# Compute the matrices for the given linear transformations

• Oct 7th 2010, 09:20 AM
Runty
Compute the matrices for the given linear transformations
The format that this question is presented in is chiefly what has me stumped. Either I'm unfamiliar with it or I just can't remember how to do it in this format.

Compute the matrices for the given linear transformations $T:F^n\rightarrow F^n$ with respect to the standard basis for $F^n$.

a) $T(a_1,...,a_n)=(a_1,...,a_1), (a_1,...,a_n)\in F^n$

b) $T(a_1,...,a_n)=(a_n,a_{n-1},...,a_1), (a_1,...,a_n)\in F^n$

c) $T(a_1,...,a_n)=(a_1+a_2,...,a_{n-1}+a_n,0), (a_1,...,a_n)\in F^n$
• Oct 7th 2010, 12:17 PM
HappyJoe
Let me consider part b).

The notation just means that $T$ maps a typical vector $(a_1,...,a_n)$ in $F^n$ to the vector $(a_n,...,a_1)$, also in $F^n$.

The matrix of this transformation with respect to the standard basis for $F^n$ (which is (1,0,0,....,0), (0,1,0,....,0), ...., (0,0,...,0,1)) is the matrix, whose first column is the coordinates of T applied to the first basis vector, and similarly for the other columns. To be concrete, T applied to the first basis vector is:

T(1,0,...,0) = (0,0,...,1).

Hence the first column of the matrix is

0
0
.
.
.
0
1

Similarly for the other columns, so the matrix will look something like this:

0 0 . . . 0 1
0 0 . . . 1 0
0 0 . . . 0 0
. . . . . . . .
. . . . . . . .
. . . . . . . .
0 1 . . . 0 0
1 0 . . . 0 0
• Oct 8th 2010, 07:10 AM
HallsofIvy
It might help to look at simple examples first.

Suppose n= 3. Then
a) T(a, b, c)= (a, a, a) so, looking at the standard basis, T(1, 0, 0)= (1, 1, 1) , T(0, 1, 0)= (0, 0, 0), T(0, 0, 1)= (0, 0, 0) and those are the columns of the matrix representation of T: $\begin{bmatrix}1 & 0 & 0 \\1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix}$.

b) T(a, b, c)= (c, b, a) so, looking at the standard basis, T(1, 0, 0)= (0, 0, 1), T(0, 1, 0)= (0, 1, 0), T(0, 0, 1)= (1, 0, 0) and those are the columns of the matrixs representation of T: $\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0\\ 1 & 0 & 0\end{bmatrix}$

c) T(a, b, c)= (a+ b, b+ c, 0) so, looking at the standard basis, T(1, 0, 0)= (1, 0, 0), T(0, 1, 0)= (1, 1, 0), T(0, 0, 1)= (, 1, 0) and those are the columns of the matrixs representation of T: $\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}$
• Oct 8th 2010, 02:58 PM
Danrio
I think your a) is wrong i'm pretty sure the answer is the transopse of what you said.
• Oct 9th 2010, 03:38 AM
HallsofIvy
If you are refering to my solution (in 3 dimensions) of $\begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 0 \\1 & 0 & 0\end{bmatrix}$, then it is sufficient to note that
$\begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 0 \\1 & 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b\\ c\end{bmatrix}= \begin{bmatrix} a \\ a\\ a\end{bmatrix}$

while the transpose $\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}$ gives
$\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b\\ c\end{bmatrix}= \begin{bmatrix}a+ b+ c \\ 0 \\ 0\end{bmatrix}$

Or are you using a convention of writing T(a, b, c) as $\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}= \begin{bmatrix} a & a & a\end{bmatrix}$?

If so, then you are correct- but we are using different conventions.