Characteristic Polynomial on 3 by 3 matrix

Question

The given matrix is

$$A =\begin{pmatrix} 2 & 4 & 3 \\ -4 & -6 & -3 \\ 3 & 3 & 1\end{pmatrix}.$$ Any alternative idea than computing $|A-xI|$? $$A - xI = \begin{pmatrix} 2-x &4 & 3 \\ -4 & -6-x & -3 \\ 3 & 3& 1-x \end{pmatrix}$$ How to find characterstics equation ??

Answer 1

In artificial exercises such as this one, you can sometimes save yourself having to grind through the generic method by checking some simple linear combinations of rows and columns. In this case, the sum of the first two rows is $(-2,-2,0)$, so one eigenvalue is $-2$, and the sum of all of the rows is $(1,1,1)$, so another eigenvalue is $1$. The eigenvalues of a matrix and its transpose are identical, so you now have two of the eigenvalues of $A$. You can always get the last one “for free” from the trace, which is the sum of the eigenvalues. Here, we get $(2-6+1)-1-(-2)=-2$ for the third eigenvalue. The characteristic polynomial is therefore $(\lambda+2)^2(\lambda-1)$.

You might also notice that if you subtract the second column from the first, you get $(-2,2,0)^T$, and then add the last column to this to get $(1,-1,1)^T$, which gives you the same two eigenvalues as above, but I found it a bit easier to spot the interesting row sums for this matrix.

Incidentally, the characteristic polynomial is usually defined as being monic (not that this affects its roots), so you should probably be computing $\det(\lambda I-A)$ instead, which is guaranteed to produce a monic polynomial for both odd- and even-order matrices. $\det(A-\lambda I)$ is monic only for even orders. On the other hand, subtracting $A$has more potential for making a sign error, so watch out for that.

Answer 2

Find the determinant $\det |A-xI|=0$ which gives you a cubic of it is called the characteristic equation. So in your case it comes to $$x^3+3x^2-4=0$$ its roots are $-2,-2,1$ whose sum should be equal to the trace of the matrix (sum of the diagonal elements) and their product equals $-\det|A|$.

Answer 3

You can make some linear row or column combinations and use multilinearity: \begin{align} &\bigl|A - xI\bigr| = \begin{vmatrix} 2-x &4 & 3 \\ -4 & -6-x & -3 \\ 3 & 3& 1-x \end{vmatrix}=\begin{vmatrix} 2-x &4 & 3 \\ -2-x & -2-x & 0 \\ 3 & 3& 1-x \end{vmatrix}\\[1ex] &=(-2-x)\begin{vmatrix} 2-x &4 & 3 \\ 1& 1 & 0 \\ 3 & 3& 1-x \end{vmatrix}=(-2-x)\begin{vmatrix} -2-x &4 & 3 \\ 0 & 1 & 0 \\ 0 & 3 & 1-x \end{vmatrix}\\ &=(2+x)^2\begin{vmatrix} 1 &4 & 3 \\ 0 & 1 & 0 \\ 0 & 3 & 1-x \end{vmatrix}=(2+x)^2\cdot 1\bigl(1\cdot(1-x)-0\cdot3\bigr)=(2+x)^2(1-x). \end{align}