Is it allowed to scale a matrix? I have a system matrix
$$
\mathbf{A} = 
\begin{pmatrix}
-1 & 3 \\
-3 & -1
\end{pmatrix}
$$
The characteristic matrix is $K_A = A-\lambda I$ so it is
$$
K_A =
\begin{pmatrix}
-\lambda - 1 & 3 \\
-3 & -\lambda - 1
\end{pmatrix}
$$
but am I allowed to set the following equal sign?:
$$
K_A=
\begin{pmatrix}
-\lambda - 1 & 3 \\
-3 & -\lambda - 1
\end{pmatrix}
=
\begin{pmatrix}
\lambda + 1 & -3 \\
3 & \lambda + 1
\end{pmatrix}
$$
I have just multiplied the matrix with $-1$. Am I allowed to do so? To me the two matrices is not equal, so I guess I should use $\sim$ instead?
$$
K_A=
\begin{pmatrix}
-\lambda - 1 & 3 \\
-3 & -\lambda - 1
\end{pmatrix}
\sim
\begin{pmatrix}
\lambda + 1 & -3 \\
3 & \lambda + 1
\end{pmatrix}
$$
 A: Yes, given that you skip the first notation. The $\sim$ symbol is not standard when relating matrixes so you should be allowed to define that. Let's say $A\sim B$ means that $A = TB$ for $T$ in some group $G$ of matrices such that:


*

*$I \in G$

*if $T\in G$ then $T$ is invertible and $T^{-1}\in G$

*if $T_1\in G$ and $T_2\in G$ then $T_1T_2 \in G$


(for example the group of scaling matrices where the scale factor is non-zero fulfills these criteria, but there are larger groups, fx the group of invertible matrices)
Then $\sim$ will form an equivalence relation between matrices:


*

*It's reflexive since $A=IA$ since the identity matrix is in the group.

*It's symmetric since $A\sim B$ means that $A = T B$, but $T$ being in the group means that $T^{-1}$ being that to so $B = T^{-1}A$ (ie $B\sim A$)

*It's transitive because if $A\sim B$ and $B\sim C$ we have $A = T_{AB}B$ and $B = T_{BC}C$ so $A = T_{AB}T_{BC}C$ where $T_{AB}T_{BC}$ is also in the group.


Now in each equivalence class under this relation either all matrices have zero determinant or all have non-zero determinant. If $A$ and $B$ are two matrices in the same class we have that $\det(A) = (\prod det{T_j})\det(B)$, but $\det{T_j} \ne 0$ always (rotations have determinant $1$ scaling has determinant $c^n$ where $c$ is the scaling factor and mirrorings have determinant $\pm 1$).
So $\det(A-\lambda I)=0 \Leftrightarrow \det(B)=0$ if $A\sim B$.
Another recent question suggested that two matrices should be called similar if $A = PBP^{-1}$, which also is a viable option (it's an equivalence relation and within each equivalence class either all determinants are zero or all determinants are non-zero, but doesn't include your transform).
A: Simple answer:
You can pull a factor out of the columns or rows. This still gives the same characteristic polynomial, so this is allowed.  
To be safe, just write it with $| \ |$ around it and write the factor in front of it:
$$
\left|
\begin{pmatrix}
-\lambda - 1 & 3 \\
-3 & -\lambda - 1
\end{pmatrix}\right|
=-1\cdot \left|
\begin{pmatrix}
\lambda + 1 & -3 \\
3 & \lambda + 1
\end{pmatrix}\right|
$$
If you are interested in finding eigenvectors:
$$K_A \vec v = \vec 0$$
You can do any operation which can be written by a multiplication with an invertible matrix from the left, since this doesn't change neither $\vec v$ nor the right side.
Here this would be multiplication with
$$ \begin{pmatrix}-1 & 0 \\ 0 & 1 \end{pmatrix}$$
