If $A$ is a positive definite real symmetric matrix and $\lambda$ its eigenvalue then $A^r v = \lambda^r v$ for all real $r>0$ .

Question

Suppose $A$ is a positive definite real symmetric matrix with eigen value $\lambda$ and $v$ an eigenvector corresponding to $\lambda$. How to prove that

$$A^rv= \lambda^r v$$ holds for all real $r>0$. So far I have manage to prove that $A^n v = \lambda^n v$ holds for all natural number $n>0$ by induction on $n$. From this point I hope I have to prove that it holds for all positive rational number $r>0$ then go on to prove for the real case. But the problem is how to proceed from natural number to rational number then to real number.

Let alone the equality how did they even define rational power and real power of matrices. Can anyone help me.

How is $(SDS^{-1})^r$ defined? Basically, how is exponentiation for matrices defined? — Sarvesh Ravichandran Iyer, Jul 29 '20 at 18:24
It depends on how you define such a thing. For example: https://math.stackexchange.com/questions/16739/arbitrary-non-integer-power-of-a-matrix — Alex R., Jul 29 '20 at 18:56

score 1 · Accepted Answer · answered Jul 29 '20 at 18:38

Let $\lambda_1,\cdots,\lambda_n>0$ be eigenvalues of $A$. Then there is an orthonormal matrix $S$ (namely $S^TS=1$) such that $$ S^TAS=\text{diag}(\lambda_1,\cdots,\lambda_n). $$ So $$ S^TA^rS=\text{diag}(\lambda_1^r,\cdots,\lambda_n^r) $$ which gives $$ A^rS=S\text{diag}(\lambda_1^r,\cdots,\lambda_n^r). $$ Let $S=(v_1,\cdots,v_n)$ and then $$ A^rv_i=\lambda_i^rv_i. $$

If $A$ is a positive definite real symmetric matrix and $\lambda$ its eigenvalue then $A^r v = \lambda^r v$ for all real $r>0$ .

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