# Can a value other than eigen values satisfy the characteristic polynomial?

Well the title says it all. I am confused since the characteristic polynomial is used to derive eigen values so all the roots are eigen values. Now if I have another set of values which satisfy the characteristic polynomial then those should also be a eigen values right?

I am looking at this matrix.

 3 -1 -1
-1  3 -1
-1 -1  3


The eigen values come out as 1,4,4 but if I use 1,1,1 as values of lambda and put it in the characteristic equation, it still satisfies the characteristic polynomial. This is all very confusing. Maybe some of my concepts are wrong.

Edit

Let me explain my confusion a little more, and where I was thinking wrong

If I have my eigenvalues then the characteristic equation is given by |A - \lambda *I| = 0,right? But in my not so well conditioned mind I made a huge mistake. Let me first tell you that this was taken from a multiple choice question where 1,1,1 and 1,4,4 were two answers. So I thought I would just put eigenvalues in the characteristic equation and see if it satisfies. Calculator can do the determinant calculation(casio Fx-991MS) and save on a lot of time (I could solve the 3 variable equation as well but deriving that equation seems like a long and error prone process). But in my mighty dumbness I put all the three values in the characteristic equation at the same time. So of course 1,1,1 seemed to satisfy as eigenvalues. What I should have done was checked each number separately. Sorry for kind of a foolish question.

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The eigenvalues of the matrix are precisely the roots of the characteristic polynomial. –  JavaMan Jul 29 '11 at 18:30
How can "$1,4,4$" be "the eigenvectors"? Eigenvectors are vectors, so they should be elements of $\mathbb{R}^3$. –  Arturo Magidin Jul 29 '11 at 18:34
@arturo sorry I meant to say eigenvalues. –  Rick_2047 Jul 29 '11 at 18:51
Certainly if $1$ is a zero of the characteristic polynomial, then every time you plug in $1$ into the characteristic polynomial you will get $0$, whether you plug it once or you plug it three times; I don't understand what you mean when you say that you "use $1,1,1$ as values of lambda". –  Arturo Magidin Jul 29 '11 at 18:52
@Rick_2047: Evaluating the characteristic polynomial at a scalar will tell you whether the scalar is an eigenvalue or not, but it will not give you the multiplicity by itself. If your options had been "1,4,4" and "1,1,4", simply evaluating the characteristic polynomial to see if you got $0$ with each number would not have allowed you to distinguish the correct answer from the incorrect one. –  Arturo Magidin Jul 29 '11 at 19:23

For an $n\times n$ matrix $A$, a scalar $\lambda$ is an eigenvalue of $A$ if and only if it is a zero of the characteristic polynomial of $A$.

Why is this? Remember that $\lambda$ is an eigenvalue of $A$ if and only if there is a nonzero vector $\mathbf{v}$ such that $A\mathbf{v}=\lambda\mathbf{v}$; this is equivalent to the existence of a nonzero vector $\mathbf{v}$ such that $(A-\lambda I)\mathbf{v}=\mathbf{0}$. That means that the nullspace of the matrix $A-\lambda I$ (remember, $I$ is the identity matrix) is not just the zero vector, which means, necessarily, that $A-\lambda I$ is not invertible. Since it is not invertible, that means that its determinant is $0$; its determinant happens to equal the characteristic polynomial evaluated at $\lambda$, so this shows that if $\lambda$ is an eigenvalue of $A$, then $\lambda$ is a zero of the characteristic polynomial.

Conversely, if $\lambda$ is a zero of the characteristic polynomial of $A$, then the determinant of $A-\lambda I$ is zero, which means that $A-\lambda I$ is not invertible, which means there is a nonzero vector $\mathbf{w}$ such that $(A-\lambda I)\mathbf{w}=\mathbf{0}$. This shows that $\mathbf{w}$ is an eigenvector of $A$ with eigenvalue $\lambda$, so $\lambda$ is an eigenvalue.

For the matrix you have, $$A = \left(\begin{array}{rrr} 3 & -1 & -1\\ -1 & 3 & -1\\ -1 & -1 & 3 \end{array}\right).$$ The characteristic polynomial $p(t)$ is: \begin{align*} p(t)=\det(A-tI) &= \left|\begin{array}{ccc} 3-t & -1 & -1\\ -1 & 3-t & -1\\ -1 & -1 & 3-t \end{array}\right|\\ &= (3-t)\left|\begin{array}{cc} 3-t & -1\\ -1 & 3-t \end{array}\right| +\left|\begin{array}{cc} -1 & -1\\ -1 & 3-t \end{array}\right| - \left|\begin{array}{cc} -1 & -1\\ 3-t & -1 \end{array}\right|\\ &= (3-t)\Bigl((3-t)^2-1\Bigr) + (t-4) - (4-t)\\ &= (3-t)\Bigl(t^2 -6t +8\Bigr) +2(t-4)\\ &= (3-t)(t-4)(t-2) + 2(t-4)\\ &= (t-4)\Bigl(2 - (t-2)(t-3)\Bigr) \\ &= -(t-4)(t^2-5t+6-2)\\ &= -(t-4)(t^2-5t+4)\\ &= -(t-4)^2(t-1). \end{align*} Since $\lambda$ is an eigenvalue of $A$ if and only if $p(\lambda)=0$, this says that the $A$ matrix has two distinct eigenvalues: $\lambda=4$, with algebraic multiplicity $2$, and $\lambda=1$.

What are the corresponding eigen vectors?

For $\lambda=1$, you want vectors $(a,b,c)^t$ such that $A(a,b,c)^t = (a,b,c)^t$ ($t$ is the transpose). Equivalently, you want the nullspace of $A-I$, except for $\mathbf{0}$. It is not hard to verify that $(1,1,1)^t$ is an eigenvector corresponding to $\lambda=1$, and that every eigenvector corresponding to $\lambda=1$ is a nonzero scalar multiple of $(1,1,1)^t$ (is this where you got confused? This is a vector, not a list of eigenvalues).

For $\lambda=4$, you want vectors $(a,b,c)^t$ that le in the nullspace of $A-4I$. Here, you want $a+b+c=0$, so the nullspace is spanned by the vectors $(1,0,-1)^t$ and $(0,1,-1)^t$; you can verify that each of these is an eigenvector corresponding to $\lambda=4$ and they are linearly independent, so the eigenvectors corresponding to $\lambda=4$ are the nonzero linear combinations of these two.

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The characteristic Polynomial of your matrix is

$$p(\lambda)=\lambda^3 - 9 \lambda^2 + 24 \lambda - 16 =(\lambda-1)(\lambda-4)(\lambda-4)$$

Now you can directly read the eigenvalues from the factorized polynomial, namely $1,4,4$. That you have $p(1)=0$ shows you that 1 is an eigenvalue but for $1,1,1$ to be all eigenvalues of your matrix you would need $p(\lambda)=(\lambda-1)^3$ which is not the case here.

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If it's "$p(x)$", then the variable should be $x$; if the variable is $\lambda$, shouldn't it be "$p(\lambda)$"? (-; –  Arturo Magidin Jul 29 '11 at 18:35
Yes :) I changed this afterwards because the OP seems to use lambda rather than x so I thought he understands this better. –  Listing Jul 29 '11 at 18:36