Does a scalar outside of a matrix have any effect on the eigenvalues? Suppose A is 3x3 matrix with eigenvalue e1, e2 and e3 and B= (1/sqrt2) A. Will eigenvalues of A & B will be the same?
 A: The characteristic polynomial of $A$ is $\det(A-tI)$, while the characteristic polynomial of $\alpha A$ is $\det(\alpha A-tI) = {\alpha}^{n} \det \left( A- {\frac{t}{\alpha}}I \right)$ (supposing $A$ is $n \times n$). Therefore, the eigenvalues of $\alpha A$ are $\alpha$ times the eigenvalues of $A$.
A: No. Consider when $A$ is the identity matrix. Then $Ax=x$ (eigenvalue $1$), but $Bx= \frac{1}{\sqrt{2}}Ax=\frac{1}{\sqrt{2}}x$ (eigenvalue $\frac{1}{\sqrt{2}}$). The eigenvectors will not change.
A little more generally, if $x$ is a eigenvector for A with eigenvalue $e$ and $B = bA$, then $Ax=ex$, but $Bx=bAx=bex$.
A: No. They will be proportional. In particular, if $\{e_i\}$ is the set of eigenvalues of A and $\{d_i\}$ is that of B ($i=1,2,3$), one has $d_i = e_i/\sqrt{2}$. 
You can see it by explicit diagonalization of $A$ and $B$. If A is non-defective $A = U\,E\,U^{-1}$, where $E = \mathrm{diag}\{e_i\}$ is the diagonal matrix of the eigenvalues and U a non singular matrix, then 
$B =  A/\sqrt{2} = (1/\sqrt{2}) U\,E\,U^{-1} = U\,(1/\sqrt{2})E\,U^{-1} := U \, D\,U^{-1}$,
where $ D  = (1/\sqrt{2}) \mathrm{diag}\{e_i\} \equiv\mathrm{diag}\{d_i\} $ is still a diagonal matrix whose elements can be identified with the eigenvalues of $B$.
A: We can generalize this question: if $ c $ is a scalar constant, and $A$ is any $n\times n$ square matrix with eigenvalues $(\lambda_1, \lambda_2, ..., \lambda_n)$, and $B = cA$, what are the eigenvalues of $B$? 
There are two answers to this question:
If $c = 0$, then $B$ is the 0 matrix, then $\lambda_i = 0$ for $i = 1,...,n$. So if $c = 0$, then the eigenvalues DO change. 
If $c \neq 0$, then for some eigenvalue $\lambda_i$, it has a corresponding eigenvector $v_i$, then $Av_i = \lambda_i v_i$ by definition, and then using the rules of scalar multiplication, we see that 
$$
Bv_i = (cA)v_i = c(Av_i) = c(\lambda_i v_i) = (c\lambda_i) v_i = (\lambda_i c) v_i = \lambda_i (c v_i)
$$
Thus for $c \neq 0$, the eigenvalues do not change. 
Alternatively, we can use the same manipulations to show that $Bv_i = (\lambda_i c) v_i$ and simply conclude that $\lambda_i$ becomes $c \lambda_i$, in which case we do not need to separate the cases where $c = 0$ and $c \neq 0$. 
