# Trouble finding eigenvector for corresponding eigenvalue -1

For a matrix of coefficients that is:

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

I find the eigenvalues to be -1 and 1.

Eigenvalue 1 gives an eigenvector of

$$\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix}$$

but I'm having trouble finding the eigenvector for the corresponding eigenvalue -1, where I seem to get

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

Wolfram Alpha is saying it should be

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

Any ideas on how it found that?

Thank you.

EDIT - the eigenvector corresponding to an eigenvalue of 1 was the wrong way round.

• If you show your work leading up to $(0,0)$ we might be able to help you. – B. Goddard Apr 17 '17 at 13:52
• Just use the definition of eigenvalue and write down the equation system you are supposed to solve if $\lambda =-1$. – mathreadler Apr 17 '17 at 13:54
• And, by the way, $[{}^1_2]$ is not an eigenvector; multiplied by the matrix it gives $[{}^{\;1}_{-1}]$. So perhaps you should also show how you got that. – hmakholm left over Monica Apr 17 '17 at 13:58
• The second column of the matrix is $-1\cdot(0,1)^T$, so... – amd Apr 17 '17 at 17:57

You need to solve the equation:

$Av = -v$

Where $A$ is your matrix and $v$ is the eigenvector you're looking for. So (using $x$ for the first coordinate and $y$ for the second):

$1x + 0y = -x$

$1x - 1y = -y$

From the first expression we have that $x = 0$. So:

$-y = -y$

That's a tautology, so any $y$ will satisfy the equation. So we may as well pick the normalized one, which has $y = 1$.

As an aside, the $0$ vector will always satisfy the eigenvalue equation, but it's excluded from the definition of eigenvectors.

• Thank you, the idea that any y will satisfy the equation was what I found online, but I never came across that before so was unsure if it was valid to write. – freja Apr 18 '17 at 19:05

You are on the right track. The matrix $$A= \left( \begin{array}{cc} 1 & 0 \\ 1 & -1 \\ \end{array} \right)$$ has characteristic function $$p(\lambda) = \lambda^{2} - 1$$ whose roots are the eigenvalues $\lambda_{\pm} = \pm 1$.

For the eigenvalue $\lambda_{-} =-1$, $$A v_{-} = \left( \begin{array}{rr} 1 & 0 \\ 1 & -1 \\ \end{array} \right) \left( \begin{array}{cc} 0 \\ 1 \\ \end{array} \right) = \left( \begin{array}{r} 0 \\ -1 \\ \end{array} \right) = - \left( \begin{array}{r} 0 \\ 1 \\ \end{array} \right) = \lambda_{-} v_{-}$$