Eigenvalues from determinant calculation

Going through Dr Strang's textbook on Linear Algebra, I am trying to understand one of the sample questions to calculate the eigenvalues of a matrix. Using

$$\det(A-\lambda I)=0$$

with $\det(A)$ as the product of the pivots. Therefore for the given matrix A

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

the pivots would be

$$\begin{bmatrix} 2-\lambda&-1\\ -1&2-\lambda \end{bmatrix} = \lambda^2-4\lambda +4 = (2-\lambda)(2-\lambda)$$

giving a single eigenvalue of $2$.

However the book says $\lambda^2-4\lambda +3$ giving eigenvalues of $1$ and $3$. I have checked a later edition of the textbook which has the same content and do not find this listed in any errata online. Therefore I am not sure if my understanding is incorrect or if this is really is an error.

• There is no error. $\det$ is not the product of the pivots. – DHMO Apr 16 '17 at 22:10
• $\det$ is $(2-\lambda)^2 - (-1)^2 = (2-\lambda -1)(2-\lambda +1)$ – mathreadler Apr 16 '17 at 22:15
• I see, I need to use $\det=ad-bc$. Thanks – clicky Apr 16 '17 at 22:19
• However product of pivots can be a kind of estimate of determinant. How reliable depends on the Geršgorin disc radii. – mathreadler Apr 16 '17 at 22:22

is $\lambda^2-4\lambda +4 -(-1)(-1) = (\lambda-1)(\lambda-3)$, so your book is correct.