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I have a confusion about orthogonal matrix.

  1. If columns of a square matrix are orthonormal to each other, is the matrix orthogonal?
  2. If yes, then are the rows of the matrix also orthonormal? Why?
  3. Why is it that QQ'=I? I get Q'Q=I but why QQ' is also I?

Thanks, Tom

Tom Bennett
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3 Answers3

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  1. Yes, that's the definition.
  2. From $QQ^\top=I$.
  3. Since $Q^\top Q=I$, $Q^\top=Q^{-1}$. So $QQ^\top=QQ^{-1}=I$.
Joe Li
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The key to your last question is that if a matrix $A$ has inverse $B$, then $$ AB = BA = I $$ Where $I$ is the identity matrix. That is, a finite square matrix always commutes with its inverse. From there, it's clear that if $Q'Q=I$ (that is, $Q'$ is the inverse of $Q$), then $QQ'=I$

The easy answer to your second question is that $Q$ is an orthogonal matrix whenever $Q'Q=I$, which means that $QQ'=I$. Now, if $Q$ is an orthogonal matrix, then $$(Q')'Q'=QQ'=I$$ which means that $Q'$ is orthogonal. This in turn means that $Q'$ has orthonormal columns, which means that $Q$ has orthonormal rows.

Ben Grossmann
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If the columns of a square matrix are orthonormal, then the matrix is said to be orthogonal.

Note: The following are equivalent for a $n \times n$ matrix $P$.

$1)P$ is invertible and $P^{-1}=P^{T}$

$2)$ The rows of $P$ are orthonormal

$3)$ The columns of $P$ are orthonormal.

Hint:

First recall that condition $(1)$ is equivalent to $PP^{T}=I$ because of the fact that a square matrix is invertible if and only if $A^{T}$ is invertible. Let $x_1,x_2,\ldots,x_n$ denote the rows of $P$. Then $x^T_j$ is the $j$th column of $P^{T}$, so the $(i,j)$-entry of $PP^{T}$ is $x_i \cdot x_j$. Thus $PP^{T}=I$ means that $x_i \cdot x_j=0$ if $i \neq j$ and $x_i \cdot x_j=1$ if $i=j$. Hence condition $(1)$ is equivalent to $(2)$. The proof of the equivalence of $(1)$ and $(3)$ is similar.

If the columns of $Q$ are an orthonormal set, then that is equivalent to "the matrix is orthogonal", and since $I = Q^TQ = QQ^T$ and $(Q^T)^T = Q$, it follows that if $Q$ is orthogonal then so is $Q^T$, hence the columns of $Q^T$ (i.e., the rows of $Q$) form an orthonormal set as well.

Sujaan Kunalan
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  • For #2, I was asking why would the rows be orthonormal if the columns are orthonormal. Thanks. – Tom Bennett Aug 08 '13 at 16:23
  • The conditions of (1) is equivalent to P'P=I. It is not equivalent to PP'=I. We want to prove PP'=I, knowing that P'P=I and P is a square matrix. – Tom Bennett Aug 08 '13 at 16:53
  • If $AB=I$, then $B=A^{-1}$. Then, $AB=BA=I$. So, if you know that $PP^{-1}=I$, then $P^{-1}P=I$ as well, assuming $P$ is an invertible square matrix. – Sujaan Kunalan Aug 08 '13 at 17:01