# What you can conclude about signs of number $\lambda$ and $\omega$

Let matrix $A\in M_n(\mathbb R)$ is symetric matrix such that $a_{11}=-1,a_{22}=-1,a_{33}=1$ and spectrum of $A=\{\lambda,\omega\}$ such that $\lambda>\omega$. $N(A-\lambda I)=L(v_1,v_2)$ and $N(A-\omega I)=L(v_3)$ where is $v_1=(1,1,0),v_2=(1,1,1),v_3=(1,-1,0)$ .

What can conclude about signs of number \lambda and omega?And explain the answer.

I have more question about this matrix, but I know other question, since this matrix is not positive definite then does not mean that eigenvalues is positive, it is not even negative definite because for $e_3^TAe_3=a_{33}>0$, so does not mean that eigenvalue is negative, but only on what I doubt is that $\lambda>\omega$, but in spectrum it put first $\lambda$ then $\omega$ maybe that does not mean anything but, in some set you put first smaller number then bigger,bit since $A=A^T$ using spectral theorema $A=Q\Lambda Q^T$, there exist orthogonal eigenvectors so it mean that matrix A have three linear independent column, so eigenvalues is not zero, do you see more information?

• Well, there is no such thing as "orthogonal eigenvalues," and a matrix may be orthogonally diagonalizable without having three independent columns. (But not in this case). However, given your information you can calculate $A$. What is the matrix $\Lambda$ -- precisely? Now to diagonalize $A$ orthogonally you need to produce an orthogonal basis out of your given three eigenvectors, but that is a routine process. Commented Sep 3, 2018 at 18:50
• Sorry orthogonal eigenvectors Commented Sep 3, 2018 at 19:13
• @JohnBrevik but what i can tell about signs of eigenvalue Commented Sep 3, 2018 at 19:15
• If a matrix is similar to a diagonal one, do you know how to read off the eigenvalues? Commented Sep 3, 2018 at 21:11

Clearly $n=3$ because $A$ diagonalizes so the eigenvectors span the space.
We have $$N(A - \lambda I) = \operatorname{span}\left\{\pmatrix{1 \\ 1 \\ 0}, \pmatrix{1 \\ 1 \\ 1}\right\} = \operatorname{span}\left\{\frac1{\sqrt2}\pmatrix{1 \\ 1 \\ 0}, \pmatrix{0 \\ 0 \\ 1}\right\}$$ $$N(A - \omega I) = \operatorname{span}\left\{\pmatrix{1 \\ -1 \\ 0}\right\} = \operatorname{span}\left\{\frac1{\sqrt2}\pmatrix{1 \\ -1 \\ 0}\right\}$$
so $A$ diagonalizes in the orthonormal basis $\left\{\frac1{\sqrt2}\pmatrix{1 \\ 1 \\ 0}, \pmatrix{0 \\ 0 \\ 1}, \frac1{\sqrt2}\pmatrix{1 \\ -1 \\ 0}\right\}$ to $D = \begin{bmatrix}\lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \omega \end{bmatrix}$.
Therefore if we set $P = \begin{bmatrix} \frac1{\sqrt2} & 0 & \frac1{\sqrt2} \\ \frac1{\sqrt2} & 0 & -\frac1{\sqrt2} \\ 0 & 1 & 0\end{bmatrix}$ we have
$$\begin{bmatrix} -1 & * & * \\ * & -1 & * \\ * & * & 1 \\ \end{bmatrix} = A = PDP^T = \begin{bmatrix} \frac{\lambda}{2}+\frac{\omega}{2} & \frac{\lambda}{2}-\frac{\omega}{2} & 0 \\ \frac{\lambda}{2}-\frac{\omega}{2} & \frac{\lambda}{2}+\frac{\omega}{2} & 0 \\ 0 & 0 & \lambda \\ \end{bmatrix}$$
If follows that $\lambda = 1$ and $\omega = -3$.