Let $ M = \begin{bmatrix} 1 &1 & 3 \\[0.3em] 1 & 5 & 1 \\[0.3em] 3 & 1 & 1 \end{bmatrix} $ a matrix. I am stuck on solving this question from my book. Find eigenvalues, characteristic polynomial and diagonalize this matrix.
Characteristic polynomial of a $3\times 3$ matrix
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0$Charlie. My question is not as easy as it seems. Someone (not you) hurried up and gave me negative vote. For you +1 from me. – 2012-11-10
1 Answers
You have that the characteristic polynomial is given by: $\det(M-\lambda I)$, so we have: $M = \begin{bmatrix} 1-\lambda &1 & 3 \\[0.3em] 1 & 5-\lambda & 1 \\[0.3em] 3 & 1 & 1-\lambda \end{bmatrix}$
Find the determinant. $\begin{align*} \det(M-\lambda I) &=(1-\lambda)(5-\lambda)(1-\lambda)+3+3-9(5-\lambda)-(1-\lambda)-(1-\lambda) \\ &=(1-\lambda)(5-\lambda)(1-\lambda)-2(1-\lambda)+6. \end{align*}$
This is the characteristic polynomial. To find the eingenvalues, make it equal to zero then find the roots of such polynomial.Then you're done. Now, for the diagonal matrix :$M$ is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix $P$ such that $P^{−1}MP$ is a diagonal matrix. You have to find the eigenvector related to the eigenvalues you found.Suppose they are $\lambda _1,\lambda _2, \lambda _3.$ They will appear in the diagonal of the diagonal matrix you look for.The proccess to find such matrix is the following:
- You find the eigenvectors related to the eigenvalues you found;
Suppose they are:
$v_1=[a_1, a_2, a_3],\ v_2=[b_1, b_2, b_3],\ v_3=[c_1, c_2,c_3].$
Now let $P$ be the matrix of these eigenvectors as its columns:
$\begin{bmatrix} a_1 &b_1 & c_1 \\[0.3em] a_2 & b_2 & c_2 \\[0.3em] a_3 & b_3 & c_3 \end{bmatrix}$
Now you will have to find $P^{-1}$.Once you found it you have to compute:
$P^{-1}MP$, which is equal to your diagonal matrix with the eigenvalues in the main diagonal.It's a good exercise to check this way.
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0edited the answer, @AdiDani – 2012-11-10