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Consider a matrix $A$ of dimension $n$X$n$ whose eigen vectors are $y_1,y_2,y_3,...,y_n$ and are linearly independent. What are the properties of the eigen vectors of the matrix $P$ whose columns are $y_1$,$y_2$,$y_3$,....,$y_n$ ? Of course, we know that $AP=PK$ where $K$ is the diagonal matrix containing the eigen values of $A$. Can we say anything about the linear independence of the eigen vectors of $P$ for its diagonalizability?

Thanks, Aravind.

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$P$ can be any invertible matrix.

Let $P\in GL_n(\mathbb{F})$, and call the columns of $P$ $y_1,\ldots,y_n$. Then $\{y_1,\ldots,y_n\}$ is a basis of $\mathbb{F}^n$, and we can define a linear transformation $T:\mathbb{F}^n\to\mathbb{F}^n$ by $y_i\mapsto\lambda_iy_i$, where $\lambda_i$ are scalars. Represent $T$ by a matrix $A$, and by the above process one can obtain $P$ (among many other matrices).

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  • $\begingroup$ Could you please elaborate a bit more? The basis vectors are multiplied by $\lambda_i$, So a vector y=$a_1y_1+a_2y_2+....+a_ny_n$ is mapped to y=$a_1\lambda_1y_1+....+a_n\lambda_ny_n$. Assume we use the same basis {$y1$,$y2$,....} for both. So the transformation matrix is going to be a diagonal matrix with $\lambda_i=A_ii$.I understood till this. What are you trying to conclude from this? $\endgroup$ – AravindRajesh Aug 11 '14 at 21:15
  • $\begingroup$ No, we represent the transformation with respect to the standard basis, and obtain a matrix with eigenvectors $y_1,\ldots,y_n$. Since this can be done for any basis $y_1,\ldots,y_n$, any invertible matrix can be obtained this way. $\endgroup$ – Amitai Yuval Aug 11 '14 at 21:24
  • $\begingroup$ Hey, I don't understand the solution. Can you explain how to get P? $\endgroup$ – AravindRajesh Aug 17 '14 at 14:20
  • $\begingroup$ Hey, can you explain how you can get P? I still don't understand. $\endgroup$ – AravindRajesh Aug 17 '14 at 14:21

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