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Assume $A\in M_{n}(\mathbb{R})$ is diagonalizable.

For practical reasons it is sometimes for convenient to work with $\alpha A$ instead of A ($\alpha\in\mathbb{R}$),

For example if $$A=\begin{pmatrix}0 & 1\\ -\frac{1}{8} & \frac{3}{4} \end{pmatrix}$$ then working with $$8A=\begin{pmatrix}0 & 8\\ -1 & 6 \end{pmatrix}$$ can be more convenient.

If I wish to find $P,D$ s.t. $P^{-1}AP=D$ then I can do so by calculating the eigenvectors and eigenvalues of $A$.

My question is this: If I work with $B=\alpha A$ and find the eigenvectors and eigenvalues of $B$ then I have it that the same $P$ will diagonalize $A$ and $B$ and the eigenvalues of $B$ are the eigenvalues of $A$ multiplied by $\alpha$ ?

My reasoning is that if $Av=\lambda v$ then $\alpha Av=\alpha\lambda v$ so it's the same $v$ but the eigenvalue of $B$ is $\alpha\lambda$, am I correct ?

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You had better not take $\alpha=0$. But otherwise, its fine. –  Marc van Leeuwen Oct 16 '12 at 8:26
    
Why the downvote ? –  Belgi Oct 19 '12 at 10:12
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1 Answer

up vote 2 down vote accepted

Yes, what you said is correct.

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