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Let $T:\mathbb{R^4}\rightarrow \mathbb{R^{4}}$ be defined by $T(x,y,z,w)=(x+y+5w,x+2y+w,-z+2w,5x+y+2z)$ then what would be the dimension of the eigenspace of $T$?

One approach may be to find out eigenvalues and then eigenvectors. Is there any other approach that will consume less amount of time and calculation?

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  • $\begingroup$ Which eigenspace are you talking about? There's an eigenspace associated to each eigenvalue. $\endgroup$
    – Chris Eagle
    May 10, 2012 at 13:38
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    $\begingroup$ Number of linearly independent eigen vectors? $\endgroup$
    – Srijan
    May 10, 2012 at 13:39

1 Answer 1

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A nonsingular real $n\times n$ symmetric matrix has $n$ linearly independent eigenvectors.

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  • $\begingroup$ I didnt get your point? Did you mean given linear transformation is symmetric nonsingular? $\endgroup$
    – Srijan
    May 10, 2012 at 14:00
  • $\begingroup$ Yes... did you hope to get through without examining the actual transformation? :) Write it down fast! $\endgroup$
    – rschwieb
    May 10, 2012 at 14:04
  • $\begingroup$ It is clear that symmetric matrices are orthogonally diagonalizable. Then no need to find out domension of eigen space it would be 4. I wanted to know in case of ordinary matrix? thanks $\endgroup$
    – Srijan
    May 10, 2012 at 14:07
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    $\begingroup$ If you want to ask the question about general matrices, then no, there is no easy answer. After all, for $n\geq 5$ you are trying to solve a polynomial in a single variable of degree 5 or more! This is by no means simple... $\endgroup$
    – rschwieb
    May 10, 2012 at 14:13
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    $\begingroup$ @DevendraSinghRana I thought the poster intended to find the dimension of each eigenspace, from the comments they included. $\endgroup$
    – rschwieb
    Jan 6, 2018 at 16:46

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