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A is 5×5 matrix, all of whose entries are 1, then

(a) A is not diagonalizable (b) A is idempotent (c) A is nilpotent

(d) The minimal polynomial and the characteristics polynomial of A are not equal.

by the given condition matrix must be singular and hence it is not diagonalizable. am i correct.

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    No. A counterexample is the matrix $\begin{bmatrix} 1&0\\0&0\end{bmatrix}$. It is singular and trivially diagonalisable.2012-11-17
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    There is no direct correlation between *singular* and *non-diagonalizable*: there are diagonalizable matrices that are singular and ones that are regular.2012-11-17
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    @Daryl : How is your example a counterexample to an assertion about the $5\times 5$ matrix in which every entry is $1$?2012-11-17
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    @MichaelHardy Hehehe2012-11-17
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    @Michael: It is a counterexample to the implication "singular and hence... not diagonalizable", isn't it?2012-11-17
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    OK . . . . ${{{{{{{{}}}}}}}}$2012-11-17
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    @MichaelHardy It is as Rahul said, singular doesn't imply non-diagonalisable.2012-11-17
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    @poton do you mean that all entries are 1 or that the non-zero entries are 1?2012-11-17
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    @Daryl : I have been well aware of that since many years before this current thread started.2012-11-17
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    @Michael: I feel like I am missing something that is glaringly obvious to you. Please enlighten me. (this is not sarcasm)2012-11-17
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    @RahulNarain : The question seemed to be about a certain $5\times5$ matrix. I initially missed the fact that the poster thought that singular matrices can't be diagonal.2012-11-17

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$A$ is symmetric (or selfadjoint, if your matrices are complex), so it is diagonalizable.

It is not idempotent, because $A^2=5A$.

It is not nilpotent, because $A^n=5^{n-1}A$.

The minimal polynomial of $A$ is $p_m(t)=t(t-5)$, while the characteristic polynomial is $p_c(t)=t^{4}(t-5)$. So they are different.

The characteristic polynomial can be obtained this way: we have $$ A=\begin{bmatrix}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 1&0&\cdots&0 \end{bmatrix}\,\begin{bmatrix}1&\cdots&1\\0&\cdots&0\\ &\ddots\\0&\cdots&0 \end{bmatrix}, $$ so its spectrum agrees with that of $$ \begin{bmatrix}1&\cdots&1\\0&\cdots&0\\ &\ddots\\0&\cdots&0 \end{bmatrix}\,\begin{bmatrix}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 1&0&\cdots&0 \end{bmatrix}=\begin{bmatrix}5&0&\cdots&0\\0&0&\cdots&0\\ & & \ddots \\0& & \cdots &0 \end{bmatrix}. $$ The eigenvalues are then $5,0,0,0,0$, and so $p_c(t)=t^4(t-5)$. As $A^5-5A=0$, the minimal polynomial is at most $t(t-5)$. But it also requires $0$ and $5$ as roots, so $p_m(t)=t(t-5)$.

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    Minor correction: $A^n=5^{n-1}A$.2012-11-17
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    Good point. Thanks!2012-11-17