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I am a graduate student.In CMI PHD entrance $2011$ there was a true/false question which is as follows:

For $n\geq 2$ there exists an $n\times n$ matrix $A$ such that $A^2=A$ and $\text{tr}(A)=n+1$.

My solution to this question is as follows:

Any such matrix $A$ is similar to $A'=\begin{pmatrix}I_{r\times r} & O_{r\times(n-r)}\\O_{(n-r)\times r}&O_{(n-r)\times (n-r)}\end{pmatrix}$ and change of basis does not affect trace of a matrix.So,$A'$ has trace $n+1$ but trace of $A'$ is $r\leq n$.Which is a contradiction.

Is this ok or is there any easier way?

Thomas Andrews
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Kishalay Sarkar
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  • I think this is the intended solution! An alternate way to phrase the same idea: the only eigenvalues of projections are $0$ and $1$ with some multiplicities that are between $0$ and $n$, but the trace is the sum of the eigenvalues and thus can't be as large as $n+1$. – Greg Martin Jul 11 '22 at 04:22
  • @GregMartin write this as an answer! I'll upvote if you do :) – Elliot Herrington Jul 11 '22 at 04:26

2 Answers2

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Projections satisfy $P^2=P$, implying that the only eigenvalues are $0$ and $1$. Thus up to similarity, the trace is at most $n$.

But trace is an invariant. So no.

suckling pig
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No. There doesn't exist any such projection matrix $A$ such that $\text{tr}(A)=n+1$.

Hint: For a projection matrix $A$ ,$\text{tr}(A) =\text{rank}(A) $

See here for the proof.

SoG
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