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Let$$A=\frac{1}{3}\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}$$ Which of the following statements are false?
(a) $A$ has only one real eigenvalue.
(b) $\operatorname{Rank}(A) = \operatorname{Trace}(A)$.
(c) Determinant of $A$ equals the determinant of $A^n$ for each integer $n > 1$.
I proceed in this way:
(a) let $\lambda$ be the eigenvalue of $A$. So,$[A-\lambda I]=\begin{pmatrix} 2/3-\lambda & -1/3 & -1/3 \\ -1/3 & 2/3-\lambda & -1/3 \\ -1/3 & -1/3 & 2/3-\lambda \end{pmatrix}=\begin{pmatrix} -\lambda & -\lambda & -\lambda \\ -1/3 & 2/3-\lambda & -1/3 \\ -1/3 & -1/3 & 2/3-\lambda \end{pmatrix}$$[R_1'\to R_1+R_2+R_3]$$$=\begin{pmatrix} -\lambda & 0 & 0 \\ -1/3 & 1-\lambda & 0 \\ -1/3 & 0 & 1-\lambda \end{pmatrix}[C_2'\to C_2-C_1, C_3'\to C_3-C_1]$$ Hence we get $$|A-\lambda I|=0$$$=>\lambda(1-\lambda)^2=0$ $=>\lambda= $ $0$ or $1$ $=>$There are two real eigenvalue.

(b) Note that $|A|=0$ and $$\begin{vmatrix} 2/3 & -1/3 \\ -1/3 & 2/3 \end{vmatrix}=5/9 \not= 0$$ So , $\operatorname{Rank}(A) =2$ and $\operatorname{Trace}(A)=\frac{1}{3}(2+2+2)=2$ Hence $\operatorname{Rank}(A) = \operatorname{Trace}(A)$

(c) $A^2=AA=A$ and $A^3=A^2A=AA=A$ so determinant of $A$ equals the determinant of $A^n$ for each integer $n > 1$. So only first statement is wrong.

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interesting. Where does the matrix come from? –  draks ... Aug 3 '12 at 6:57
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@draks: Compute the matrix of the orthogonal projection of the 3-space to the plane $x+y+z=0$. –  Jyrki Lahtonen Aug 3 '12 at 7:15
    
@JyrkiLahtonen very interesting. Did you read Ranabir's thoughts or what makes you so confident? Are there anymore interpretations? –  draks ... Aug 3 '12 at 7:39
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@draks: Picking a non-zero vector in 3-space and calculating the matrix of either this orthogonal projection, or the reflection about the plane, is a rich source of examples of matrices with predictable eigenvalues. At least for we the united linear algebra lecturers :-) –  Jyrki Lahtonen Aug 3 '12 at 8:02
    
Without the $\frac{1}{3}$ it is the Cartan matrix of the affine Lie algebra $A_2^{(1)}$. –  PAD Aug 3 '12 at 8:16
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2 Answers

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When looking at this matrix, the first thing came into my mind is: Laplacian of graphs. A matrix is Laplacian if its row sums are zero and off-diagonal entries are non-negative. Of course, this matrix is more special than a generic Laplacian, but the statements in your question can be easily answered.

Statement 1: here A is symmetric, of course, all of its eigenvalues are real.

Statement 2: you may notice that $x=[1,1,1]^T$ is a right eigenvector of A, and the associate eigenvalue is zero. For any graph with a spanning tree, the zero eigenvalue of the Laplacian is simple. Here the matrix A represents a complete graph with three vertices and three edges. Of course, the zero eigenvalue is also simple here. Thus the rank of A is 3-1=2. The fact $rank A=tr A$ I think perhaps is a coincidence.

Statement 3: it seems $\det A=0$.

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your answer is very informative . thakhs –  Argha Aug 3 '12 at 10:09
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You are totally right, except the third part. You do not have $A^2=A$, but instead you have that $det(A^n)=(det(A))^n=0^n=0=det(A)$.

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:-( Given that the matrix represents the orthogonal projection of the space $\mathbf{R}^3$ to the plane with normal $\mathbf{n}=(1,1,1)$ s/he does have $A^2=A$. It is, of course, not necessary to have that relation, but nothing wrong with OP's solution! –  Jyrki Lahtonen Aug 3 '12 at 7:14
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