linear map with n+1 eigenvectors

eugene · Yang-Mills Theory Offline Joined: 16 May 2004 Posts: 970

$V$ is an $n$ -dimensional vector space. Can we find a linear map $A : V \to V$ with $n+1$ eigenvectors, any $n$ of which are linearly independent, which is not a scalar multiple of the identity?

**Posted:** Sat Apr 23, 2005 10:05 pm

grobber

Of course not Smile

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Let $x_1,\ldots,x_n$ be $n$ of those vectors, with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n$ . The $n+1$ 'th vector must have the form $x=\sum_{i=1}^n a_i x_i$ , where all the $a_i$ 's are non-zero. Let $\lambda$ be the eigenvalue corresponding to $x$ . We have $Tx=\sum a_i\lambda_i x_i=\sum a_i\lambda x_i$ , meaning that $\sum a_i(\lambda_i-\lambda)x_i=0$ , which is equivalent to $\lambda_i=\lambda,\ \forall i$ . This means that all the eigenvalues of $T$ are equal. Since $T$ is diagonalizable, we get the conclusion: $T=\lambda I$ .

**Posted:** Sat Apr 23, 2005 10:28 pm