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deusexmachina.


Dec
16
accepted Square root of this $2x2$ matrix
Dec
16
comment Square root of this $2x2$ matrix
@CameronWilliams Indeed you are right, post an answer and I will accept.
Dec
16
comment Square root of this $2x2$ matrix
@CameronWilliams Do you not get the inverse via having an indentity matrix beside the matrix you wish to change perform row operations until the left side is the identity matrix (leaving the right side the inverse)?
Dec
16
asked Square root of this $2x2$ matrix
Dec
16
accepted How to calculate $R=S\sqrt{\Lambda}S^{-1}$
Dec
16
comment How to calculate $R=S\sqrt{\Lambda}S^{-1}$
Add up the square roots of the diagonal?
Dec
16
comment How to calculate $R=S\sqrt{\Lambda}S^{-1}$
Oh, when I do take the values, what do I do with them, add them up?
Dec
16
revised How to calculate $R=S\sqrt{\Lambda}S^{-1}$
deleted 169 characters in body; edited title
Dec
16
comment How to calculate $R=S\sqrt{\Lambda}S^{-1}$
@Amzoti Oh jeeze, It's only the square root of $\Lambda $? Hows does one compute that anyway?
Dec
16
comment How to calculate $R=S\sqrt{\Lambda}S^{-1}$
@5xum Updated! It's on diagonalization. Specifically finding the square root of A.
Dec
16
asked How to calculate $R=S\sqrt{\Lambda}S^{-1}$
Dec
16
accepted Square root of these $2\times2$ matrices
Dec
16
comment Square root of these $2\times2$ matrices
Derp! Thanks pal, I wrote down the formula incorrect and did incorrect gauss-jordan calculations. I'm back on track!
Dec
16
revised Square root of these $2\times2$ matrices
edited body
Dec
16
comment Square root of these $2\times2$ matrices
@MarkBennet Indeed! I am however stuck at trying to find $S$ :/, I get a zero matrix from the second eigenvalue.
Dec
16
asked Square root of these $2\times2$ matrices
Dec
15
comment This answer is confusing $4\times 4$ eigenvalue calculation
@AlexKruckman So $(2,2,2,2)$ is the basically the scalar(eigenvalue) times the direction(eigenvector)? Are there any particular types of matrices this is done for or not done for, and what is this type of process/technique called(so I can read more on it!)?
Dec
15
comment This answer is confusing $4\times 4$ eigenvalue calculation
$\bigl(\begin{smallmatrix} 1+0+1+0\\ 0+1+0+1\\ 1+0+1+0\\ 0+1+0+1 \end{smallmatrix}\bigr)=(2,2,2,2)$, two however disappear because there are only $2$ independent vectors, meaning $(2,2,0,0)$?
Dec
15
comment This answer is confusing $4\times 4$ eigenvalue calculation
@AlexKruckman I apologise if I am too dumb to understand, I normally require examples to figure the relationships between things. From what yous are saying this is what I'm getting in my head at the moment: $\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 & 1\\ \end{bmatrix} =\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1 \end{bmatrix}$
Dec
15
accepted This answer is confusing $4\times 4$ eigenvalue calculation