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How do I compute the left and right polar decompositions of a matrix by hand? I understand the definition of the decomposition but not how it is calculated.

For instance are the left and right polar decompositions of this matrix different, and how do I compute them?

$$ \begin{bmatrix} 2 & -3\\ 1 & 6\\ \end{bmatrix} $$

James
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John
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2 Answers2

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The polar decomposition is strongly related with the Singular Value Decomposition.
The SVD looks as follows:
$$A=U\mathrm{\Sigma}V^T$$ $$V\ contains\ the\ eigenvectors\ of\ A^TA\ as\ columns$$ $$U\ contains\ the\ eigenvectors\ of\ AA^T\ as\ columns$$ You do not need to calculate both $U$ and $V$, since they can be derived from the other: $$(1)\ U=\left[\begin{matrix}|&|\\u_1&u_2\\|&|\\\end{matrix}\right]\ with\ u_i=\frac{Av_i}{\sigma_i}$$

$$(2)\ V=\left[\begin{matrix}|&|\\v_1&v_2\\|&|\\\end{matrix}\right]\ with\ v_i=\frac{A^Tu_i}{\sigma_i}$$

The polar decomposition looks as follows:

$$(3)\ A=\ \left(U\mathrm{\Sigma}U^{T}\right)\ \left(UV^T\right)=S_A\ R_\theta$$ $$(4)\ A=U\mathrm{\Sigma}V^T=\left(UV^T\right)\left(V\mathrm{\Sigma}V^T\right)={R_\theta} S_{AT}$$

So to calculate the polar decomposition you

  1. Calculate the eigenvalues $\lambda_i$ and eigenvectors $v_i$ of $A^TA$
  2. The matrix $\Sigma$ contains $\sigma_i=\sqrt{\lambda_i}$ on its diagonal
  3. Derive eigenvectors $u_i$ of $AA^T$ using (1)
  4. Now calculate $UV^T$
  5. if I am not mistaken $\left(V\mathrm{\Sigma}V^{T}\right)=A^TA$, so (3) can be completed
  6. Now you need to calculate $AA^T$ or $\left(U\mathrm{\Sigma}U^{T}\right)$, to complete (4)

    A geometric analysis of polar decomposition can be found on Heaviside's Dinner.
    An interactive geogebra plot can be found on the geogebra website: polar decomposition.
Bart Vanderbeke
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A brute force (numerical) method for obtaining the right polar decomposition has been presented in this answer at MSE. When applied to the OP's problem (Test4) we get an outcome like: $$ \begin{bmatrix} 2.000000 & -3.000000 \\ 1.000000 & 6.000000 \end{bmatrix} = \\ \begin{bmatrix} 0.894427 & -0.447214 \\ 0.447214 & 0.894427 \end{bmatrix} \begin{bmatrix} 2.236068 & 0.000000 \\ 0.000000 & 6.708204 \end{bmatrix} $$ However, when searching for the number $0.447214$ on the internet, we find that it's approximately equal to What is 1 over square root of 5?. So let's do an educated guess: $$ \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} = \begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix} \begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix} $$ It is noticed that the above question is similar to another one.

EDIT. But the above is, of course, overkill.
Here comes a much simpler solution. First form the transpose times the original matrix: $$ \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}^T\begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -3 & 6 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 45 \end{bmatrix} $$ Then make the following Ansatz: $$ \begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}^2 = \begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix} = \begin{bmatrix} a^2 & 0 \\ 0 & c^2 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 45 \end{bmatrix} $$ So the square root of a diagonal matrix is extremely simple; just take the square roots of the diagonal elements: $$ \begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix} = \begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix} $$ At last, the orthogonal matrix is found with: $$ \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix}^{-1} = \\ \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} 1/\sqrt{5} & 0 \\ 0 & 1/(3\sqrt{5}) \end{bmatrix} = \begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix} $$ Same conclusion: $$ \begin{bmatrix} 2 & -3 \\ 1 & 6 \end{bmatrix} = \begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix} \begin{bmatrix} \sqrt{5} & 0 \\ 0 & 3\sqrt{5} \end{bmatrix} $$

Han de Bruijn
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