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

1 Answers 1

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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?2012-05-10
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    Yes... did you hope to get through without examining the actual transformation? :) Write it down fast!2012-05-10
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    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? thanks2012-05-10
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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...2012-05-10
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    But isn't it that we have to mention , eigenspace is corresponding to which eigenvalue, I mean does the question makes sense without mentioning to which eigenvalue??2018-01-06
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    @DevendraSinghRana I thought the poster intended to find the dimension of each eigenspace, from the comments they included.2018-01-06
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    Well I also have a feeling that it might be a possiblity but I am not cent percent sure that's why I asked. @rschweib2018-01-07
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    @DevendraSinghRana You saw the comment by the OP, right? *Number of linearly independent eigen vectors? – srijan May 10 '12 at 13:39* Sounds like that's what they were looking for, in the end.2018-01-07
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    @DevendraSinghRana Whether *what* is true?2018-01-07
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    @rschwieb whether it is true that if we are not given to corresponding which eigenvaluer we have to find the eigenvector then we find all the eigenvector possibly LI.???2018-01-07
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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