2

For fun, I set myself the exercise of understanding the matrix \begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}

geometrically.

What I am looking for is a decomposition of this matrix into simpler linear transformations like reflections, shears, scalings and rotations.

Things I tried:

Constructing the image of the unit square. Constructing the image of the axes. Computing eigenvalues/eigenvectors.

Things I know:

Based on the above, the linear transformation represented by the matrix transforms lines into lines and reverses orientation. Also it is not an isometry.

Adam
  • 3,422
  • 1
  • 33
  • 50
  • 1
    This matrix is simple, so a good way to get a geometric understanding is by looking at its elementary matrix decomposition, and looking at its action on the unit vectors in $\mathbb{R}^2$. –  Oct 23 '17 at 11:41
  • Would you be happy with the following description? Suppose $\lambda_1,\lambda_2$ and $v_1,v_2$ are the corresponding eigenvalues and eigenvectors of the matrix. $A=\begin{bmatrix}1&2\3&4\end{bmatrix}$. Then, for a point $(x,y)$, the image $A\cdot \begin{bmatrix} x\ y\end{bmatrix}$ can be obtained by: (1) Project the vector $(x,y)$ on the line generated by $v_1$, in the direction of $v_2$. (2) Project the vector $(x,y)$ on the line generated by $v_2$, in the direction of $v_1$. – Darío G Oct 23 '17 at 11:44
  • (3) Multiply the first vector by $\lambda_1$, and the second vector by $\lambda_2$. (scaling) (4) Add the vectors obtained in (3). – Darío G Oct 23 '17 at 11:44

2 Answers2

4

IMO the singular-value decomposition is a much better way to visualize the geometrical behavior of a matrix, than the eigenvalue decomposition.

The decomposition for a matrix $M∈ℝ^{m×n}$ is given by: $$M=UΣV^\top$$ with $U∈ℝ^{m×m}$ orthogonal, $V∈ℝ^{n×n}$ orthogonal and $$Σ=\left(\begin{array}{ccc|ccc}ο_1 & && & \vdots & \\ &\ddots &&\cdots& 0 & \cdots\\ &&σ_r & & \vdots & \\ \hline & \vdots & & & \vdots & \\ \cdots& 0 & \cdots & \cdots& 0 & \cdots\\ & \vdots & & & \vdots & \\ \end{array}\right)∈ℝ^{m×n}$$

With this decomposition the mapping $$\tilde{M}:ℝ^n→R^m\\x↦Mx$$ can be seen as the composition of three mappings $\tilde{M}=\tilde{U}\tilde{Σ}\tilde{V^\top}$, with $$\tilde{V^\top}:ℝ^n→R^n\\x↦V^{\top}x$$ $$\tilde{Σ}:ℝ^n→R^m\\x↦Σx$$ $$\tilde{U}:ℝ^m→R^m\\x↦Ux$$

The mappings $V^\top$ and $U$ define coordinate transformations in $ℝ^n$ / $ℝ^m$, and $Σ$ is a scaling, as it is diagonal.

You can see that in the following picture, that illustrates the transformation of the unit disc. The arrows represent basis vectors.

enter image description here

Source

For your matrix it is $$A=\begin{pmatrix} -0.4046 &-0.9145 \\ -0.9145 & 0.4046 \end{pmatrix} \begin{pmatrix} 5.4650 \\ &0.3660\end{pmatrix} \begin{pmatrix} -0.5760 & -0.8174 \\ 0.8174 &-0.5760 \end{pmatrix}$$

P. Siehr
  • 3,672
0

Based on the comment of 정준환, I found a following decomposition of A into elementary matrices.

$ \begin{pmatrix} 1 & 1/2 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} -1/2 & 0 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 4 \\ \end{pmatrix} = A $

These correspond to horizontal shear, horizontal scaling, vertical shear and vertical scaling.

Adam
  • 3,422
  • 1
  • 33
  • 50