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.

  • 2
    $\begingroup$ There is no error. $\det$ is not the product of the pivots. $\endgroup$
    – DHMO
    Apr 16, 2017 at 22:10
  • 1
    $\begingroup$ $\det$ is $(2-\lambda)^2 - (-1)^2 = (2-\lambda -1)(2-\lambda +1)$ $\endgroup$ Apr 16, 2017 at 22:15
  • $\begingroup$ I see, I need to use $\det=ad-bc$. Thanks $\endgroup$
    – clicky
    Apr 16, 2017 at 22:19
  • $\begingroup$ However product of pivots can be a kind of estimate of determinant. How reliable depends on the Geršgorin disc radii. $\endgroup$ Apr 16, 2017 at 22:22

2 Answers 2


The determinant of

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

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


Chenyang is right in his answer about the determinant. However the product of pivots are the product of the midpoints of it's Geršgorin discs. Depending on how large diagonal elements are compared to off-diagonal elements the matrix is this can give a more or less accurate estimate of the determinant.


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