How to say if Eigenvectors of A are orthogonal or not? without computing eigenvectors

I am give matrix : $$A=\begin{bmatrix} 0&-1 & 2 \\ -1 & -1 & 1 \\ 2 & 1 &0 \end{bmatrix}$$

• 1. Without finding the eigenvalues and eigenvectors, determine whether the eigenvectors are orthogonal or not. Justify your answer
• 2. Express matrix $A$ in the form $A=UDU^T$ where $D$ is a diagonal matrix and $U$ is an orthogonal matrix. What are $U$ and $D$ ?

• I can check if a vectors are orthogonal or not, by dot product = 0
• I know that if $B^T=B^{-1}$ so that $B$ be can be said orthogonal, and $B^TB=I$
• I also can find the eigenvalues and eigenvectors, but the question asks without finding them..

How to check whether eigenvalues are orthogonal or not without finding? and how to express $A=UDU^T$?

• Do you know the Spectral theorem? – Huy Jan 9 '15 at 10:40
• The matrix is symmetric. – Git Gud Jan 9 '15 at 10:42
• If the eigenvalues are not distinct then some of the eigenvectors are not orthogonal. However, there is an orthogonal basis for the vector space. – Ofir Schnabel Jan 9 '15 at 10:44
• Spectral theorem is like $D$ diagonal matrix is just the main diagonal elements are $\lambda$s ? @Huy If so how i can find the orthogonal matrix $U$? – Andy Jan 9 '15 at 10:46
• @Andy Do you know how to diagonalize a matrix? Then do it. When you find a basis of eigenvectors, simply apply G.S. to it and you'll get your $U$. – Git Gud Jan 9 '15 at 10:53

Let $v$ be an eigenvector correspond to $\lambda$ and let $w$ be an eigenvector correspond to $\delta$. Then $$\lambda \langle v,w \rangle= \langle \lambda v,w \rangle=\langle Av,w \rangle=\langle v,A^tw \rangle=\langle v,\delta w \rangle= \delta \langle v,w \rangle\Rightarrow (\lambda-\delta)\langle v,w \rangle \Rightarrow \langle v,w \rangle=0$$