# How to deduce that A has an eigenvalue -1 with algebraic multiplicity 2 without calculating characterstic polynomial.

$A = \begin{pmatrix} 0 & 1 &1 \\ 1 & 0 &1 \\ 1& 1 &0 \end{pmatrix}$

The matrix $(A+I)$ has rank $1$ , so $-1$ is an eigenvalue with an algebraic multiplicity of at least $2$ .

I was reviewing my notes and I don't understand how the first statement implies the second one.

Can anyone please explain how rank 1 of $(A + I)$ implies -1 is an eigenvalue with an algebraic multiplicity of 2?

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Hmm, $-1$ is a multiplicity $2$ eigenvalue, sure... – J. M. Jan 3 '12 at 16:34
When we compute eigenvalues we observe the rank of $A - \lambda I$, not $A + \lambda I$. Your observation tells you that $-1$ is an eigenvalue with multiplicity of at least 2. – Patrick Da Silva Jan 3 '12 at 16:35
Somthing is wrong here. $A+I$ has the eigenvalue $3,$ and the eigenvalue $0$ with multiplicity $2$ ( the latter because $A+I$ has rank one). So I'm pretty sure that the question should say that $A$ itself has the eigenvalue $-1$ with multiplicity $2.$ – Geoff Robinson Jan 3 '12 at 16:37
@J.M. yeah sorry it's suppose to be -1 with algebraic multiplicity 2. – Rashmirathi Jan 3 '12 at 16:40

It is well known that the algebraic multiplicity of an eigenvalue is greater or equal than the geometric multiplicity of that eigenvalue (i.e. the dimension of its eigenspace). Therefore, knowing that the rank of $(A-(-1)I)$ is $1$, you know that the algebraic multplicity of $-1$ is at least $2$ because if we denote the geometric multiplicity by $n$, $rank + n = 3$ in this case (dimension of the kernel + dimension of the image is $3$ for a $3 \times 3$ matrix). Therefore $n = 2$ which is the lower bound for the algebraic multiplicity.

EDIT : After reading Pierre-Yves's answer, I had a little flash : you can actually deduce more from this. Using the argument above gives you that the algebraic multiplicity of $-1$ is at least $2$, and you know that $2$ is an eigenvalue since $(1,1,1)$ is an eigenvector of $A$. The sum of the geometric multiplicities is $3$. Since $-1$ has a geometric multiplicity of two and $2$ has a geometric multiplicity of $1$ (it can't have more), we know that there are no other eigenvalues. (I never computed the polynomial! Yay =D)

Hope that helps,

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isn't rank of -1 as an eigenvalue 1? – Rashmirathi Jan 3 '12 at 17:12
Sorry, that's a typo. Hm. My answer looks weird. I'll think a little. – Patrick Da Silva Jan 3 '12 at 17:17
Thanks, I figured it out in the same way as you. – Rashmirathi Jan 3 '12 at 17:45
Dear Patrick, Symmetry of the matrix plays no role here. For any matrix, eigenspaces corresponding to distinct eigenvalues are linearly independent. Thus, if you have a $2$-dimensional eigenspace with eigenvalue $-1$, and a $1$-dimensional eigenspace with eigenvalue $2$, then $\mathbb R^3$ is the direct sum of these eigenspaces, the matrix is diagonalizable, and has char. poly equal to $(x-2)(x+1)^2$. (I didn't need to invoke symmetry so as to know a priori that $A$ was diagonalizable; the fact that $A$ is diagonalizable just fell out.) Regards, – Matt E Jan 3 '12 at 20:01
Hm. Yeah, true that. I should remove that line. I just thought I needed the fact that the sum of the geometric multiplicities was $3$ but I already have it, as you made me notice. Thanks, removed it! – Patrick Da Silva Jan 3 '12 at 20:06

It is clear that $A+I$ is diagonalizable with $(0,0,3)$ on the diagonal.

Thus, $A$ is diagonalizable with $(-1,-1,2)$ on the diagonal.

Justification of the first claim:

A vector is in the kernel of $A+I$ if and only if the sum of its coordinates is zero.

The vector $(1,1,1)$ is an eigenvector for $A+I$ with eigenvalue $3$.

EDIT. We can apply the following observation to $B:=A+I$:

If $B$ is a rank one $n$ by $n$ matrix with entries in a field, then

$\bullet$ either the trace of $B$ is zero, and $B$ is similar to the direct sum of $(\begin{smallmatrix}0&1\\ 0&0\end{smallmatrix})$ and a zero matrix,

$\bullet$ or the trace $t$ of $B$ is nonzero, and $B$ is similar to the direct sum of the scalar $t$ and a zero matrix.

Indeed, either the kernel (which is a hyperplane) contains the image (which is a line), or it doesn't.

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An other way to see that is the following (maybe less clear, but I try anyway) : $A+I$ is a $3\times 3$ matrix with rank one, and thus has for eigenvalues $0$ with multiplicity two and an other one which is given by ${\rm Tr}(A+I)$ [is that clear for you ?].

Then, since $A+I$ and $I$ trivially commutes, they have common basis for diagonalization, so that the eigenvalues of $A$ are the one of $A+I$ minus $I$, namely $-1$ with multiplicity two, and ${\rm Tr}A$.

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Here is my trying (which may be less simple than the previous answers):

we note that $A+I=ee^t\text{ or }$ $$A=ee^t-I$$ post multiplication by $e$ readily shows that $e$ is an eigenvector of $A$ with eigenvalue $2$. Since we can find two other vectors orthogonal to $e$, post multiplications by those show that $-1,-1$ are the remaining eigenvalues of $A$. Note that this happens for any such $n\times n$ $A$.

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This is basically the same as Patrick's answer, just worded differently.

First, we don't need to know what A is, other than the fact that it's $3\times 3$.

Assume $rank(A+I)=1$. Then $-1$ is an eigenvalue of $A$ because $(A+I)=(A-(-1)I)$. By the rank-nullity theorem we know that $rank(A)+nullity(A)=3$. Therefore the dimension of the eigenspace of $A$ corresponding to the eigenvalue $-1$ is $3-rank(A)=2$, the eigenspace just being the null space of $A-I$. Since the algebraic multiplicity is always greater than or equal to the geometric multiplicity we know that $-1$ has multiplicity of at least 2.

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