# Help with finding eigenvectors and polynomial factorization

I have a real and invertible matrix

$$A = \left( \begin{matrix} 5 & 4 \\ 4&5 \end{matrix} \right)$$

and I need to find the eigenvectors.

The characteristic polynomial is $(5-\lambda)^2 - 16$. According to Wolfram Alpha this factors to $(1-\lambda)(9-\lambda)$ (How does one get there, are there any good tutorials on polynomial factorization?), so the eigenvalues are $1$ and $9$.

Calculating the eigenvector to the eigenvalue $1$ given by $(A-I)\vec x = \vec 0$.

I get that $\vec x$ is the null vector $=\left( \begin{array}{c} 0\\ 0\\ \end{array} \right)$(can that be right?).

For the second eigenvector $(A-9I)\vec x = 0$ I get

$$x_{1} = 4x_{2}$$

$$-x_{2} = 5x_{1}$$

Is this right, and how can I write the second eigenvector with this information??

• @ AWerheim: is there an other definition for eigenvector?, see here: en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors wher it's defined by a non zero vector such that ... – Mohamed Jun 6 '14 at 18:46
• @Mohamed You are right- $\vec 0$ is not an eigenvector. – beep-boop Jun 6 '14 at 19:07
• Oops, my mistake. Thanks for the correction! – Alex Wertheim Jun 6 '14 at 20:00

## 6 Answers

Everyone else has focussed on the polynomial side of things, so I'll focus on the linear algebra part.

Firstly, the null vector, $\vec x=\vec 0$, is not an eigenvector of any matrix.

Why? By definition, $\vec x$ is an eigenvector of $A \in \mathbb{M}_{m,n}$ iff $\vec x$ is a nonzero vector such that $A \vec x = \lambda \vec x$ (for some scalar $\lambda$).

Also, I note that you've asked "is this the eigenvector for this eigenvalue?"

For any eigenvalue, there is no unique eigenvector. Indeed, if $\left( \begin{array}{c} 1\\ 1\\ \end{array} \right)$ is an eigenvector of $A$, then so is $\left( \begin{array}{c} 2\\ 2\\ \end{array} \right)$, $\left( \begin{array}{c} 3\\ 3\\ \end{array} \right)$ or any scalar multiple of $\left( \begin{array}{c} 1\\ 1\\ \end{array} \right)$.

• You can add than the eigenvectors associated to the same eigenvalue form a vector field ;-) – Ant Jun 6 '14 at 21:58

Well $(5-\lambda)^2-4^2$ is a difference of squares, but you really need to learn factoring.

$$\begin{pmatrix} 4&4\\4&4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$ gives $x=-y$ so $\begin{pmatrix}1\\-1\\\end{pmatrix}$ is one eigenvector, and $$\begin{pmatrix} -4&4\\4&-4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$ gives $x=y$ so $\begin{pmatrix}1\\1\\\end{pmatrix}$ is the other eigenvector,

The characteristic polynomial is, as you correctly state, $(5 - \lambda)^2 - 16$. Expanding this we get $25 - 10\lambda + \lambda^2 - 16$ i.e. \begin{eqnarray} p(\lambda) = \lambda^2 - 10 \lambda + 9 \end{eqnarray} Now, this is a quadratic polynomial in $\lambda$, so you can use the quadratic formula to find the values for $\lambda$ such that $p(\lambda) = 0$. But, in this case it is "easy" (when you get the hang of it) to see directly from the coefficients what $\lambda$ should be, i.e. $\lambda$ should be either $9$ or $1$. To see this, consider the product \begin{eqnarray} (\lambda - a)(\lambda - b) = \lambda^2 - (a + b)\lambda + ab \end{eqnarray}

now which $a$ and $b$ should we choose to get the coefficients in $p(\lambda)$? Well $ab$ should be $9$ and $a + b$ should be $10$. So $a = 9$ and $b = 1$ works.

To find the eigenvector of $\lambda = 1$ we must find $x$ such that \begin{eqnarray} (A - 1I)x = 0 \end{eqnarray} So lets write out what $A - 1I$ is, namely the matrix \begin{eqnarray} A - 1I = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} \end{eqnarray} Thus let $x= (a,b)$ be such that $(A - 1I)x = 0$ then \begin{eqnarray} x = 4 \begin{pmatrix} a + b \\ a + b \end{pmatrix} \end{eqnarray} so for this vector to be $0$ we must have that $b = - a$, so $x$ is a multiple of the vector $(1,-1)$.

Now, try to do the same for $\lambda = 9$.

For the polynomial factorization, expand it all then factor.

\begin{align}&(5-\lambda)^2-16\\=&25-10\lambda+\lambda^2-16\\=&9-10\lambda+\lambda^2\\=&(1-\lambda)(9-\lambda)\end{align}

The two vectors are not correct.

The first: Since $A-I=4J$ where $J=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ you have: $$(A-I)x=0 \iff x_1+x_2=0$$ $\vec0$ is never an egen vector. An egen vector is by definition one vector $x$ such that $x \neq 0$ and $(A-I)x=0$

The second: Since $A-9I=4K$ where $K=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}$:$$(A-9I)x=0 \iff x_1-x_2=0$$

Observe that the sum of each row is same so $A\begin{pmatrix}1\\1\end{pmatrix}=9\begin{pmatrix}1\\1\end{pmatrix}$. Hence $\lambda=9$ is an eigen value and $\begin{pmatrix}1\\1\end{pmatrix}$ is an eigen vector. To get the other eigen value, you may use determinant is product of eigen values (or trace is sum of eigen values). Hence the other eigen value is $\lambda=1$.