I have
$m$ of column vectors in $x_i \in \mathbb{R}^{n\times 1}$ as
$$\mathbf{X} = \begin{bmatrix}x_1 & x_2 & \ldots x_m \end{bmatrix},$$
Then I multiply with a diagonal matrix, note that each element does not have to be equal, other wise a scalar matrix is too easy,
$$\mathbf{\Lambda} = \begin{bmatrix}\lambda_1 & & \\
& \ddots & \\
& & \lambda_n\end{bmatrix}$$
What is the geometric meaning of $\mathbf{X^\prime} = \mathbf{\Lambda X}$?
What is the geometric relationship between $\mathbf{X}^\prime$ and $\mathbf{X}$?
It seems to be pretty difficult to describe its behavior. It does not preserve the anlge between column vector in $\mathbf{X}$.
Best Answer
Suppose that all $\lambda_i$ are non-zero. One way to think of this action is to consider the fact that $$ \Lambda X = (\Lambda X \Lambda^{-1}) (\Lambda \Lambda\Lambda^{-1}) = [X]_{\mathcal B} [\Lambda]_{\mathcal B} $$ The transition $X \mapsto \Lambda X \Lambda^{-1}$ can be thought of geometrically as a change in basis. That is, we reinterpret $X$ as a transformation relative to the basis $\mathcal B = \{\lambda_1 e_1, \dots, \lambda_n e_n\}$, where $e_i$ is the $i$th standard basis vector.
Multiplying from the right by $\Lambda$, as before, amounts to scaling the columns.
Or, if you prefer: multiply first, then change basis, since $$ \Lambda X = \Lambda [X \Lambda]\Lambda^{-1} $$