Matrix to the Power of T: Solving for AT in A=[2 4 6 8]

[Daaniyaal]

Homework Statement

If A= [2 4 6 8] find A^T

Homework Equations

The Attempt at a Solution

Would I simply distribute the T across the matrix like this: [2^T 4^T 6^T 7^T 8^T]

 

[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.

 

[Daaniyaal]

Okay, thank you so much!

 

[Ray Vickson]

Homework Statement

If A= [2 4 6 8] find A^T

Homework Equations

The Attempt at a Solution

Would I simply distribute the T across the matrix like this: [2^T 4^T 6^T 7^T 8^T]

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].

 

[Daaniyaal]

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].

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!

 

[Ray Vickson]

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!

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.