$A= \begin{bmatrix}-1 & 1 & 1 &1\\ 1 & -1 & 1 &1\\ 1 & 1 & -1 &1\\ 1 & 1 & 1 &-1\\ \end{bmatrix}$ $A$ is a symmetric matrix.Find its eigenvalues?

Question

$A= \begin{bmatrix}-1 & 1 & 1 &1\\ 1 & -1 & 1 &1\\ 1 & 1 & -1 &1\\ 1 & 1 & 1 &-1\\ \end{bmatrix}$

$A$ is a symmetric matrix.Find its eigenvalues ?

I saw this question in my book. I know that it can be solved using the cofactor definition of determinant and its eigenvalues are $+2,-2$

However, this process is cumbersome . I wonder that whether there is any trick or shortcut for finding its eigenvalues .

Answer 1

Hints: If the eigenvalues of $A+2I$ are $\lambda_1,\lambda_2,\lambda_3,\lambda_4$, then the eigenvalues of $A$ are $\lambda_1-2,\lambda_2-2,\lambda_3-2,\lambda_4-2$. Do you see why this is true?

The matrix $A+2I$ is a $4 \times 4$ matrix of all ones. Hence, the rank of $A+2I$ is one, and it has at most one non-zero eigenvalue (so three of its eigenvalues are $0$). You can find the non-zero eigenvalue by finding the corresponding eigenvector by inspection like J. W. Tanner suggested.

Answer 2

There is a cool argument borrowed from representation theory to calculate the eigenvectors of a circulant matrix $A$: in these matrices, each column is obtained from the previous one by shifting to the right its elements. In the shift, the last element goes back to the beginning. Formally, let $R$ by the permutation matrix that sends $e_i$ to $e_{i+1}$ for $i=1, 2,3$ and $e_4$ to $e_1$. Let also $A_i$ be the $i$-th column of $A$. The fact that $A$ is circulant formally writes as $A_{i+1} = RA_i$ for $i=1,2,3$ and $A_1 = R A_3$.

Recipe. Suppose $A$ is a 4x4 circulant matrix. Suppose the first row is $(v_1, v_2, v_3, v_4)$. If $i^2 = -1$, then its complex eigenvalues are: $$ v_1+v_2+v_3+v_4$$ $$(v_1-v_3) +i(v_2-v_4) $$ $$v_1-v_2+v_3-v_4$$ $$(v_1- v_3) -i(v_2-v_4) $$

In your case you get $2,-2,-2,-2$.

Upshot. By the end of the answer you will know how to generalize this to higher dimensions, and I highly recommend you to do this exercise.

Proof. The key observation is the following: being circulant implies $AR = RA$. Indeed, for any $i = 1,2,3$ we have

$$ A R e_i = A e_{i+1} = A_{i+1} = R A_i = RA e_i $$

and the same holds for $i=4$. Since $AR,RA$ coincide on a basis, they are equal. How the hell this fits into the eigenvalue problem? Firstly, notice that $R$ is diagonalizable, and its eigenvectors are $$z_1 =(1,1,1,1)$$ $$z_2 = (1,i,i^2, i^3)$$ $$z_3 =(1,-1, 1, -1)$$ $$z_4 = (1,-i,-1, i)$$

where $i^2=-1$.Indeed:

$$ Rz_1 = R(1,1,1,1) = (1,1,1,1) = \lambda_1 z_1 = z_1$$ $$ Rz_2 = R(1,i,-1,-i) = (-i,1,i,-1) = -i(1,i,-1,-i) = \lambda_2 z_2 = -i z_2 $$ $$ R z_3 = R(1,-1,1,-1) = (-1,1,-1,1) = - (1,-1,1,-1) = \lambda_3 z_3 = - z_3 $$ $$ R z_4 = R(1,-i,-1, i) = (i,1,-i,-1) = i(1,-i,-1,i) = \lambda_4 z_4 = i z_4$$

Secondly, if we compute $AR =RA$ in $z_i$ we get that $\lambda_i Az_i = R(A z_i)$. This means that $Az_i$ is an eigenvector of $R$ relative to $\lambda_i$; but there is essentially just one eigenvector relative to $\lambda_i$, so $A z_i$ must be a multiple scalar of $z_i$! This is a very tricky passage. We conclude that $z_i$ is also an eigenvector of $A$, and since we have found a basis of eigenvectors we are done with our search.

We finish with the explicit computation. Since $z_i$ has a 1 in the first component, if we have that $A z_i = \mu_i z_i$, then by looking at the first component we get that $v \cdot z_i = \mu_i$, where $v=(v_1, v_2, v_3, v_4)$. If you look closer to the expressions in the recipe, you will see that the formulas are exactly $v \cdot z_1, v \cdot z_2, v \cdot z_3, v \cdot z_4$.