Polar Decomposition

Definition Let $A \in \mathbb{F}^{m \times n}$. The matrix $A^{*}A \in \mathbb{F}^{n \times n}$ is Hermitian, since $(A^{*}A)^{*} = A^{*}A$, and positive semidefinite, because $x^{*}A^{*}Ax = \lVert Ax \rVert^2 \ge 0$ for every $x \in \mathbb{F}^n$. The absolute value of $A$ is the unique Hermitian positive semidefinite square root of $A^{*}A$,

It acts on the input space $\mathbb{F}^n$, and it is real symmetric when $A$ is real. For every $x \in \mathbb{F}^n$

so $\lvert A \rvert$ and $A$ give every input the same length. Hence $\ker \lvert A \rvert = \ker A$, and since $\lvert A \rvert$ is Hermitian, its image is $(\ker A)^{\perp}$. Suppose $r = \operatorname{rank} A \ge 1$ and $A = L_r\Sigma_r V_r^{*}$ is a compact SVD, where $L_r = [u_1 \ \cdots \ u_r]$ and $V_r = [v_1 \ \cdots \ v_r]$ have orthonormal columns and $\Sigma_r = \operatorname{diag}(\sigma_1, \dots, \sigma_r)$ holds the positive singular values. Then

because this matrix is Hermitian positive semidefinite and its square is $V_r\Sigma_r^2 V_r^{*} = A^{*}A$. So the positive eigenvalues of $\lvert A \rvert$ are the positive singular values of $A$, with eigenvectors $v_1, \dots, v_r$. The left factor of the compact SVD is written $L_r$ here rather than $U_r$, so that the letter $U$ stays free for the polar factor below.

Definition Let $A \in \mathbb{F}^{m \times n}$. There is exactly one matrix $U \in \mathbb{F}^{m \times n}$ such that

This factorization is the polar decomposition of $A$, and $U$ is its canonical polar factor. For existence, define $U$ on the image of $\lvert A \rvert$ by $U(\lvert A \rvert x) = Ax$. This is well defined, since $\lvert A \rvert x = \lvert A \rvert y$ gives $\lVert A(x - y) \rVert = \lVert \lvert A \rvert (x - y) \rVert = 0$. It is linear and keeps lengths on $\operatorname{im}\lvert A \rvert = (\ker A)^{\perp}$, and then $U$ is set to $0$ on $\ker A$. For uniqueness, any $W$ with $A = W\lvert A \rvert$ satisfies $W(\lvert A \rvert x) = Ax$, so it agrees with $U$ on $(\ker A)^{\perp}$, and the kernel condition fixes it on $\ker A$. In a compact SVD as above,

and $U\lvert A \rvert = L_r V_r^{*}V_r\Sigma_r V_r^{*} = L_r\Sigma_r V_r^{*} = A$. Since $L_r\Sigma_r$ is one to one, $\ker A = \ker V_r^{*}$, so this $U$ also vanishes on $\ker A$. If $A = 0$ both factors are $0$. Equivalently $U = A\lvert A \rvert^{+}$, with the pseudoinverse of $\lvert A \rvert$. The positive factor is forced as well. If $A = WP$ with $P \in \mathbb{F}^{n \times n}$ Hermitian positive semidefinite and $W^{*}WP = P$, for instance with $W$ an isometry, then $A^{*}A = PW^{*}WP = P^2$, so $P = \lvert A \rvert$. When $m = n$ and $A$ is invertible, $\lvert A \rvert$ is positive definite, the kernel condition holds automatically, and

is unitary, since $U^{*}U = \lvert A \rvert^{-1}A^{*}A\lvert A \rvert^{-1} = I$. Over $\mathbb{R}$ it is orthogonal. It is then the only matrix of any kind with $A = U\lvert A \rvert$. For real $A$ all factors are real.

Definition A matrix $W \in \mathbb{F}^{m \times n}$ is a partial isometry if

Its initial space is $(\ker W)^{\perp}$ and its final space is $\operatorname{im} W$. By the polarization identity $W$ preserves inner products on the initial space, so it maps that space one to one onto the final space, and the two have the same dimension. Equivalently, $W^{*}W$ is the orthogonal projector onto the initial space, and then $WW^{*}$ is the orthogonal projector onto the final space. The canonical polar factor $U$ of $A$ is a partial isometry with initial space $(\ker A)^{\perp}$ and final space $\operatorname{im} A$. With the pseudoinverse,

So $U^{*}U = I_n$ exactly when $\ker A = \{0\}$, and $UU^{*} = I_m$ exactly when $\operatorname{im} A = \mathbb{F}^m$.

Definition A matrix $W \in \mathbb{F}^{m \times n}$ is an isometry if $\lVert Wx \rVert = \lVert x \rVert$ for every $x \in \mathbb{F}^n$. Equivalently $W^{*}W = I_n$, which says that the columns of $W$ are orthonormal. An isometry is one to one, so it needs $n \le m$, and $WW^{*}$ is the orthogonal projector onto $\operatorname{im} W$, equal to $I_m$ only when $m = n$. A square isometry is unitary, with $W^{*}W = WW^{*} = I$, and a real one is orthogonal. A matrix with $WW^{*} = I_m$, that is with orthonormal rows, is a coisometry. It needs $m \le n$, and its adjoint $W^{*}$ is an isometry. Isometries and coisometries are partial isometries, with initial space $\mathbb{F}^n$ for an isometry and final space $\mathbb{F}^m$ for a coisometry. For $A \in \mathbb{F}^{m \times n}$ of rank $r$, the canonical polar factor $U$ is

• an isometry exactly when $r = n$, that is when $A$ has full column rank,
• a coisometry exactly when $r = m$, that is when $A$ has full row rank,
• unitary exactly when $m = n$ and $A$ is invertible.

When $r < \min(m, n)$, neither $U^{*}U$ nor $UU^{*}$ is an identity.

Definition Let $A \in \mathbb{F}^{n \times n}$ have rank $r$ and canonical polar factor $U$. A unitary extension of $U$ is a unitary matrix $W \in \mathbb{F}^{n \times n}$, orthogonal when $\mathbb{F} = \mathbb{R}$, that agrees with $U$ on $(\ker A)^{\perp}$. Because $A = U\lvert A \rvert$ and $\operatorname{im}\lvert A \rvert = (\ker A)^{\perp}$, a unitary $W$ satisfies $A = W\lvert A \rvert$ exactly when it agrees with $U$ there, so the unitary extensions are exactly the unitary matrices with $A = W\lvert A \rvert$. Such a $W$ maps $(\ker A)^{\perp}$ onto $\operatorname{im} A$, so it maps $\ker A$ onto $(\operatorname{im} A)^{\perp}$. Conversely, every isometry $\Phi$ from $\ker A$ onto $(\operatorname{im} A)^{\perp}$, both of dimension $n - r$, combined with $U$ on $(\ker A)^{\perp}$, gives a unitary extension. So unitary extensions always exist, and

• if $r = n$, the only one is $U$ itself,
• if $r < n$ and $\mathbb{F} = \mathbb{C}$, there are infinitely many, since $\Phi$ can be multiplied by any $e^{i\theta}$,
• if $r < n$ and $\mathbb{F} = \mathbb{R}$, there are at least two, from $\Phi$ and $-\Phi$, exactly two when $n - r = 1$, and infinitely many when $n - r \ge 2$, since $\Phi$ can be composed with any rotation of $\ker A$.

A unitary matrix has zero kernel, so when $r < n$ no unitary extension vanishes on $\ker A$, and the uniqueness of $U$ is not contradicted. For a rectangular $A \in \mathbb{F}^{m \times n}$ with $m \ge n$ the same recipe, with an isometry from $\ker A$ into $(\operatorname{im} A)^{\perp}$, whose dimensions are $n - r \le m - r$, gives an isometric extension $W$ with $W^{*}W = I_n$ and $A = W\lvert A \rvert$. When $m < n$ no isometry on all of $\mathbb{F}^n$ exists. A coisometric extension with $WW^{*} = I_m$ and $A = W\lvert A \rvert$ can be built instead, by mapping an $(m - r)$-dimensional subspace of $\ker A$ isometrically onto $(\operatorname{im} A)^{\perp}$ and sending the orthogonal complement of that subspace inside $\ker A$ to $0$.

Definition Let $A \in \mathbb{F}^{m \times n}$ and let

be the Hermitian positive semidefinite square root of $AA^{*}$, which is $\lvert A^{*} \rvert$. With the same canonical polar factor $U$,

This is the polar form with the positive factor on the left. In a compact SVD $K = L_r\Sigma_r L_r^{*}$, and since $L_r^{*}L_r = V_r^{*}V_r = I_r$,

and multiplying out $U\lvert A \rvert U^{*}$ the same way gives $K$. The uniqueness condition now sits on the output side. Among all $W$ with $A = KW$, the canonical $U$ is the only one with $\operatorname{im} W \subseteq \operatorname{im} A$, because $K$ is one to one on $\operatorname{im} K = \operatorname{im} A$. For a nonzero $A$ whose image is not all of $\mathbb{F}^m$, vanishing on $\ker A$ does not pin $W$ down in this form. Adding to $U$ a nonzero map that is zero on $\ker A$ and takes values in $\ker K = (\operatorname{im} A)^{\perp}$ gives another such $W$. When $A$ is square and invertible, $U = K^{-1}A$ is the same unitary as in $A = U\lvert A \rvert$. The two positive factors act on different sides, $\lvert A \rvert$ on the input space $\mathbb{F}^n$ and $K$ on the output space $\mathbb{F}^m$, and for square $A$ they are equal exactly when $A^{*}A = AA^{*}$, that is when $A$ is normal.