Determinant of a matrix times constant

Question

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?

Answer 1

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 $\frac16$.

Answer 2

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!

Answer 3

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.

Answer 4

$$ det(A-\lambda I)=0 \implies$$

$$ det( 6[A/6-\lambda /6 I])=0 \implies$$

$$ 6^3det( A/6-\lambda /6 I)=0 \implies$$

$$ det( A/6-\lambda /6 I)=0 $$

Thus your eigenvalues are scaled by a factor of $1/6$