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for 1.9 $trace(A)=1+w+w^2=0$, $A$ is not diagonalizable over reals as it has no three distinct real eigen values, $\lambda=1$ is an eigen value that is true.

for 1.10, a) let $\lambda$ be an eigen value, $Ax=\lambda x$ so $x^TAx=x^T\lambda x\ge 0$ so $\lambda\ge 0$ so a) is true, b) is false if some $\lambda=0$ then $A$ is surely not invertible as detA=product of eigen values. c) i have no idea please help.

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    If $A$ is $2\times 2$, how does it have three eigenvalues?2012-12-16
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    oooooops.........:(2012-12-16
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    @ThomasAndrews what will be the trace?2012-12-16
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    What can the two eigenvalues be? The have to be cube roots of $1$. Can they both be $1$?2012-12-16
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    The minimal polynomial of $A$ must divide $x^3-1$. Since $A\not=I$ (because it is not diagonal by assumption), it must be $x^2+x+1$. But the characteristic polynomial $p(x)$ of $A$ is also of degree 2, so $p(x)=x^2+x+1$. Now you may answer 1.9(a)-(c) based on this.2012-12-16
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    $trace(A)=w+w^2=-1$, A is not diagonalizable over reals, $\lambda=1$ is not an eigen value of $A$2012-12-16

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Hint: Show that each leading principal submatrix $A_k = A(1\ldots k|1\ldots k)$ is positive semi-definite. Then by 1.10(a), the eigenvalues of $A_k$ are nonnegative.

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    could you precisely help me to understand about $1.10 c$2013-06-17
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    @TaxiDriver If $A$ is positive semidefinite, all its principal submatrices are also positive semidefinite. (Why?) If a principal submatrix is positive semidefinite, its determinant (i.e. the minor) is nonnegative. (Why?)2013-06-17