# Find the eigen values and vectors of the matrix

Find the eigenvalues and eigenvectors of the matrix $A$:

$$A = \begin{bmatrix}-2 & 2 & -3\\2 & 1 & -6\\-1 & -2 & 0\\\end{bmatrix}.$$

$$A - \lambda I = \begin{bmatrix}-2-\lambda & 2 & -3\\2 & 1-\lambda & -6\\-1 & -2 & -\lambda\\\end{bmatrix}\\ \det(A-\lambda I)=(-2-\lambda)[(-1-\lambda)(-\lambda)-12]-2(-2\lambda-6)-3(-4-(-1+\lambda))\\ =(-2-\lambda)[(-\lambda+\lambda^2)-12]+4\lambda+12+9+3\lambda\\ =2\lambda-2\lambda^2+\lambda^2-\lambda^3+24+12\lambda+4\lambda+12+9+3\lambda\\ =-\lambda^3-\lambda^2+21\lambda+45$$

After factoring....... $\lambda=5,\lambda=-3$

When $\lambda=5$,

$$\begin{bmatrix}-7 & 2 & -3\\2 & -4 & -6\\-1 & -2 & -5\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

$$-7x+2y-3z=0,\\ 2x-4y-6z=0,\\ -x-2y-5z=0.$$

I am stuck here, i have no idea what to do next. I hope someone can help Please and thanks

• Row reduce the matrix to find the solution set. Have you studied row reduction? – user139388 Apr 26 '14 at 9:02
• The eigenvectors are not unique (i.e. there are an infinite number of eigenvectors, scalar multiples of each other, for each eigenvalue). You have a $3\times3$ matrix which means you hope to find three eigenvalues. If you only found two then this means there is a multiplicity which means one of the eigenvalues has two different (orthogonal) eigenvectors. – Jared Apr 26 '14 at 9:08
• The multiplicity here is with $\lambda = -3$, so $\lambda = 5$ shouldn't be a real issue (it's the other that's more difficult). – Jared Apr 26 '14 at 9:12

$$\pmatrix{2& -7& -3\\ -4& 2& -6\\ -2&-1&-5} \rightarrow \pmatrix{2& -7& -3\\ 0& -12& -12\\ 0&-8&-8}$$
$$-12x - 12z = 0 \rightarrow z = -x \\ \text{using the first equation, }2y - 7x - 3z = 0 \rightarrow y = \frac{7x + 3z}{2} = \frac{7x - 3x}{2} = 2x$$
This gives the eigenvector as $\langle x, 2x, -x\rangle \leadsto \langle1, 2, -1\rangle$, or as a unit vector: $\left\langle \pm\sqrt{\frac{1}{6}}, \pm \sqrt{\frac{2}{3}}, \mp \sqrt{\frac{1}{6}}\right\rangle$