Proof that a certain number is eigenvalue without performing diagonalization

Question

Suppose we have the following square matrix of order 3:

$$A = \begin{pmatrix} 1 & 3 & 3\\ -3 & -5 & -3 \\ 3 & 3 & 1 \end{pmatrix}$$

How can I prove that $\lambda=-2$ is eigenvalue without having to calculate the characteristic polynomial and diagonalization?

I know the definition of eigenvalue and I know how to diagonalize. But, I wouldn't know how to do that.

Answer 1

Here's a really goofy answer: observe that $$ (A - I) (A + 2I) = 0 $$ hence the minimal polynomial of $A$ must be $$ m(x) = (x-1)(x+2), $$ so that $-2$ must be an eigenvalue. (The question, as written, is wrong, and $+2$ is not an eigenvalue at all).

This meets the letter of the question without really teaching much, alas. How could one possibly guess that this product was going to be zero? Well, you could look at $I, A,$ and $A^2$, \begin{align} I &= \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\\ A &= \pmatrix{1 & 3 & 3 \\ -3 & -5 & -3 \\ 3 & 3 & 1}\\ A^2 &= \pmatrix{1 & -3 & -3 \\ 3 & 7 & 3 \\ -3 & -3 & 1} \end{align} and notice, from the regularity of the entries, that $A^2 + A - 2I$ is the zero matrix, and this factors as $(A-I)(A+2I)$. But why would you even think that the minimal polynomial was quadratic rather than linear? Well, $A + 2I$ is so 'repetitive' that you might guess, as I did, that it had only rank $1$ rather than rank $2$.

In other words, if this is obvious to you, it's easy. :)

Answer 2

By definition, 2 is an eigenvalue iff there exists v such that Av=2v. If you have such a v, you can prove that 2 is an eigenvalue simply by calculating Av. If you don't know such a v, then you'll have to calculate one, such as by taking the equation (A-2I)v=0 and solving for v.

Answer 3

There's a mistake as pointed out by John: it must be $\;\lambda =-2\;$ , not $\;\lambda=2\;$

Just check that $\;A+2I\;$ is singular, since that'd mean that there exists a vector $\;v\neq0\;$ s.t. $\;(A+2I)v=0\iff Av+2v=0\implies Av=-2v\;$. In our case we get at once:

$$A+2I=\begin{pmatrix}3 & 3 & 3\\ -3 & -3 & -3 \\ 3 & 3 & 3\end{pmatrix}\implies \det (A+2I)=0\\$$

Answer 4

This is not true. $2$ is not an eigenvalue of $A$. Rather, $-2$ is an eigenvalue of $A$.

It is obvious that $$ A+2I= \pmatrix{3&3&3\\ -3&-3&-3\\ 3&3&3} $$ is singular. Since $\operatorname{tr}(A)=-3$ is odd, $A$ has an eigenvalue other than $-2$. As $A+2I$ has rank one, only one eigenvalue of $A$ can differ from $-2$, i.e. $-2$ is an eigenvalue of multiplicity two. It follows that the other eigenvalue of $A$ is $\operatorname{tr}(A)-2(-2)=1$. Indeed, $$ A-I = \pmatrix{0&3&3\\ -3&-6&-3\\ 3&3&0} $$ is singular because its column sums are zero.