# Show non-singularity of orthogonal matrix

I'm given a question that says:

A matrix $Q$ of size $n \times n$ is called orthogonal if its columns are orthogonal to each other and all columns have length $1$.

a) Show that the matrix is non-singular

b) Show that $Q^{-1}$ = $Q^T$

I've seen a concrete example on how to determine whether an orthogonal matrix is non-singular, but I'm struggling to figure out how to apply that to this question. For part b as well, I'm not sure how to properly show the two are equal without just writing the same matrix twice.

• Think about $Q^TQ$. Feb 11, 2018 at 7:26
• The way you show $Q^{-1}= Q^T$ is to show $QQ^T=I$. Feb 11, 2018 at 7:28

Suppose $A$ is a orthogonal matrix (as by your definition). Then
\begin{eqnarray} (AA^T)_{i,j} &=& \sum_{k=1}^n A_{ik}A^T_{kj}\\ &=& \sum_{k=1}^n A_{ik}A_{jk}.\\ \end{eqnarray} Notice that that expression is simply the inner product between the $i$-th and $j$-th row of $A$. These are orthogonal to each other by definition (well actually the columns are, but you can simply consider the transpose to get orthogonal rows or consider $A^TA$).
It follows that $AA^T=Id$ and thus $A$ is invertible and both statements follow. Notice that a one-sided inverse is enough when dealing with square matrices to conclude that this is a two-sided inverse.