two bases of these vector space exist, such that I can find an associative matrix (i)
I know that for the bases, $(1,X,X^2)$, respectively $(e_1,e_2,e_3)$ of the two vector spaces, the associative matrix is: begin{bmatrix}1&0&0\0&1&2\1&2&4end{bmatrix}
Is there maybe a general rule, or a linear system to do, to find what I'm looking for?
linear-algebra matrices linear-transformations
asked Dec 4 '18 at 13:42
DadaDada
7510
$endgroup$
add a comment |
$begingroup$
Linear transfomation $T: mathbb{R}[X]_{≤2} rightarrow mathbb{R}^3$ with $T(f)=(f(0),f'(1),f(2))$
Find the bases such that the associative matrix is:
(i)begin{bmatrix}0&0&0\0&0&0\0&0&0end{bmatrix} (ii)begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix} (iii)begin{bmatrix}1&0&0\0&1&0\0&0&0end{bmatrix} (iv)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}
Probably the (iv) is the easiest, but my real question is, if do two bases of these vector space exist, such that I can find an associative matrix (i)
I know that for the bases, $(1,X,X^2)$, respectively $(e_1,e_2,e_3)$ of the two vector spaces, the associative matrix is: begin{bmatrix}1&0&0\0&1&2\1&2&4end{bmatrix}
Is there maybe a general rule, or a linear system to do, to find what I'm looking for?
linear-algebra matrices linear-transformations
asked Dec 4 '18 at 13:42
DadaDada
7510
$endgroup$
add a comment |
$begingroup$
Linear transfomation $T: mathbb{R}[X]_{≤2} rightarrow mathbb{R}^3$ with $T(f)=(f(0),f'(1),f(2))$
Find the bases such that the associative matrix is:
(i)begin{bmatrix}0&0&0\0&0&0\0&0&0end{bmatrix} (ii)begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix} (iii)begin{bmatrix}1&0&0\0&1&0\0&0&0end{bmatrix} (iv)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}
Probably the (iv) is the easiest, but my real question is, if do two bases of these vector space exist, such that I can find an associative matrix (i)
I know that for the bases, $(1,X,X^2)$, respectively $(e_1,e_2,e_3)$ of the two vector spaces, the associative matrix is: begin{bmatrix}1&0&0\0&1&2\1&2&4end{bmatrix}
Is there maybe a general rule, or a linear system to do, to find what I'm looking for?
linear-algebra matrices linear-transformations
asked Dec 4 '18 at 13:42
DadaDada
7510
$endgroup$
Linear transfomation $T: mathbb{R}[X]_{≤2} rightarrow mathbb{R}^3$ with $T(f)=(f(0),f'(1),f(2))$
Find the bases such that the associative matrix is:
(i)begin{bmatrix}0&0&0\0&0&0\0&0&0end{bmatrix} (ii)begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix} (iii)begin{bmatrix}1&0&0\0&1&0\0&0&0end{bmatrix} (iv)begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}
Probably the (iv) is the easiest, but my real question is, if do two bases of these vector space exist, such that I can find an associative matrix (i)
I know that for the bases, $(1,X,X^2)$, respectively $(e_1,e_2,e_3)$ of the two vector spaces, the associative matrix is: begin{bmatrix}1&0&0\0&1&2\1&2&4end{bmatrix}
Is there maybe a general rule, or a linear system to do, to find what I'm looking for?
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Dec 4 '18 at 13:42
DadaDada
7510
asked Dec 4 '18 at 13:42
DadaDada
7510
asked Dec 4 '18 at 13:42
DadaDada
7510
asked Dec 4 '18 at 13:42
DadaDada
7510
asked Dec 4 '18 at 13:42
DadaDada
7510
7510
add a comment |
add a comment |
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