# Something with eigenvalue and eigenvectors.

Suppose the matrix $$A=\begin{pmatrix}0&0&1\\x&1&y\\1&0&0\end{pmatrix}$$ has three independent eigenvectors. Prove that $x+y=0$

Becasue $A$ has three independent eigenvectors, and its rank is three, so it must have three different eigenvalues right? But why if I calculate the eigenvalues of $A$, I got $1,1,-1$. There are two repeated eigenvalues why? Or I calculate the eigenvalues wrongly? And how to prove that $x+y=0$

• The same eigenvalue can correspond with several orthogonal eigenvectors. – Peter Jan 16 '14 at 21:42
• Think of an identity matrix. It has only $1$ as eigenvalue but every vector in the space is an eigenvector. – xavierm02 Jan 16 '14 at 21:47
• @Mico but $x+y=2$ – user10444 Jan 16 '14 at 21:52
• Oh, I see. If there are n different eigenvalues, then there are n independent eigenvectors, but the inverse is not correct. Thanks. – user121819 Jan 16 '14 at 21:53
• @user10444 - Sorry, I had overlooked the main concern seems to be with the fact that the matrix has a repeated (non-zero) eigenvalue. The eigenvectors associated with repeated eigenvalue are linearly independent (and orthogonal to the eigenvector associated with the distinct, third eigenvalue), but they needn't be orthogonal to each other. – Mico Jan 16 '14 at 22:09

The characteristic polynomial:

$$p_A(t)=\det(tI-A)=t^2(t-1)-(t-1)=(t-1)^2(t+1)$$

Let's see what's the dimension of the eigenspace corresponding to $\;t=1\;$ :

$$\begin{cases}\;\;\;\;\,x_1-x_3&=0\\-xx_1-yx_3&=0\\\;\,-x_1+x_3&=0\end{cases}$$

Clearly equations $\;I,III\;$ are linearly dependent, and since the matrix is diagonalizable iff there's only one single, independient equation (i.e., iff the eigenspace's dimension is $\;2\;$) (why?), we must have

$$I\;\;:\;\; x_1=x_3\implies II\;\;:\;\;x_1(-x-y)=0\implies x+y=0$$



[ 0 0 1 ] [ u ] [ w ]

[ x 1 y ] [ v ] = [ xu + v + yw ]

[ 1 0 0 ] [ w ] [ u ]

u = w

(x+y)u = 0

So, $u\ne 0$, hence x+y=0.