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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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Which eigenspace are you talking about? There's an eigenspace associated to each eigenvalue. –  Chris Eagle May 10 '12 at 13:38
    
Number of linearly independent eigen vectors? –  srijan May 10 '12 at 13:39

1 Answer 1

up vote 4 down vote accepted

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

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I didnt get your point? Did you mean given linear transformation is symmetric nonsingular? –  srijan May 10 '12 at 14:00
    
Yes... did you hope to get through without examining the actual transformation? :) Write it down fast! –  rschwieb May 10 '12 at 14:04
    
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 –  srijan May 10 '12 at 14:07
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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... –  rschwieb May 10 '12 at 14:13

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