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Consider the following matrix A:

$$ A= \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{pmatrix} $$

I have to find the characteristic polynomial $P_A(\lambda)$ using the following approach:

$$P_A(\lambda)=\det(A-\lambda I)$$

I worked out the first part:

$$\begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{vmatrix} $$

But then I get stuck calculating the determinant with all those $\lambda$ floating around.

Help? :( The answer is supposed to be $P_A(\lambda)=-(\lambda-1)(\lambda+2)^2$

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You could use properties of determinants to avoid having to factor a cubic afterwards; for example:

  • subtract the last column from the first two;
  • add the first two rows to the third:

$$\begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{vmatrix}=\begin{vmatrix} -2-\lambda & 0 & 1 \\ 0 & -2-\lambda & 1 \\ 2+\lambda & 2+\lambda & -1-\lambda \\ \end{vmatrix}=\begin{vmatrix} -2-\lambda & 0 & 1 \\ 0 & -2-\lambda & 1 \\ 0 & 0 & 1-\lambda \\ \end{vmatrix}$$ This is the determinant of a diagonal matrix, so it is the product of the diagonal elements: $$\left( -2-\lambda \right)^2\left( 1-\lambda \right) \color{blue}{ = 0 \iff \lambda = -2 \;\vee\; \;\lambda = 1}$$

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So the characteristic polynomial is: $$ p(\lambda) = \begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{vmatrix}$$ Notice that if you put $\lambda=-2$ in the polynomial you get a root (since all the rows are the same and therefore dependent and the determinant is 0). So the first eigenvalue is $\lambda_1 = -2$.

Secondly, notice that the sum of each row is constant in this matrix is constant (1). This means that the vector \begin{pmatrix}1\\1\\1\end{pmatrix} is an eigenvector (it sums each row), and: $$ \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{pmatrix} \begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}1\\1\\1\end{pmatrix} = 1\begin{pmatrix}1\\1\\1\end{pmatrix} $$ So we get from here that the second eigenvalue is $\lambda_2 = 1$.

Now, the trace of the matrix is the sum of the eigenvalues. So we can deduce: $\lambda_1 + \lambda_2 + \lambda_3 = -3$ meaning $-1 + \lambda_3 = -3$ and therefore $\lambda_3 = -2$. So the characteristic polynomial is: $$p(\lambda) = (\lambda+2)^2(\lambda - 1)$$

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Since you're using a $3\times3$ matrix you can use this system: enter image description here

It follows directly from the definition of the determinant which is quite the hairy function so bear with me.

There are generally 2 ways to introduce determinants, one is by defining the formula and stating the properties, the other is by stating the properties and deriving the formula. I prefer the latter, but entire chapters have been dedicated to this, so we'll be skipping most steps and I'll just introduce the formula. Hoffman and Kunze's Linear Algebra has a great chapter on determinants.

First, let's call $\sigma$ a permutation, and we'll define $sgn(\sigma)=(-1)^z$ where $z$ stands for the amount of "switches" that have to happen in order to arrive at a permutation. Example:

If we consider a series $(1,2,3)$, then if we change this to $(1,3,2), \quad sgn(\sigma)=-1$. Change it to $(3,1,2)$ and $sgn(\sigma)=1$.

Now the determinant is defined as:

$$det(A)=\sum_\sigma(sgn(\sigma))A_{(1,\sigma1)},...,A_{(n,\sigma n)}$$ That is to say the sum of all permutations where the horizontal positions of the terms within our product get determined by the specific permutation.

Let us consider a $3\times3$ matrix. Then all possible permutations are: $$(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)$$ (It is no coincidence that this coincides with $3!$)

Can you use this definition to show that $$det \begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}=0$$ ?

This method, though clunky, will really deepen your understanding of what a determinant is and how it works. Cofactor expansion will generally be easier when it comes to a quick calculation, but the reason it works can be found in this formula.

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Let's compute step by step expanding along the first column

$$\begin{align}{\begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{vmatrix}}&=-(1+\lambda)((1+\lambda)^2-1)-(-(1+\lambda)-1)+(1+(1+\lambda)\\ &=-(1+\lambda)\lambda(\lambda+2)+2(\lambda+2)\\ &=(\lambda+2)(-\lambda^2-\lambda+2)\\ &=(\lambda+2)^2(\lambda-1)\end{align} $$

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You can also try to "guess" the eigenvalues, it is in this precise case fairly easy. Indeed the sum of all lines is equal to $1$, meaning that $(1,1,1)$ is an eigen-vector for the eigenvalue $1$. Besides, you can see that the rank of matrix $A+2I$ is $1$, so $-2$ is another eigenvalue of $A$ (and it has multiplicity $2$). Since the characteristic polynomial has degree $3$ here, it is $-(\lambda-1)(\lambda+2)^2$.

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Hint:

use the Sarrus' rule. It is really easy in this case.

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When you use the 3x3 formula, you get: (−1−λ)^3+1+1-(-1-λ)-(-1-λ)-(-1-λ).

Which is simplified to: -λ^3+3λ+2.

Factor it and you get: −(λ−1)(λ+2)^2

λ1 = 1 with multiplicity 1 and λ2 = -2 with multiplicity 2

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