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zeion
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Linear Algebra - Change of basis matrices and RREF question what in the world??
Suppose the linear transformation T: P3 -> P2, over R has the matrix
<br /> <br /> A = \begin{bmatrix}1&2&0&0\\0&1&2&1\\1&1&1&1 \end{bmatrix} <br /> <br />
relative to the standard bases of P3 and P2.
Find bases A' pf P3 and B' of P2 such that the matrix A' of T relative to A' and B' is the reduced row-echelon form for A.
Okay, so the relation is:
<br /> T_{B'A'} = I_{B'B}T_{BA}I_{AA'}<br />
Where A is the standard basis for for P3 and B is the standard basis for P2.
I also have the relation:
R = VA, where R is the RREF of A and V is an invertible matrix that maps A to R.
If I take the basis A' of A to be equal to A, then
<br /> I_{AA'} = <i>_{4x4}<br /> </i>
and <br /> I_{B'B} = V<br />
Then I form an augmented matrix to find V:
<br /> \begin{bmatrix}1&2&0&0&|&1&0&0\\0&1&2&1&|&0&1&0\\1&1&1&1&|&0&0&1\end{bmatrix} <br />
which row reduces to
<br /> \begin{bmatrix}1&0&0&2/3&|&-1/3&2/3&4/3\\0&1&0&-1/3&|&2/3&-1/3&-2/3\\0&0&1&2/3&|&-1/3&1/3&1/3\end{bmatrix} <br />
Then V is the augmented part..
After this I will find the inverse of V, then find B'.. but it doesn't look right, and the final answer I got was wrong ~_~
Can someone confirm if this is correct track or if I made a mistake somewhere? Thanks.
Homework Statement
Suppose the linear transformation T: P3 -> P2, over R has the matrix
<br /> <br /> A = \begin{bmatrix}1&2&0&0\\0&1&2&1\\1&1&1&1 \end{bmatrix} <br /> <br />
relative to the standard bases of P3 and P2.
Find bases A' pf P3 and B' of P2 such that the matrix A' of T relative to A' and B' is the reduced row-echelon form for A.
Homework Equations
The Attempt at a Solution
Okay, so the relation is:
<br /> T_{B'A'} = I_{B'B}T_{BA}I_{AA'}<br />
Where A is the standard basis for for P3 and B is the standard basis for P2.
I also have the relation:
R = VA, where R is the RREF of A and V is an invertible matrix that maps A to R.
If I take the basis A' of A to be equal to A, then
<br /> I_{AA'} = <i>_{4x4}<br /> </i>
and <br /> I_{B'B} = V<br />
Then I form an augmented matrix to find V:
<br /> \begin{bmatrix}1&2&0&0&|&1&0&0\\0&1&2&1&|&0&1&0\\1&1&1&1&|&0&0&1\end{bmatrix} <br />
which row reduces to
<br /> \begin{bmatrix}1&0&0&2/3&|&-1/3&2/3&4/3\\0&1&0&-1/3&|&2/3&-1/3&-2/3\\0&0&1&2/3&|&-1/3&1/3&1/3\end{bmatrix} <br />
Then V is the augmented part..
After this I will find the inverse of V, then find B'.. but it doesn't look right, and the final answer I got was wrong ~_~
Can someone confirm if this is correct track or if I made a mistake somewhere? Thanks.
