# Find the eigenvalues of the following matrix where $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$ as well as the associated eigenvector

Find the eigenvalues of the following matrix $$\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix}$$

The eigenvalues are $$\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$$ . Find the associated eigenvector for $$\lambda_1, \lambda_2, \lambda_3, \lambda_4$$

So should I start out by doing $$A - I_m \lambda$$ where $$I_m$$ is the identity matrix? So I would have $$\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix} - \begin{bmatrix}\lambda&0&0&0\\0&\lambda&0&0\\0&0&\lambda&0\\0&0&0&\lambda\end{bmatrix} = \begin{bmatrix}-12-\lambda&-6&0&0\\8&2-\lambda&0&0\\0&0&-14-\lambda&-9\\0&0&42&25-\lambda\end{bmatrix}$$ then would it be correct to then put the matrix in either Upper Triangular Form or Lower Triangular Form so that I can then take the product of the diagonal which is the determinant and then solve for the eigenvalues $$\lambda$$?

• Yes this is correct. Be careful though, some row operations change the determinant and you need to know which ones they are. If you want don't want to deal with that you can also use co-factor expansion to find the determinant.
– CEH
Jul 14, 2016 at 21:28
• Since the matrix is block diagonal, the determinant is the product of the determinants of the two blocks. Jul 14, 2016 at 21:31
• so then would I have $(-12-\lambda) \begin{bmatrix}2-\lambda&0&0\\0&-14-\lambda&9\\0&42&25-\lambda\end{bmatrix}$ ? @CEH Jul 14, 2016 at 21:32
• @carmichael561 Oh, I think I know what you're saying so you are saying that I can take the determinant of that first $2 \times 2$ matrix in the top left and then multiply that by the $2 \times 2$ matrix in the bottom right to get the determinant correct? Jul 14, 2016 at 21:33
• That's correct. Jul 14, 2016 at 21:34

Note that your matrix is the block diagonal matrix $$M = \left[\begin{array}{rr} A & 0\\ 0 & B \end{array}\right]$$ where \begin{align*} A &= \left[\begin{array}{rr} -12 & -6 \\ 8 & 2 \end{array}\right] & B &= \left[\begin{array}{rr} -14 & -9 \\ 42 & 25 \end{array}\right] \end{align*} Next, note that the eigenvalues and corresponding eigenvectors of $A$ are \begin{align*} \lambda_1 &= -4 & v_1 &= (3,-4) & \lambda_2 &= -6 & v_2 &= (1,-1) \end{align*} and the eigenvalues and corresponding eigenvectors of $B$ are \begin{align*} \lambda_3 &= 7 & v_3 &= (3,-7) & \lambda_4 &= 4 & v_4 &= (1,-2) \end{align*} The eigenvalues and corresponding eigenvectors of $M$ are thus \begin{align*} \lambda_1 &= -4 & v_1 &= (3,-4,0,0) & \lambda_2 &= -6 & v_2 &= (1,-1,0,0) \\ \lambda_3 &= 7 & v_3 &= (0,0,3,-7) & \lambda_4 &= 4 & v_4 &= (0,0,1,-2) \end{align*}