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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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    Number of linearly independent eigen vectors?2012-05-10

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

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    @DevendraSinghRana The comment by the person who wrote the question left would seem to indicate *no*, which was the point I was trying to make earlier. Also they accepted the answer, so it must have been what they wanted.2018-01-07