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Find the eigenvalues and eigenspaces for the matrix $\begin{bmatrix}4&2&2\\2&4&2\\2&2&4\end{bmatrix}$

I know there is a trick to this one with out doing a $3 \times 3$ determinant to find it's eigenvalues (which is horrendous in this case) (i.e: $i - j + k$)

I started with $\begin{bmatrix}4-\lambda&2&2\\2&4-\lambda&2\\2&2&4-\lambda\end{bmatrix}$ then thought it would be best to seek advice before going deep into the rabbit hole. Can I put this matrix into a form where the 2's are all 0s? What is this form called? And will that change my determinant? Because if I can do that, I can take the product of the diagonal and find my eigenvalues much easier!

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    $\begingroup$ What if you factor a $2$? $\endgroup$ – Moo Jul 15 '16 at 2:14
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    $\begingroup$ You can write your matrix as $2I+2X$ where all entries of $X$ are $1$. Since the rank of $X$ is $1$, its eigenvectors should be easy to find. $\endgroup$ – stewbasic Jul 15 '16 at 2:17
  • $\begingroup$ After I Get the determinant do I need to multiply it back by 2 again since i factored out a 2? $\endgroup$ – Shammy Jul 15 '16 at 2:26
  • $\begingroup$ What? How does $2I + 2X$ help me? That just gives me a new matrix of $\begin{bmatrix}6&2&2\\2&6&2\\2&2&6\end{bmatrix}$ $\endgroup$ – Shammy Jul 15 '16 at 2:31
  • $\begingroup$ @Shammy all vectors are eigenvectors of the identity matrix. If you find the eigenvectors of 2X they will also be eigenvectors of 2I+2X due to the linearity of the operators. By the way you did not add 2I+2X correctly. X is a matrix with 1's in each entry, the result of 2I+2X should have 4's down the main diagonal and 2's in the off diagonal entries as in the matrix you began with. $\endgroup$ – Tucker Jul 15 '16 at 5:40
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Expanding on stewbasic's comment: Your matrix can be decomposed as $2I + 2X$, where $X$ is simply filled with $1$s. We are looking for scalars $\lambda$ and vectors $v$ satisfying $$\lambda v = (2I+2X)v = 2v + 2Xv \iff (\lambda/2 - 1)v = Xv.$$

The eigenvalues of $X$ are easy to compute: the eigenvalues of any $n \times n$ matrix of rank $1$ are given by its trace (with multiplicity $1$), and zero (with multiplicity $n-1$). In this case the trace is $3$, and so we obtain the eigenvalues $\lambda = 3,0$, with respective eigenspaces spanned by $\{(1,1,1)\}$ and $\{(-1,0,1),(-1,1,0)\}$.

Therefore the corresponding eigenvalues for $2I + 2X$ are $2(3+1)=8$ and $2(0+1) = 2$, with the same eigenvectors as above.

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  • $\begingroup$ The eigenvalues are not easy to compute at all for me. How exactly are you doing this? what is the matrix that you are taking the determinant of? $\endgroup$ – Shammy Jul 15 '16 at 2:29
  • $\begingroup$ What are you doing to this matrix $\begin{bmatrix}4&2&2\\2&4&2\\2&2&4\end{bmatrix}$ $\endgroup$ – Shammy Jul 15 '16 at 2:30
  • $\begingroup$ What? How does $2I + 2X$ help me? That just gives me a new matrix of $\begin{bmatrix}6&2&2\\2&6&2\\2&2&6\end{bmatrix}$ $\endgroup$ – Shammy Jul 15 '16 at 2:33
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    $\begingroup$ @Shammy That's not correct. (We're talking about the determinant in my comment, right?) You should get $3\lambda^2 - \lambda^3$. Check your computations. $\endgroup$ – Alex Provost Jul 15 '16 at 2:56
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    $\begingroup$ @Shammy Also maybe get some rest. You sound very tired! $\endgroup$ – Alex Provost Jul 15 '16 at 2:57

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