# Determinant of symmetric matrix $(A-\lambda I)$

If we have a matrix $(A-\lambda I)$ which is:

$\left( \begin{array}{ccc} 1-\lambda & -1 & 2 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right)$

Then it's determinant can be written as : $(-1)^n(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$. In this case what will $\lambda_1$,$\lambda_2$ and $\lambda_3$ be equal to? And how do we determine it's value given that the matrix is symmetric?

• $\lambda_1$, $\lambda_2$ and $\lambda_3$ are the three eigenvalues of the matrix. You have to compute the determinant of the matrix you have given, it is a function of $\lambda$, a polynomial of degree three, so you factor it into linear factors, to get the equivalence. – Gregory Grant May 17 '15 at 12:18
• Solve the equation $\det(A-\lambda I)=0$. – Samrat Mukhopadhyay May 17 '15 at 12:23
• @GregoryGrant yes we have to compute it via a polynomial but that is a much longer method. This method suggested that we could easily determine the eigenvalues by merely looking at the matrix but I don't know how to use this method. – Paradox 101 May 17 '15 at 12:27
• Ah probably because it is symmetric. Try to look up what is known about the eigenvalues of symmetric matrices. – Gregory Grant May 17 '15 at 12:28
• I looked it up and I don't think there's a real easy way to just read them off from the matrix. If the matrix was upper triangular then yes, but not in the form given. Unless I'm missing something. – Gregory Grant May 17 '15 at 12:32

You could simply expand the determinant, which is not that much work for a $3\times 3$.
Or you could set $\lambda=0$ and guess the eigenvalues of the resulting matrix, for example $(1\,{-1}\,0)^T$ and $(1\,1\,1)^T$ are obvious eigenvectors of eigenvalue $2$. The fact that the matrix is symmetric guarantees us that all eigenvalues will be real.
$\left( \begin{array}{ccc} 1-\lambda & -1 & 2 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right)$ = $\left( \begin{array}{ccc} 2-\lambda & -2+\lambda & 0 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) =(-2+\lambda) \left( \begin{array}{ccc} -1 & 1 & 0 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) = (-2+\lambda) \left( \begin{array}{ccc} -1 & 0 & 0 \\ -1 & -\lambda & 2 \\ 2 & 4 & 2-\lambda \\ \end{array} \right)= (-1)(-2+\lambda) \left( \begin{array}{ccc} -\lambda & 2 \\ 4 & 2-\lambda \\ \end{array} \right) = (-1)(-2+\lambda) \left( \begin{array}{ccc} 4 -\lambda & 4 -\lambda \\ 4 & 2-\lambda \\ \end{array} \right)= (-2+\lambda)(\lambda-4) \left( \begin{array}{ccc} 1 & 1\\ 4 & 2-\lambda \\ \end{array} \right) = (-2+\lambda)(\lambda-4) \left( \begin{array}{ccc} 0 & 1\\ 2+\lambda & 2-\lambda \\ \end{array} \right)=(-2+\lambda)(\lambda-4)(2-\lambda)=(-1)(\lambda-2)(\lambda-4)(\lambda-2)$ Therefore, $\lambda_1=2,\lambda_2=2,\lambda_3=4$.
HINT : Solve $\det(A-\lambda I)=0$, with: $$\begin{vmatrix}1-\lambda&-1&2\\-1&1-\lambda&2\\2&2&2-\lambda\end{vmatrix}=(1-\lambda)\begin{vmatrix}1-\lambda&2\\2&2-\lambda\end{vmatrix}-(-1)\begin{vmatrix}-1&2\\2&2-\lambda\end{vmatrix}+2\begin{vmatrix}-1&1-\lambda\\2&2\end{vmatrix}$$