# Matrix to the power t

1. Dec 9, 2012

### Daaniyaal

1. The problem statement, all variables and given/known data
If A= [2 4 6 8] find AT

2. Relevant equations

3. The attempt at a solution
Would I simply distribute the T across the matrix like this: [2T 4T 6T 7T 8T]

2. Dec 9, 2012

### Mute

The superscript 'T' here indicates the transpose of the matrix A. For a matrix A, if the element in the ith row and jth column is $A_{ij}$, then for the transpose matrix $A^T$, the element in the ith row and jth column is $A_{ji}$.

Note that if A is an mxn matrix, $A^T$ is an nxm matrix.

3. Dec 9, 2012

### Daaniyaal

Okay, thank you so much!

4. Dec 9, 2012

### Ray Vickson

Ia A a true matrix? What you have written is a 4-dimensional row vector. Its transpose would be the 4-dimensional column vector with elements 2,4,5,8 in that order; that is:
$$[2\;4\;6\;8]^T = \left[ \begin{array}{c}{2\\4\\6\\8}\end{array}\right].$$

5. Dec 9, 2012

### Daaniyaal

It is given to me by my teacher as a matrix, but I guess if it's a vector than it is a vector.

I did transpose it that way tho :D

Thanks!

6. Dec 10, 2012

### Ray Vickson

A vector is a special case of a matrix. However, perhaps you were supposed to interpret A as a 2x2 matrix, so A = [[2,4],[6,8]], where these are the two rows; that is:
$$A = \pmatrix{2&4\\6&8}.$$
The transpose of A is
$$A^T = \pmatrix{2 & 6 \\4 & 8}.$$

On the other hand, A to the power t is considerably more complicated. It would take several pages to write out the answer.