Determinant of a matrix times constant I have a problem figuring the following out:
I am aware that this hold: $\det(kA)=k^n * \det(A)$ for A being (n×n) matrix.
However, if I wish to calculate the determinant of 
$\begin{bmatrix}5-λ&-2&1\\-2&2-λ&2\\1&2&5-λ\end{bmatrix}$
I get eigenvalues 0, and 6, 
But if I wish to calculate the determinant of: (1/6 is a constant front of the matrix, but I couldn't align it)
$1/6 \cdot \begin{bmatrix}5-λ&-2&1\\-2&2-λ&2\\1&2&5-λ\end{bmatrix}$
I get eigenvalues 0, 1. 
How's this true taking in the account that the equality at the top holds?
 A: $$ det(A-\lambda I)=0 \implies$$
$$ det( 6[A/6-\lambda /6 I])=0 \implies$$
$$ 6^3det( A/6-\lambda /6 I)=0 \implies$$
$$ det( A/6-\lambda /6 I)=0 $$
Thus your eigenvalues are scaled by a factor of $1/6$  
A: By scaling a matrix by a factor $a$ you multiply the matrix by $aI$ with $i$ being the identity of correct size. 
By this, you can easily see, how the first equality holds true. The determinant of $aI$ is $a^k$. 
Take a look at the eigenvalues and -vectors now. If you right multiply $A$ with an eigenvector, it get's scaled according to it's corresponding eigenvalue. But after that, it get's scaled by $aI$. Therefore the corresponding eigenvalue needs to be scaled too!
A: Let the first matrix be $A$.
Note that $\det(A)= 0=0\cdot t$ and and $\det(kA)=0 = 0 \cdot 1$ as well, hence there is no contradiction.
Let $A = V \operatorname{diag}\left(\lambda_1, \lambda_2, \lambda_3\right) V^T$, then we have
$$\frac{A}6 = V \operatorname{diag}\left(\frac{\lambda_1}6, \frac{\lambda_2}6, \frac{\lambda_3}6\right) V^T$$
The eigenvalues does scale down in the right ratio.
A: Let $\lambda$ be an eigenvalue of $A$ with the eigenvector $v$.
Then for $\alpha A$ we have:
$$(\alpha A)v = \alpha(Av) = \alpha(\lambda v) = (\alpha\lambda) v$$
Therefore, $v$ is the eigenvector of the eigenvalue $\alpha\lambda$.
That is why your eigenvalues got scaled by $\frac16$.
