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Mathematics LibreTexts

14.3: Relating Orthonormal Bases

Suppose \(T=\{u_{1}, \ldots, u_{n} \}\) and \(R=\{w_{1}, \ldots, w_{n} \}\) are two orthonormal bases for \(\Re^{n}\).  Then:
w_{1} &=& (w_{1}\cdot u_{1}) u_{1} + \cdots + (w_{1}\cdot u_{n})u_{n}\\
 & \vdots & \\
w_{n} &=& (w_{n}\cdot u_{1}) u_{1} + \cdots + (w_{n}\cdot u_{n})u_{n}\\
\Rightarrow w_{i} &=& \sum_{j} u_{j}(u_{j}\cdot w_{i}) \\
Thus the matrix for the change of basis from \(T\) to \(R\) is given by 
P = (P^{j}_{i}) = (u_{j}\cdot w_{i}).
We would like to calculate the product \(PP^{T}\). For that, we first develop a dirty trick for products of dot products:
(u.v)(w.z)=(u^{T} v) (w^{T} z) = u^{T} (v w^{T}) z\, . 
The object \(v w^{T}\) is the square matrix made from the outer product of \(v\) and \(w\)!  Now we are ready to compute the components of the matrix product \(PP^{T}\):
\sum_{i}(u_{j}\cdot w_{i})(w_{i}\cdot u_{k})&=&
 \sum_{i}(u_{j}^{T} w_{i}) (w_{i}^{T} u_{k})\\
&=& u_{j}^{T} \left[\sum_{i} (w_{i} w_{i}^{T}) \right] u_{k} \\
&\stackrel{(*)}=& u_{j}^{T} I_{n} u_{k} \\\
&=& u_{j}^{T} u_{k} = \delta_{jk}.
The equality \((*)\) is explained below.  Assuming \((*)\) holds, we have shown that \(PP^{T}=I_{n}\), which implies that 

The equality in the line \((*)\) says that \(\sum_{i} w_{i} w_{i}^{T}=I_{n}\).  To see this, we examine \(\left(\sum_{i} w_{i} w_{i}^{T}\right)v\) for an arbitrary vector \(v\).  We can find constants \(c^{j}\) such that \(v=\sum_{j} c^{j}w_{j}\), so that:
\left(\sum_{i} w_{i} w_{i}^{T}\right)v &=& \left(\sum_{i} w_{i} w_{i}^{T}\right)\left(\sum_{j} c^{j}w_{j}\right) \\
&=& \sum_{j} c^{j} \sum_{i} w_{i} w_{i}^{T} w_{j} \\
&=& \sum_{j} c^{j} \sum_{i} w_{i} \delta_{ij} \\
&=& \sum_{j} c^{j} w_{j} \textit{ since all terms with \(i\neq j\) vanish}\\
Thus, as a linear transformation, \(\sum_{i} w_{i} w_{i}^{T}=I_{n}\) fixes every vector, and thus must be the identity \(I_{n}\).



A matrix \(P\) is \(\textit{orthogonal}\) if \(P^{-1}=P^{T}\).


Then to summarize,



A change of basis matrix \(P\) relating two orthonormal bases is an orthogonal matrix.  \(\textit{i.e.}\), \(P^{-1}=P^T.\)


Example 123

Consider \(\Re^{3}\) with the orthonormal basis 
u_{1}=\begin{pmatrix}\frac{2}{\sqrt{6}}\\ \frac{1}{\sqrt{6}}\\ \frac{-1}{\sqrt{6}}\end{pmatrix},
u_{2}=\begin{pmatrix}0\\ \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{pmatrix},
u_{3}=\begin{pmatrix}\frac{1}{\sqrt{3}}\\ \frac{-1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{pmatrix}
Let \(E\) be the standard basis \(\{e_{1},e_{2},e_{3} \}\).  Since we are changing from the standard basis to a new basis, then the columns of the change of basis matrix are exactly the  standard basis vectors.  Then the change of basis matrix from \(E\) to \(S\) is given by:
e_{1}\cdot u_{1} & e_{1}\cdot u_{2} & e_{1}\cdot u_{3} \\
e_{2}\cdot u_{1} & e_{2}\cdot u_{2} & e_{2}\cdot u_{3} \\
e_{3}\cdot u_{1} & e_{3}\cdot u_{2} & e_{3}\cdot u_{3} \\
\end{pmatrix} \\
= \begin{pmatrix}
u_{1} & u_{2} & u_{3}
\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{3}}\\
\frac{-1}{\sqrt{6}}& \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{3}}\\
\end{pmatrix}. \\


From our theorem, we observe that:
P^{-1}=P^{T} &=& \begin{pmatrix}u_{1}^{T}\\u_{2}^{T}\\u_{3}^{T}\end{pmatrix} \\
&=& \begin{pmatrix}
\frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}& \frac{-1}{\sqrt{6}}\\
0 & \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{3}}& \frac{-1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\
\end{pmatrix}. \\

We can check that \(P^{T}P=I\) by a lengthy computation, or more simply, notice that
&=& \begin{pmatrix}u_{1}^{T}\\u_{2}^{T}\\u_{3}^{T}\end{pmatrix} \begin{pmatrix}u_{1} & u_{2}& u_{3}\end{pmatrix} \\
&=& \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}.
Above we are using orthonormality of the \(u_{i}\) and the fact that matrix multiplication amounts to taking dot products between rows and columns.  It is also very important to realize that the columns of an \(\textit{orthogonal}\) matrix are made
from an \(\textit{orthonormal}\) set of vectors.


Remark:    (Orthonormal Change of Basis and Diagonal Matrices)
Suppose \(D\) is a diagonal matrix  and we are able to use an orthogonal matrix \(P\) to change to a new basis.  Then the matrix \(M\) of \(D\) in the new basis is:
M = PDP^{-1} = PDP^{T}.
Now we calculate the transpose of \(M\).
M^{T} &=& (PDP^{T})^{T}\\
&=& (P^{T})^{T}D^{T}P^{T} \\
&=& PDP^{T}\\
&=& M
The matrix \(M=PDP^{T}\) is symmetric!