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Can we determine if the following matrix has 3 linearly independent eigenvectors without any calculations? I know that it can be determined by calculating the eigenvalues, but I was wondering if something like that can be concluded just by observing this matrix, since it's a true/false question. Thanks

$$ \begin{bmatrix} 1 && 3 && 3 \\ -3 && -5 && -3 \\ 3 && 3 && 1 \end{bmatrix} $$

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  • $\begingroup$ Note that the columns all add to the same number, that is, $1$. That tells you immediately that $1$ is an eigenvalue. This is really a follow up to @egreg answer. $\endgroup$ – B. Goddard May 20 '17 at 14:46
  • $\begingroup$ @B.Goddard okay, so whenever I have a marix whose columns all add to the same number, I can automatically conclude that one eigenvalue is 1? $\endgroup$ – ivana14 May 20 '17 at 14:47
  • $\begingroup$ Not "1", but whatever they add to. Because when you subtract $\lambda I$, you're subtracting $\lambda$ from the sum of each column, making the column sums of $A-\lambda I$ equal $0$, which means $A-\lambda I$ is singular. $\endgroup$ – B. Goddard May 20 '17 at 15:28
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If you try $A+2I$, you get $$ \begin{bmatrix} 3 & 3 & 3 \\ -3 & -3 & -3 \\ 3 & 3 & 3 \end{bmatrix} $$ which has rank $1$, so $-2$ is an eigenvalue with geometric multiplicity $2$.

Can you see another eigenvalue?

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  • $\begingroup$ As @B.Goddard stated in their comment, it's 1. But the problem is that I need to conclude if the matrix has 3 linearly independent eigenvectors just by looking at it. I am no supposed to actually calculate them. Can it be done? $\endgroup$ – ivana14 May 20 '17 at 14:49
  • $\begingroup$ @ivana14 The first thing to look for is row sums and column sums. So immediately you get one eigenvalue is $1$. A simple look shows also that an eigenvalue is $-2$. $\endgroup$ – egreg May 20 '17 at 15:00
  • $\begingroup$ so whenever a matrix has a constant row sum and column sum, I can say that one eigenvalue is $1$? or either row or column sum can be constant? How did you get -2 just by looking at the matrix? Sorry, I am just really confused. thanks $\endgroup$ – ivana14 May 20 '17 at 15:07
  • $\begingroup$ @ivana14 No: if the columns have constant sum $s$, then $s$ is an eigenvalue; similarly if the rows have constant sum. For $-2$, I observed that adding $2$ to the diagonal gives a rank one matrix. $\endgroup$ – egreg May 20 '17 at 15:29
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    $\begingroup$ @ivana14 Not only the matrix $A+2I$ has zero determinant: it has rank $1$, so its null space has dimension $3-1=2$. $\endgroup$ – egreg May 20 '17 at 16:14

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