Finding eigenvalues of $4 \times 4$ matrices

Question

$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ \end{bmatrix}$

tried to find Eigen value for this matrix $\begin{bmatrix} 1-\lambda & 0 & 0 & 1 \\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0 \\ 1 & 0 & 0 & 1-\lambda \\ \end{bmatrix}$ and got $\lambda^4-4\lambda^3+6\lambda^2-4\lambda$
so $\lambda(\lambda^3-4\lambda^2+6\lambda-4)$
but I tried to factorize $\lambda^3-4\lambda^2+6\lambda-4$ got $(\lambda-2)<br>\lambda^2-2\lambda+2$ and cannot factorize $\lambda^2-2\lambda+2$ .
is my calculation wrong? thanks!!

Answer 1

By permuting columns and rows you can immediately see the following: $$ \det\begin{pmatrix} 1-λ & 0 & 0 & 1 \\ 0 & 1-λ & 1 & 0 \\ 0 & 1 & 1 - λ & 0 \\ 1 & 0 & 0 & 1 - λ \end{pmatrix} = \det\begin{pmatrix} 1-λ&1&0&0\\ 1&1-λ&0&0\\ 0&0&1-λ&1\\ 0&0&1&1-λ \end{pmatrix} = \det\begin{pmatrix} 1-λ&1\\ 1&1-λ \end{pmatrix}^2 = ((1-λ)^2 - 1)^2 = (λ^2 - 2λ)^2 = λ^2(λ-2)^2 $$

Answer 2

as the matrix has rank=2 you have: $$\lambda_1=\lambda_2=0$$

in this case you can easily find the 2 no trivial eigenvectors by inspection

$$e_3=(1,0,0,1)$$ $$e_4=(0,1,1,0)$$

at which correspond the following eigenvalues:

$$\lambda_3=2$$ $$\lambda_4=2$$

Answer 3

I will just point out that it is quite easy to notice that:

• $(1,0,0,-1)A=(0,1,-1,0)A=0$
• $(1,0,0,1)A=(2,0,0,2)$ and $(0,1,1,0)A=(0,2,2,0)$

So with a bit of guesswork we can see that $0$ and $2$ are eigenvalues of (geometric) multiplicity two. (I think that since this is a very simple matrix with several symmetries, it is not that difficult to spot at least some eigenvector for these eigenvalues.)

Looking at some of those eigenvectors can also give you a reasonable idea which row operations could be used to calculated the charateristic polynomials. For example:

$$\det\left(\begin{array}{cccc} 1-\lambda& 0 & 0 & 1 \\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0 \\ 1 & 0 & 0 & 1-\lambda \end{array}\right)= \det\left(\begin{array}{cccc} 2-\lambda& 0 & 0 & 2-\lambda \\ 0 & 2-\lambda & 2-\lambda & 0 \\ 0 & 1 & 1-\lambda & 0 \\ 1 & 0 & 0 & 1-\lambda \end{array}\right)= (2-\lambda)^2\det\left(\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1-\lambda & 0 \\ 1 & 0 & 0 & 1-\lambda \end{array}\right)= (2-\lambda)^2\det\left(\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & -\lambda & 0 \\ 0 & 0 & 0 & -\lambda \end{array}\right)=(2-\lambda)^2\lambda^2$$

Answer 4

Check your determinant calculation. I get $\lambda^4 - 4 \lambda^3 + 4 \lambda^2$, giving the eigenvalues $0$, $0$, $2$, and $2$.

Answer 5

Let \begin{equation} A-\lambda I = \begin{bmatrix} 1-\lambda & 0 & 0 & 1 \\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0 \\ 1 & 0 & 0 & 1-\lambda \\ \end{bmatrix} \end{equation} Then \begin{equation} \det(A-\lambda I) = [(1-\lambda)^2-1]^2 \end{equation} Thus the eigenvalues are $\lambda = 0, 2$ with multiplicity 2. To be more precise, computing the determinant based on the first column gives \begin{align} \det(A-\lambda I) &= (1-\lambda)\det \begin{bmatrix} 1-\lambda & 1 & 0 \\ 1 & 1- \lambda & 0 \\ 0 & 0 & 1-\lambda \end{bmatrix} - \det \begin{bmatrix} 0 & 0 & 1 \\ 1-\lambda & 1 & 0 \\ 1 & 1-\lambda & 0 \end{bmatrix} \\ &=(1-\lambda)\{(1-\lambda)[(1-\lambda)^2-1]\} -[(1-\lambda)^2-1]\\ &=[(1-\lambda)^2-1]^2 \end{align}

For your convenience, let me compute the 3x3 matrix determinant. \begin{align} \det \begin{bmatrix} 1-\lambda & 1 & 0 \\ 1 & 1- \lambda & 0 \\ 0 & 0 & 1-\lambda \end{bmatrix} &= (1-\lambda) \det \begin{bmatrix} 1-\lambda & 0 \\ 0 & 1-\lambda \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 1 & 0 \\ 0 & 1-\lambda \end{bmatrix} \\ &= (1-\lambda)(1-\lambda)^2 - (1-\lambda) = (1-\lambda)[(1-\lambda)^2-1] \end{align}

Answer 6

The matrix can be written $A=B+I$. Then $B$ is a permutation matrix corresponding to the permutation $(14)(23)$ and so $B^2=I$. Since $B\ne I$ and $B\ne -I$, the minimal polynomial of $B$ must be $X^2-1$. Hence, the eigenvalues of $B$ are $-1$ and $1$ and so the eigenvalues of $A$ are $-1+1$ and $1+1$, or $0$ and $2$.

It easy to see that the characteristic polynomial of $B$ is $(X^2-1)^2$ and so both eigenvalues of $B$ have multiplicity $2$ and so do the eigenvalues of $A$.

Answer 7

You may try a simpler method, like Hamilton Caley method by using the traces:

$$\text{det}A = \frac{1}{24}\left\{(\text{tr} A)^4 - 6(\text{tr}A)^2\text{tr}(A^2) + 3(\text{tr}(A^2))^2 + 8\text{tr}(A^3)\text{tr}A - 6\text{tr}(A^4)\right\}$$

Anyway, as eigenvalues I got:

$$\{2; 2; 0; 0\}$$