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Linear Algebra (Similar Matrices)

  • #1

Homework Statement



Write the standard matrix representation TE for T.

Homework Equations



[tex]
T\left(ax^3+bx^2+cx+d\right)=(a-b)x^2+(c-d)x+(a+b-c)
[/tex]

The Attempt at a Solution



[tex]
[T]_E=\left(
\begin{array}{c}
a+b-c \\
c-d \\
a-b \\
0
\end{array}
\right)
[/tex]

I think this is right, but a subsequent problem asks me to show that this matrix is similar to another matrix. Similarity only makes sense for an n x n matrix, so this must be wrong.

Any ideas?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
The basis 'vectors' are x^3, x^2, x and 1. So ax^3+bx^2+cx+d=(a,b,c,d).This is exactly the same as other exercises you have already done, it's T(a,b,c,d)=(0,a-b,c-d,a+b-c). Sure, it's an 4x4 matrix.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Is T here a function from the space of polynomials of degree 3 or less to itself or to the space of polynomials of degree 2 or less? If the former, the matrix is 4 by 4. If the latter, it is 4 by 3 (but the first row is all "0"s).
For example, taking x3 as the first "basis vector", T(x3)= ax2+ 0x+ a as a polynomial of degree 2 or less or 0x3+ ax2+ 0 x+ a as a polynomial of degree 3 or less. That is:
[tex]T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ 0 \\ a\end{bmatrix}[/tex]
in the first case or
[tex]T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ a \\ 0 \\ a\end{bmatrix}[/tex]
in the second.

Do you see that those give you the first column of the matrix?
 
  • #4
Ah yes, I see. I was thinking of T as a vector instead of a transformation. Thanks!
 

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