Geometric meaning of left multiply with a diagonal matrix?

Question

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 $\mathbf{X}$.

Answer 1

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.

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Or, if you prefer: multiply first, then change basis, since $$ \Lambda X = \Lambda [X \Lambda]\Lambda^{-1} $$

Answer 2

Notice that when you multiple $\Lambda$ to X, the resultant matrix is $\left[ \lambda_1 x_1\,\,\,\,\,\lambda_2 x_2\,\,\,\ldots \lambda_n x_n\right]$. You multiply each column but the corresponding scalar.

The relationship between X$^\prime$ and X is that it stretches (or contracts) each column vector.