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.

  • 2
    $\begingroup$ Think about $Q^TQ$. $\endgroup$ Feb 11, 2018 at 7:26
  • $\begingroup$ The way you show $Q^{-1}= Q^T$ is to show $QQ^T=I$. $\endgroup$
    – saulspatz
    Feb 11, 2018 at 7:28

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.