How would I factor this equation that was calculated as the determinant of a matrix?

Question

So I have a determinant of a matrix given to be (p_0 - λ)(p_1 - λ) - (1 - p_0)(1 - p_1) which I simplified to λ^2 - λp_0 - λp_1 + p_1 + p_0 - 1 Now I need to factor this equation to determine the eigenvalues. I factored to (λ + p_1)(λ - p_0)= 1 and then λ=p_0 and λ=-p_1 Does this look correct?

Edit: P_0 and p_1 are constants, $$ det(A)= \begin{bmatrix} p_{0} - \lambda & 1 - p_{1} \\ 1 - p_{0} & p_{1} - \lambda \end{bmatrix} $$

Answer 1

It looks like your Determinantal equation is $\left|\begin{array}[cc] pp_0-\lambda & 1-p_0\\1-p_1 & p_1-\lambda\end{array}\right|$. If so the charactersitic polynomial will be

$(p_0-\lambda)(p_1-\lambda)-(1-p_0)(1-p_1)$

and if you foil(first-outer-inner-last) both products and collect terms in powers of $\lambda$, you get

$p_0p_1-\lambda p_0-\lambda p_1+\lambda^2-1+p_1+p_0-p_0p_1$

Notice the cancellation of the $p_0p_1$ term. This gives

$\lambda^2-(p_0+p_1)\lambda+p_0+p_1-1$

Answer 2

You will get a quadratic equation in $\lambda$: $$\lambda^2-\lambda(p_0+p_1)+p_0+p_1-1=0$$ We get $$\lambda_1=1$$ or $$\Lambda_2=p_0+p_1-1$$ so the factirzation is given by $$(\lambda-1)(\lambda-p_0-p_1+1)$$

Answer 3

You can find the eigenvalues of this matrix without directly involving the characteristic polynomial. The original matrix is $$P=\begin{bmatrix}p_0&1-p_1\\1-p_0&p_1\end{bmatrix}.$$ Since the determinant of a matrix and its transpose are equal, a matrix and its transpose have the same eigenvalues. Observe that the sum of each of the columns of this matrix is $1$. Summing the columns is equivalent to left-multiplying by $(1,1)$, therefore $1$ is an eigenvalue of $P^T$ and hence also of $P$. (This is a general property of stochastic matrices: they always have $1$ as an eigenvalue.)

You can always find the last eigenvalue of a matrix “for free” because the trace of a matrix is equal to the sum of its eigenvalues. Thus, we have $1+\lambda=p_0+p_1$, therefore the second eigenvalue is $p_0+p_1-1$.

With a bit more experience at this sort of thing, you might also have spotted the second eigenvalue by looking at the difference of the two columns of $P$, namely $(p_0+p_1-1,-(p_0+p_1-1))^T$. This subtraction is equivalent to right-multiplying by $(1,-1)^T$, so that’s an eigenvector of $P$ with eigenvalue $p_0+p_1-1$, just as we got from the trace, above.