# finding the eigenvalues of a matrix without the polynomial

I have a matrix

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

I already know that one eigenvector of that matrix is $(1, 1, 1)$ and one eigenvalue is $1$, based on the constant sum of rows. Now I am supposed to calculate the other eigenvalues, without using the polynomial of the matrix.. I know that there are supposed to be two more eigenvalues and that the product of the eigenvalues is equal to $detA=4$ and that the sum of eigenvalues is equal to $trA=-3$, but I just can't seem to get the right answer.

Is this even the right approach? Any hints or ideas on how to proceed are appreciated.

• The trace gives you : $1+\lambda_2+\lambda_3 = -3$ and the determinant : $1\cdot \lambda_2 \cdot \lambda_3 = -4$, can you combine those two equations somehow? May 20 '17 at 17:05
• May 20 '17 at 18:47

Let $k_1,k_2,k_3$ be the eigenvalues. As you said one of them is $1$ but $\prod k_i=det(A)=4$ (you said $-4$ but it's wrong, retry to calculate) and $\sum k_i=tr(A)=-3$. So if $k_1=1$ then $k_2k_3=4$ and $k_2+k_3=-4$ and so it's easy to see that $k_2=k_3=-2$

Hint

The trace gives you : $$1+\lambda_2+\lambda_3 = -3$$ and the determinant (thanks to StabiloBoss for smotting the sign change) : $$1\cdot \lambda_2 \cdot \lambda_3 = 4$$, can you combine those two equations somehow to find $\lambda_2, \lambda_3$?

• One more hint for the OP: $(x-\lambda_2) (x-\lambda_3)=x^2-(\lambda_2+\lambda_3)x+\lambda_2 \lambda_3$ May 20 '17 at 17:10
• @mathreadler I did try that and got some weird solutions which gave me more eigenvectors than there are supposed to be in this matrix.. I might have made a calculation error, but I did try it a few times and always get the same thing May 20 '17 at 17:12
• @ivana14 have you noted the sign error on the determinant that StabiloBoss spotted? May 20 '17 at 17:18
• @mathreadler yes, I did May 20 '17 at 17:19

If part of the specification really is "without using its polynomial" then it does seem to me that using the fact that trace is the sum of the roots and the determinant their product comes pretty close to using the polynomial. (On the whole "do this with your hands tied behind your back" questions are rather silly, so I am not criticising @mathreadler's solution at all: it's exactly what one should do in real life.) But this is an attempt to meet the spec.

We can proceed in the way the OP has started: we see that $1$ is an eigenvalue because when we subtract $1$ from the diagonal elements then we have a matrix whose singularity is guaranteed by the (eigen)vector $(1,1,1)^{T}$.

If we cast around for other things to subtract we might try $0$, then $\pm 1$ then $\pm 2$ and then Eureka, $-2$ is an eigenvalue twice over, because when we subtract it we will be able to guarantee the singularity by the looking at the obvious eigenvectors $(1,-1,0)^{T}$ and $(1,0,-1)^{T}$.

• how do you know that $(1, -1, 0)$ and $(1, 0, -1)$ are obvious eigenvectors? I am wondering how I can see the eigenvectors without actually calculating them.. It would really help if you explained a bit. Thanks May 21 '17 at 7:25
• When you subtract twice the identity from your matrix you get the all $1$-s matrix; as all columns are equal subtracting any two (which is what applying the vectors I mentioned does) gives zero. May 21 '17 at 7:46

Because the matrix is symmetric, we know that eigenvectors from different eigenspaces are orthogonal. If you guess the eigenvector $(1,-1,0)$ (you only have to guess among vectors orthogonal to $(1,1,1)$), you find that it is an eigenvector. A third eigenvector must then be $(1,1,1) \times (1,-1,0) = (1,1,-2)$. Knowing the eigenvectors, you can easily find the eigenvalues.

Here's a way to find the eigenvalues that truly does not use the characteristic polynomial at all. Note that $\DeclareMathOperator{\id}{id} A = \mathbb{1} - 2\id$, where $\mathbb{1}$ is the matrix of all $1$s and $\id$ is the identity matrix. Given an eigenvector $v$ of $A$ with eigenvalue $\lambda$, then \begin{align} \label{eig} \lambda v = Av = (\mathbb{1} - 2 \id)(v) = \mathbb{1}v - 2v \implies \mathbb{1}v = (\lambda + 2) v \tag{1} \end{align} so $v$ is also an eigenvalue of $\mathbb{1}$ with eigenvalue $\lambda + 2$. Thus we have reduced the problem to finding the eigenvalues of $\mathbb{1}$.

By the same observation you made in your question, since the row sums of $\mathbb{1}$ are constant, then $(1,1,1)^T$ is an eigenvector with eigenvalue $3$. Since $\mathbb{1}$ has only $1$ linearly independent column, then its kernel is $2$-dimensional, hence $0$ is an eigenvalue of multiplicity $2$. Thus the eigenvalues of $\mathbb{1}$ are $3, 0, 0$, so the eigenvalues of $A$ are $1, -2, -2$ by (\ref{eig}).