In the following statement I don't understand the case for $\ i = j$:

Let $\mathbf A$ be an $\ m \times \ n$ orthogonal matrix where $\ a_i$ is the $\ i^{th}$ column vector. The $\ ij^{th} $ element of $\mathbf A^{T}\mathbf A$ is

$$ \left(\mathbf A^T \mathbf A\right)_{ij} = a_i^{T} a_j = \begin{cases} 1 &\text{if} &i = j\\ 0 &\text{if} &\text{otherwise}\\ \end{cases} $$

Therefore, because $\mathbf A^{T} \mathbf A = \mathbf I $, it follows that $\mathbf A^{-1} = \mathbf A^{T}$.

  • 3
    $\begingroup$ This question seems like a better fit for math.SE. $\endgroup$
    – Jason R
    Jan 17, 2017 at 20:25

2 Answers 2


An orthogonal matrix has orthogal columns, i.e. the scalar product of two different columns is zero (the case $i\neq j$). For the case $i=j$ you have $a_i^Ta_i=\|a_i\|^2>0$.

So, all you can say about an orthogonal matrix with colums $a_i$ is:

$$ a_i^Ta_j=\begin{cases}c_i>0 & i=j\\ 0 & i\neq j\end{cases}$$

where $c_i>0$ is some constant, depending on the matrix.

Then, there are orthonormal matrices. These fulfill, in addition to the previous statement, $c_i=1$ and hence $A^TA=I$.

Finally, only if your $A$ is a square orthonormal matrix, it holds $A^{-1}=A^T$.

The task you describe has two flaws:

  • it only talks about ortogonal matrices, where we can just say $A^TA=\text{diag}(\vec{c})$ with $\vec{c}$ contains the norms of the columns.
  • It states $A$ is an $n\times m$ matrix (i.e. not square). Rectangular matrices dont have an inverse. So, the last statement is not correct.

Edit: Normally in literature, orthogonal matrices are considered to be square with unit-norm columns. Then $A^TA=I$ and $A^{-1}=A^T$. However, as the question explicitely mentions rectangular matrices, I want to point out that it really depends on your definition of what an orthogonal matrix is. And, if you just define it as orthogonal columns, the arguments above hold.

  • $\begingroup$ The usual definition seems to be that an orthogonal matrix is a square matrix with orthonormal columns. So I disagree with your flaw#1. Eg. See Gilbert Strang's Linear Algebra 4th Ed. Pg. 175: "Orthonormal matrix would have been a better name, but it is too late to change. Also, there is no accepted word for a rectangular matrix with orthonormal columns. We still write Q, but we won't call it an "orthogonal matrix" unless it is square." $\endgroup$
    – Atul Ingle
    Jan 17, 2017 at 20:06

Suppose you have a set $n$ vectors $o_k$ in $\mathbb{R}^{m,1}$ (length $m$, column-style), pairwise orthonormal, i.e. orthogonal with unit norm. Stack them side by side in a matrix $$A = \left[o_1,\ldots,o_n\right]\,.$$

You have what you call an orthogonal (rectangular) matrix, sometimes called an orthogonal column matrix. The same concept applies row-wise. First, note that you should have $m\ge n$. If not, it is not possible to find strictly more than $m$ linearly independent vectors in an $m$ dimensional linear space.

Then, each element $p_{i,j}$ of $A^T A$ is given by $o_i^T o_j$, so $1$ when $i=j$ and $0$ otherwise, by definition of the orthonormality of the $o_k$.

Note that for non-square matrices, there exists so-called generalized inverses, such as the Moore-Penrose pseudo-inverse, that coincides with the standard inverse in the square case..


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