Here is the argument for the standard procedure that computes the eigenvalues of a matrix. The equation
![{\displaystyle Av=\lambda v}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9f7a927ca654271e6607c7f9cded57fea70c1d71)
is equivalent to
![{\displaystyle (A-\lambda \mathrm {Id} )v=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7f069cc9b3081df314c6ea01425e489a7333bf1e)
This latter equation has a non-zero solution if and only if the matrix
![{\displaystyle A-\lambda \mathrm {Id} }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/252f09a417815373028bc4c87a6074824d776a40)
is
not invertible, which is equivalent to
![{\displaystyle \det(A-\lambda \mathrm {Id} )=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/93f4d47e2775509de9baa727d7d7c63e6e4fb87e)
. Therefore, we solve this last equation, since it is an equation of numbers (as opposed to an equation of vectors). We find the determinant to be
![{\displaystyle {\begin{aligned}\det(A-\lambda \mathrm {Id} )&=\det \left[{\begin{array}{ccc}-\lambda &5&-5\\1&4-\lambda &1\\0&0&5-\lambda \end{array}}\right]\\&=(5-\lambda )\det \left[{\begin{array}{cc}-\lambda &5\\1&4-\lambda \end{array}}\right]\\&=(5-\lambda )(-\lambda (4-\lambda )-5)=(5-\lambda )(\lambda ^{2}-4\lambda -5)=(5-\lambda )(\lambda -5)(\lambda +1).\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9fd713d9054ad3121825e4e353159d64d6966642)
From the factorization above, we see that there are two eigenvalues of
: -1 and 5, and 5 is called a repeated eigenvalue, since it shows up more than once in the factorization of
.