Question

If Ais any orthogonal matrix, then matrix $\mathbf{A}\mathbf{+}{\mathbf{A}}^{\mathbf{-}\mathbf{1}}$is diagonalizable (over R).

Short answer

The given statement is TRUE.

Step-by-step solution

Check whether the given statement is TRUE or FALSE

It is known that an orthogonal matrix is a square matrix whose rows and columns are the orthogonal unit vectors.

If A is orthogonal, then ${\mathrm{A}}^{-1}={\mathrm{A}}^{\mathrm{T}}$. Thus,$\mathrm{A}={\mathrm{A}}^{-\mathrm{T}}=\mathrm{A}+{\mathrm{A}}^{\mathrm{T}}$

$$\begin{gathered} {\left(\mathrm{A}+{\mathrm{A}}^{-1}\right)}^{\mathrm{T}}={\left(\mathrm{A}+{\mathrm{A}}^{\mathrm{T}}\right)}^{\mathrm{T}} \\ ={\mathrm{A}}^{\mathrm{T}}+\mathrm{A} \\ =\mathrm{A}+{\mathrm{A}}^{\mathrm{T}} \\ =\mathrm{A}+{\mathrm{A}}^{-1} \end{gathered}$$

This shows that $\mathrm{A}+{\mathrm{A}}^{-1}$is symmetric and hence diagonalizable.

Final Answer

The given statement is TRUE.