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Let $V$ be 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?

Here's a solution I found. The answer is no. Let the $n+1$ corresponding eigenvalues have sum $k$. Given any eigenvector with eigenvalue $\lambda,$ the remaining $n$ eigenvectors are linearly independent, so they form a basis.

The basis formed by the remaining $n$ eigenvectors diagonalizes the matrix of the linear transformation. The trace of the resulting matrix is $k-\lambda$. Since trace is independent of the choice of basis, all eigenvalues are equal to lambda. Hence $A$ is a scalar matrix (i.e. a scalar multiple of the identity).

I still can't convince myself that $A$ is a scalar matrix. What if $A$ is a matrix such that it that satisfies the given conditions and before it is diagonalized by the basis, it is not a scalar multiple of the identity?

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Your proof is correct. To deal with the last part, the key is that any scalar matrix commutes with all other matrices. Indeed, suppose $A$ is such a matrix, i.e. it can be diagonalized to a scalar matrix $\lambda I$. Then there is some invertible matrix $P$ such that $P\lambda I P^{-1}=A$. Furthermore, $P \lambda I P^{-1} = PP^{-1} \lambda I = \lambda I$ by commutativity of scalar matrices. Thus, $A = \lambda I$ is a scalar matrix.

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Your argument is entirely correct, and quite clever.

I will try to answer your question: 'what if it was not a scalar multiple of the identity before diagonalizing?'

The point is that this does not happen. We all know that diagonalizable matrices can look very non-diagonal when picking a different basis, but this is not true for scalar multiples of the identiy. More on point: scalar multiples of the identity look the same in every basis.

The reason for that is: if $A = \lambda I$ then every vector in $V$ is an eigenvector of $A$ with eigenvalue $\lambda$. So in particular: every vector in every basis you choose is an eigenvector of $A$ with eigenvalue $\lambda$ and hence with respect to that chosen basis your matrix looks like a diagonal matrix with $\lambda$'s on the diagonal.

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