Eigen vector proof: $cI=A$ if $A$ has all non 0 eigenvetors

A homework question I'm having difficulty with:

-Prove that if every non-zero vector is an eigen vector of $A$.

-Then $A = cI$ for some real number $c$.

-Either, solution/ hint...would be nice :)

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2 Answers

You need to show that $A$ can have only one unique eigenvalue.

Suppose $x$ and $y$ are linearly independent vectors. Then there exist $\lambda$ and $\mu$ such that $Ax = \lambda x$ and $Ay = \mu y$, and we know that $A(x + y) = \lambda x + \mu y$. Because we also know that $x + y$ is non-zero (by linear independence), it is an eigenvector of $A$, i.e., there exists $\nu$ such that $A(x + y) = \nu(x + y)$. Set this equal to the result obtained earlier: $\nu(x + y) = \lambda x + \mu y$. By linear independence of $x$ and $y$, we must have $\nu = \lambda$ and $\nu = \mu$. Therefore $\lambda = \mu$.

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if $\alpha_1,\alpha_2$ are linear independent vectors, assume $A(\alpha_1)=k_1\alpha_1,A(\alpha_2)=k_2\alpha_2$ but $\alpha_1+\alpha_2$ is eigenvector of $A$. so $A(\alpha_1+\alpha_2)=k(\alpha_1+\alpha_2)=k_1\alpha_1+k_2\alpha_2$ we get $(k-k_1)\alpha_1+(k-k_2)\alpha_2=0$. $\rightarrow k_1=k_2$ so eigenvalues of $A$ are all equal. and $A$ can be diagonalization,so $A=cI$

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