# How to use characteristic polynomial to find eigenvalue and determine diagonalization of matrix A?

If the matrix, $$A = \begin{pmatrix}-3&0&0 \\\ -4&-7&8\\-4&-4&5 \end{pmatrix}$$

• Find the characteristic polynomial of A and, hence, find all the eigenvalues of A.

• Parametrizing (as a subset of $$\mathbb{R}^3)$$ the eigenspace of A that corresponds to the eigenvalue $$\lambda = -3$$. Is it possible to diagonalize A?

So far I have taken the determinant of $$(A-\lambda I)$$ and got the characteristic polynomial $$-\lambda^{2} -5\lambda^{2} -3\lambda +9$$. Then calculated the eigenvalues to be $$\lambda = 1, -3$$. After this, I am unsure how to proceed.

• What did you do? – Bernard May 7 at 14:28
• There’s an easier way for this matrix. Hint: At least two eigenvalues are easily found by inspection, and you can always get the last one “for free” by considering the trace of the matrix. – amd May 8 at 0:02
• @Bernard I am not familiar with the trace of the matrix, how does that work? – Dylan Grimes Larkin May 8 at 7:04
• The trace of a square matrix is the sum of its diagonal coefficients. It happens that it is also the sum of its eigenvalues (counted with multiplicity). You can see more details on Wikipedia. – Bernard May 8 at 9:38

First, you should not expand the determinant of $$A-\lambda I$$, but try to factor it: \begin{align}\begin{vmatrix} -3-\lambda &&0 \\ -4&-7-\lambda &8 \\ -4&-4&5-\lambda\end{vmatrix}&=\begin{vmatrix} -3-\lambda &&0\\-4&-7-\lambda &8 \\ 0&3+\lambda&-3-\lambda\end{vmatrix}=(3+\lambda)\begin{vmatrix} -3-\lambda &&0\\-4&-7-\lambda &8 \\ 0&1&-1\end{vmatrix}\\[1ex] &=(3+\lambda)\Bigl[(-3-\lambda)(\lambda+7-8)\Bigr]=-(\lambda+3)^2(\lambda-1) \end{align}
For $$1$$, which is a simple eigenvalue, there is no problem. For the double eigenvalue $$-3$$, this means you have to check whether the corresponding eigenspace $$\ker(A+3I)$$ has dimension $$2$$.