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Equivalence Conditions for Orthogonal Matrices 📂Matrix Algebra

Equivalence Conditions for Orthogonal Matrices

Theorem

For a real matrix AA, the following propositions are all equivalent.

(a) AA is an orthogonal matrix.

(b) The set of row vectors of AA forms a normal orthogonal set in Rn\mathbb{R}^n.

(c) The set of column vectors of AA forms a normal orthogonal set in Rn\mathbb{R}^n.

(d) AA preserves inner product, i.e., for all x,y∈Rn\mathbf{x},\mathbf{y}\in \mathbb{R}^{n}, the following holds:

(Ax)⋅(Ay)=x⋅y (A \mathbf{x}) \cdot (A\mathbf{y}) = \mathbf{x} \cdot \mathbf{y}

(e) AA preserves length, i.e., for all x∈Rn\mathbf{x}\in \mathbb{R}^{n}, the following holds:

∥Ax∥=∥x∥ \left\| A \mathbf{x} \right\| = \left\| \mathbf{x} \right\|

Proof

(a)  ⟺  \iff(b) and (a)  ⟺  \iff(c) are proven in the same way, so the latter is omitted.

(a)   ⟺  \iff (b)

Let’s denote AA as a n×nn\times n matrix and its row vectors by ri\mathbf{r}_{i}.

A=[r1r2⋮rn] A = \begin{bmatrix} \mathbf{r}_{1} \\ \mathbf{r}_{2} \\ \vdots \\ \mathbf{r}_{n} \end{bmatrix}

Then AATA A^{T} is as follows:

AAT=[r1r2⋮rn][r1Tr2T⋯rnT]=[r1⋅r1r1⋅r2⋯r1⋅rnr2⋅r1r2⋅r2⋯r2⋅rn⋮⋮⋱⋮rn⋅r1rn⋅r2⋯rn⋅rn] AA^{T} = \begin{bmatrix} \mathbf{r}_{1} \\ \mathbf{r}_{2} \\ \vdots \\ \mathbf{r}_{n} \end{bmatrix} \begin{bmatrix} \mathbf{r}_{1}^{T} & \mathbf{r}_{2}^{T} & \cdots & \mathbf{r}_{n}^{T} \end{bmatrix} = \begin{bmatrix} \mathbf{r}_{1} \cdot \mathbf{r}_{1} & \mathbf{r}_{1} \cdot \mathbf{r}_{2} & \cdots & \mathbf{r}_{1} \cdot \mathbf{r}_{n} \\ \mathbf{r}_{2} \cdot \mathbf{r}_{1} & \mathbf{r}_{2} \cdot \mathbf{r}_{2} & \cdots & \mathbf{r}_{2} \cdot \mathbf{r}_{n} \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{r}_{n} \cdot \mathbf{r}_{1} & \mathbf{r}_{n} \cdot \mathbf{r}_{2} & \cdots & \mathbf{r}_{n} \cdot \mathbf{r}_{n} \end{bmatrix}

  • (a)   ⟹  \implies (b)

    Assume AA is an orthogonal matrix. Then AAT=IAA^{T}=I, which leads to:

    ri⋅rj={1,i=j0,i≠j \mathbf{r}_{i} \cdot \mathbf{r}_{j} = \begin{cases} 1, & i=j \\ 0, & i\ne j \end{cases}

    Hence, the set of row vectors of AA {ri}\left\{ \mathbf{r}_{i} \right\} forms a normal orthogonal set.

  • (a) ⟸\Longleftarrow (b)

    Suppose {ri}\left\{ \mathbf{r}_{i} \right\} forms a normal orthogonal set. Then,

    ri⋅rj={1,i=j0,i≠j \mathbf{r}_{i} \cdot \mathbf{r}_{j} = \begin{cases} 1, & i=j \\ 0, & i\ne j \end{cases}

    leads to

    AAT=[r1⋅r1r1⋅r2⋯r1⋅rnr2⋅r1r2⋅r2⋯r2⋅rn⋮⋮⋱⋮rn⋅r1rn⋅r2⋯rn⋅rn]=[10⋯001⋯0⋮⋮⋱⋮00⋯1]=I AA^{T} = \begin{bmatrix} \mathbf{r}_{1} \cdot \mathbf{r}_{1} & \mathbf{r}_{1} \cdot \mathbf{r}_{2} & \cdots & \mathbf{r}_{1} \cdot \mathbf{r}_{n} \\ \mathbf{r}_{2} \cdot \mathbf{r}_{1} & \mathbf{r}_{2} \cdot \mathbf{r}_{2} & \cdots & \mathbf{r}_{2} \cdot \mathbf{r}_{n} \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{r}_{n} \cdot \mathbf{r}_{1} & \mathbf{r}_{n} \cdot \mathbf{r}_{2} & \cdots & \mathbf{r}_{n} \cdot \mathbf{r}_{n} \end{bmatrix} = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} =I

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(a)   ⟺  \iff (d)   ⟺  \iff (e)

  • (a)   ⟹  \implies (d)

    Assume AA is an orthogonal matrix and x\mathbf{x}, y∈Rn\mathbf{y} \in \mathbb{R}^{n}. Then by the property of the transpose matrix and assumption, the following equation holds:

    (Ax)⋅(Ay)=(Ax)T(Ay)=xTATAy=xTy=x⋅y \begin{align*} \left( A \mathbf{x} \right) \cdot \left( A \mathbf{y} \right) &= \left( A \mathbf{x} \right)^{T} \left( A \mathbf{y} \right) \\ &= \mathbf{x}^{T} A^{T} A \mathbf{y} \\ &= \mathbf{x}^{T} \mathbf{y} \\ &= \mathbf{x} \cdot \mathbf{y} \end{align*}

  • (d)   ⟹  \implies (e)

    Assuming AA preserves the inner product, the following equation by assumption:

    ∥Ax∥=(Ax)⋅(Ax)=x⋅x=∥x∥ \begin{align*} \left\| A \mathbf{x} \right\| &= \sqrt{(A \mathbf{x}) \cdot (A \mathbf{x})} \\ &= \sqrt{\mathbf{x} \cdot \mathbf{x}} \\ &= \left\| \mathbf{x} \right\| \end{align*}

  • (e)   ⟹  \implies (a)

    Assuming ∥Ax∥=∥x∥\left\| A \mathbf{x} \right\| =\left\| \mathbf{x}\right\| holds, the following equation is true:

    ∥Ax∥=∥x∥  ⟹  ∥Ax∥2=∥x∥2  ⟹  Ax⋅Ax=x⋅x \begin{align*} && \left\| A \mathbf{x} \right\| &= \left\| \mathbf{x}\right\| \\ \implies && \left\| A \mathbf{x} \right\|^{2} &= \left\| \mathbf{x}\right\|^{2} \\ \implies && A \mathbf{x} \cdot A \mathbf{x} &= \mathbf{x} \cdot \mathbf{x} \end{align*}

    By the property of the inner product Au⋅v=u⋅ATvA \mathbf{u} \cdot \mathbf{v} = \mathbf{u} \cdot A^{T} \mathbf{v}, the above equation is:

    Ax⋅Ax=x⋅ATAx=x⋅x A \mathbf{x} \cdot A \mathbf{x} = \mathbf{x} \cdot A^{T} A \mathbf{x} = \mathbf{x} \cdot \mathbf{x}

    Rearranging gives:

    x⋅(ATAx−x)=0 \mathbf{x} \cdot \left( A^{T}A\mathbf{x} -\mathbf{x}\right) = 0

    This must hold for all x\mathbf{x}, hence (ATAx−x)=0\left( A^{T}A\mathbf{x} -\mathbf{x}\right) = 0. Therefore,

    (ATAx−x)=0  ⟹  (ATA−I)x=0 \begin{align*} && \left( A^{T}A\mathbf{x} -\mathbf{x} \right) &= 0 \\ \implies && \left( A^{T}A -I\right) \mathbf{x} &= 0 \end{align*}

    This also must satisfy for all x\mathbf{x}, leading to:

    ATA−I=0  ⟹  ATA=I A^{T}A-I = 0 \implies A^{T}A=I

    Hence, AA is an orthogonal matrix.

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