I have this 4×4 matrix:
$$A=
\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
$$
I read that it's easy to calculate it by converting the matrix to upper diagonal. I tried that using line operations but I couldn't make it upper diagonal. Is this the best/easiest method? If so can anyone help me with the process?
Then I have to calculate the eigenvalues and eigenvectors. Any suggestion on how to find them? Do I have to calculate the det(A -λI) to find the characteristic equation? Is there an easy way to find it for such a matrix like A?
Any help will be much appreciated.
Thanks in advance!
Best Answer
As suggested in the comments, Gauss elimination is usually the way to go, and the fastest in this case, too:
$$\det A= \det\begin{pmatrix} 2 & 3 & 1 & 0 \\ 4 & -2 & 0 & -3\\ 8 & -1 & 2 & 1\\ 1 & 0 & 3 & 4\\ \end{pmatrix} = \det\begin{pmatrix} 0 & 3 & -5 & -8 \\ 0 & -2 & -12 & -19\\ 0 & -1 & -22 & -31\\ 1 & 0 & 3 & 4\\ \end{pmatrix} = (-1)^{4+1}\cdot 1\cdot\det\begin{pmatrix} 3 & -5 & -8 \\ -2 & -12 & -19\\ -1 & -22 & -31\\ \end{pmatrix} = -\det\begin{pmatrix} 0 & -71 & -101 \\ 0 & 32 & 43\\ -1 & -22 & -31\\ \end{pmatrix} = -1\cdot(-1)^{3+1}\cdot(-1)\cdot\det\begin{pmatrix} -71 & -101 \\ 32 & 43\\ \end{pmatrix} = (-71)\cdot 43-(-101)\cdot 32=179 $$
(Wolfram Alpha-verified result; I never could remember the 3x3-formula, so I don't use it)
If you absolutely want an upper diagonal matrix, you can do this, but it's only a restriction of the normal algorithm:
$$\det A= \det\begin{pmatrix} 2 & 3 & 1 & 0 \\ 4 & -2 & 0 & -3\\ 8 & -1 & 2 & 1\\ 1 & 0 & 3 & 4\\ \end{pmatrix} = \det\begin{pmatrix} 2 & 3 & 1 & 0 \\ 0 & -8 & -2 & -3\\ 0 & -13 & -2 & 1\\ 0 & -\frac12 & \frac52 & 4\\ \end{pmatrix} = \det\begin{pmatrix} 2 & 3 & 1 & 0 \\ 0 & -8 & -2 & -3\\ 0 & 0 & ? & ?\\ 0 & 0 & ? & ?\\ \end{pmatrix} = \det\begin{pmatrix} 2 & 3 & 1 & 0 \\ 0 & -8 & -2 & -3\\ 0 & 0 & ? & ?\\ 0 & 0 & 0 & ?\\ \end{pmatrix} $$
(I'm too lazy to calculate the $?$ now, just continue with the Gaussian Elimination. The determinant will then be the product of the entries on the diagonal.)
For the eigenvalues: yes, you have to calculate the characteristic polynomial.