Question

If A is an $\mathbf{n}\mathbf{\times }\mathbf{n}$matrix such that role="math" localid="1659514225617" ${\mathbf{AA}}^{\mathbf{T}}\mathbf{=}{\mathbf{I}}_{\mathbf{n}}$, then A must be an orthogonal matrix.

Short answer

The given statement is true.

Step-by-step solution

Recollect the properties of the orthogonal a matrix.

According to the properties of a orthogonal matrix, a $\mathrm{n}\times \mathrm{n}$matrix Ais an orthogonal matrix, if$\left|\left|\mathrm{A}\stackrel{\rightharpoonup }{\mathrm{x}}\right|\right|=\left|\left|\stackrel{\rightharpoonup }{\mathrm{x}}\right|\right|\mathrm{for}\mathrm{all}\stackrel{\rightharpoonup }{\mathrm{x}}\in {\mathrm{R}}^{\mathrm{n}}$.

The relation between dot and matrix product is defined as$\stackrel{\rightharpoonup }{\mathrm{v}}.\stackrel{\rightharpoonup }{\mathrm{w}}={\stackrel{\rightharpoonup }{\mathrm{v}}}^{\mathrm{T}}\stackrel{\rightharpoonup }{\mathrm{w}}.$

Check whether the given statement is a true or false.

The given statement is if A is an $n\times n$matrix such that$A{A}^T=I_n$ , then A must be orthogonal matrix.

To prove A is orthogonal, prove that$\left|\left|A\stackrel{\rightharpoonup }{x}\right|\right|=\left|\left|\stackrel{\rightharpoonup }{x}\right|\right|$.

$$\begin{gathered} {\left|\left|A\stackrel{\rightharpoonup }{x}\right|\right|}^2=\left(A\stackrel{\rightharpoonup }{x}\right).\left(A\stackrel{\rightharpoonup }{x}\right) \\ ={\left(A\stackrel{\rightharpoonup }{x}\right)}^T\left(A\stackrel{\rightharpoonup }{x}\right) \\ ={\left(\stackrel{\rightharpoonup }{x}\right)}^T{A}^T\left(A\stackrel{\rightharpoonup }{x}\right) \\ ={\left(\stackrel{\rightharpoonup }{x}\right)}^T{A}^{T}A\stackrel{\rightharpoonup }{x} \end{gathered}$$

Solve further as,

$$\begin{gathered} {\left|\left|A\stackrel{\rightharpoonup }{x}\right|\right|}^2={\left(\stackrel{\rightharpoonup }{x}\right)}^T\left(A^{T}A\right)\stackrel{\rightharpoonup }{x} \\ ={\left(\stackrel{\rightharpoonup }{x}\right)}^T\left(I_n\right)\stackrel{\rightharpoonup }{x} \\ ={\left(\stackrel{\rightharpoonup }{x}\right)}^T\stackrel{\rightharpoonup }{x} \\ =\stackrel{\rightharpoonup }{x.}\stackrel{\rightharpoonup }{x} \end{gathered}$$

Solve further as,

$$\begin{gathered} {\left|\left|A\stackrel{\rightharpoonup }{x}\right|\right|}^2={\left|\left|\stackrel{\rightharpoonup }{x}\right|\right|}^2 \\ \left|\left|A\stackrel{\rightharpoonup }{x}\right|\right|=\left|\left|\stackrel{\rightharpoonup }{x}\right|\right| \end{gathered}$$

So,$\left|\left|A\stackrel{\rightharpoonup }{x}\right|\right|=\left|\left|\stackrel{\rightharpoonup }{x}\right|\right|$

This means, A is orthogonal.

Then the given statement is true.