7D Isometries, Unitary Operators, and QR Factorization

Let $S \in \mathscr{L}(V,W)$, $(e_1, \ldots, e_n)$ an orthonormal basis for $V$ and $(f_1, \ldots, f_m)$ an orthonormal basis for $W$. Then the following are equivalent: a) $S$ is an isometry. b) $S^*S = I.$ c) $\langle S(u), S(v) \rangle = \langle u, v \rangle$ for every $u, v \in V$. d) $(S(e_1), \ldots, S(e_n))$ is an orthonormal list in $W$. e) The columns of $\mathcal{M}(S, (e_1, \ldots, e_n), (f_1, \ldots, f_m))$ form an orthonormal list of vectors in $\mathbb{F}^m.$

a) $\Rightarrow$ b): We will use the fact that $I - S^*S$ is self-adjoint to show that it is the zero operator as follows:

Therefore, $I - S^*S$ must be the zero operator. b) $\Rightarrow$ c):

c) $\Rightarrow$ d): Because the inner product does not change after applying $S$, $(S(e_1), \ldots, S(e_n))$ must also be orthonormal. d) $\Rightarrow$ e): This follows from the last result because the coordinates of the columns are from an orthonormal basis. So the inner product becomes the standard inner product on $\mathbb{F}^m.$ e) $\Rightarrow$ a): Because we are working with an orthonormal basis for $W$, the fact that the columns are orthonormal mean the images of $e_1$ through $e_n$ form an orthonormal list in $W$. Now, because the inner product is biliniar, we only needed to show that this property held for the basis vectors, so we're done.

Suppose $S \in \mathscr{L}(V)$. Suppose $(e_1,\ldots, e_n)$ is an orthonormal basis of $V$. Then the following are equivalent: a) $S$ is a unitary operator. b) $S^*S = SS^* = I$ c) $S$ is invertible with inverse $S^*$. d) $(S(e_1), \ldots, S(e_n))$ is an orthonormal basis for $V$. e) The rows of $\mathcal{M}(S, (e_1, \ldots, e_n))$ form an orthonormal basis for $\mathbb{F}^n$ with respect to the standard inner product on $\mathbb{F}^n.$ f) $S^*$ is a unitary operator.

a) $\Rightarrow$ b): Since $S$ is unitary, it is invertible and an isometry (and hence $S^*S = I$). So

Therefore $SS^* = SS^{-1} = I.$ b) $\Rightarrow$ c): Because $\ker(S) \subseteq \ker(S^*S) = \ker(I) = \{ 0 \}$, $S$ must be injective. And because it is an operator, $S$ must therefore be invertible. Furthermore,

c) $\Rightarrow$ d): We know from the last result in the previous section that $S^*S = I$ is equivalent to $\left( S(e_1), \ldots, S(e_n) \right)$ being an orthonormal basis (parts b and d in particular). So this just follows from that equivalence. d) $\Rightarrow$ e): The main idea behind this proof is that the adjoint is the conjugate transpose and thus has columns that are conjugates of an orthonormal basis, which also form an orthonormal basis. The full proof is longer: Suppose $(S(e_1), \ldots, S(e_n))$ is an orthonormal basis for $V$. Then from the last result of the previous section, this is equivalent to $S$ being an isometry. Because $S$ is an injective operator on a finite-dimensional vector space, $S$ must be invertible, so $S$ is a unitary operator (an invertible isometry). Now, from part e) of the last result in the previous section, this means that the columns of our matrix with respect to this basis must form an orthonormal basis for $\mathbb{F}^n$. This finally means that the rows also form an orthonormal basis, since they are just the conjugates of the matrix of the adjoint, which means they are also an orthonormal basis. e) $\Rightarrow$ f): Suppose the rows of $\mathcal{M}(T)$ form an orthonormal basis for $\mathbb{F}^n$. Then the columns of $\mathcal{M}(S)$ also form an orthonormal basis for $\mathbb{F}^n$ and thus $S$ is an isometry. So $S$ and $S^*$ are inverses of each other. f) $\Rightarrow$ a): Suppose $S^*$ is unitary and apply all of the previous implications to $S^*$, showing that $(S^*)^*$ is also unitary. And since $(S^*)^*=S$, we are done.

Suppose $\lambda$ is an eigenvalue of a unitary operator. Then $| \lambda | = 1$.

This follows from teh preservation of the inner product, and hence the norm: If $v \in V$ is an eigenvector with eigenvalue $\lambda \in \mathbb{F}$, then

Suppose $\mathbb{F} = \mathbb{C}$ and $S \in \mathscr{L}(V)$. Then the following are equivalent: a) $S$ is a unitary operator. b) There is an orthonormal basis of $V$ consisting of eigenvectors of $S$ whose corresponding eigenvalues all have absolute value 1.

$\Rightarrow$: If $S$ is unitary then it is normal. So by the complex spectral theorem, it is diagonalizable with respect to an orthonormal basis. We also know that its eigenvalues must have an absolute value of 1, so we're done. $\Leftarrow$: Let $(e_1, \ldots, e_n)$ be an orthonormal basis of eigenvectors of $S$ with eigenvalues with absolute value 1. Then the $S$-images of these vectors are also orthonormal, as

Now, this means that $S$ is unitary from the first result of this chapter.

Suppose $Q$ is an $n \times n$ matrix. Then the following are equivalent. a) $Q$ is a unitary matrix. b) The rows of $Q$ form tan orthonormal list in $\mathbb{F}^n$. c) $\Vert Q(v) \Vert = \Vert v \Vert$ for every $v \in \mathbb{F}^n.$ d) $Q^*Q = QQ^* = I$, where $I$ represents the identity matrix.

(Proof idea) The matrix $Q$ is a linear operator from $\mathbb{F}^n$ to $\mathbb{F}^n$. All of these conditions have been proven to be equivalent for operators.

Suppose $A$ is a square matrix with linearly independent columns. Then there exist unique matrices $Q$ and $R$, where $Q$ is unitary, $R$ is upper-triangular, with only positive numbers on its diagonal, and $A = QR$.

Apply the Gram-Schmidt procedure to the columns of $A$. $Q$'s columns consist of these orthonormal basis vectors. Now, $R$ is the right-inverse of the matrix whose columns would apply the linear combinations resulting in the Gram-Schmidt procedure on $A$ to result in $Q$. Specifically, each entry of $R$ can be found as $R_{j,k} = \langle v_k, e_j \rangle,$ where $v_k$ is the $k$th column of $A$ and $e_j$ is the $j$th column of $Q$. Then $Q$ is unitary because its columns are orthonormal. Furthermore, $R$ is upper-triangular because all of the changes in the Gram-Schmidt procedure add scalar multiples of previous vectors only, making values below the diagonal all zero. Furthermore, the diagonals are positive because the last step is dividing by the norm, which is positive.