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I was looking at the definition of an orthogonal matrix, which is as follows:

Square matrix $Q$ is orthogonal if its columns are pairwise orthonormal, i.e.,

$$Q^TQ = I$$

Hence also

$$Q^T = Q^{−1}$$

I understood what it means for two vectors to be orthonormal, they basically need to be orthogonal, and, in addition, they have length $1$.

I don't understand why we have:

$$Q^TQ = I$$

Could you please explain this to me?

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Let the matrix $M$ be an $n \times n$ matrix with components

$$M =\begin{pmatrix} v_1 & v_2 & \cdots &v_n \end{pmatrix}$$

where $v_i$ is the $i$th column vector with $n$ components.

Now, consider $M \times M$

$$M^T \times M = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\v_n \end{pmatrix} \times \begin{pmatrix} v_1 & v_2 & \cdots &v_n \end{pmatrix} \\ = \begin{pmatrix} v_1 \cdot v_1 & v_1 \cdot v_2 & \ldots &v_1 \cdot v_n \\ v_2 \cdot v_1 & v_2 \cdot v_2 & \ldots &v_2 \cdot v_n \\ v_3 \cdot v_1 & v_3 \cdot v_2 & \ldots &v_2 \cdot v_n \\ \vdots& \vdots & \ddots &\vdots \\ v_n \cdot v_1 & v_n \cdot v_2 & \ldots &v_n \cdot v_n \end{pmatrix}$$

However, remember that since the vectors $v_i$ that form the matrix $M$ are pairwise orthogonal,

$$v_i \cdot v_j = 1 \ \text{if} \ i = j \\ v_i \cdot v_j = 0 \ \text{if} \ i \neq j $$

Using this to simplify the matrix product,

$$M^T \times M = \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 &\cdots & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} = I $$

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We need to recall how we compute each entry in a product $AB$ of matrices $A$ and $B$: The entry at row $i$, column $j$, is the scalar product of row $i$ from $A$, and column $j$ from $B$. Now, recall that the $i$'th row of $Q^T$ is the $i$'th column of $Q$ (draw up a matrix and its transpose to convince yourself of this). Thus the entry at row $i$, column $j$ in $Q^TQ$ is the scalar product of row $i$ in $Q^T$ and column $j$ in $Q$, or columns $i$ and $j$ from $Q$. Then if $i\neq j$ the entry is zero, and if $i=j$ the entry is one, by the orthonormality of the columns of $Q$.

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That is because, if we denote $C_i$ the column vectors of $Q$, the coefficient $a_{ij}$ in ${}^{\mathrm t\mkern-2mu}Q\, Q$ is precisely $\langle C_i,C_j\rangle$.

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Let $Q$ be a matrix with columns $C_1$,$C_2$,$C_3$, ..... $C_n$. Then $Q^T$ would a matrix with rows $C_1^T$,$C_2^T$,$C_3^T$, ..... $C_n^T$. Now $Q^TQ$, means we have to multiply these matrices. The 1st row and 1st column element of this result ($Q^TQ$) is the multiplication of the first row of $Q^T$ with the first column of the $Q$. That will be scalar one because we are multiplying $C_1^T$ with $C_1$ and $C_1$ is an orthonormal vector.

The 1st row and 2nd column element of the product of the matrices above ($Q^TQ$) is the multiplication of the first row of $Q^T$ with the second column of the $Q$. That will be scalar zero because we are multiplying $C_1^T$ with $C_2$ and $C_1$ $C_2$is are orthogonal to each other.

Reasoning in this way, you can conclude that $Q^TQ=I$

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  • $\begingroup$ The idea I think I was missing is that if vectors $v$ and $u$ are orthogonal, then $v^T * u = 0$.. I knew about $v * u = 0$, but this statement is actually equivalent to the previous one. $\endgroup$ – nbro May 1 '16 at 12:57
  • $\begingroup$ @nbro The $*$ symbol in your comment should be avoided as it means two different things. Two (column) vectors of the same size cannot be multiplied, but you can take their dot product, which is zero if they are orthogonal: $u \cdot v = 0$. I think what you were missing is that the dot product can be expressed as $u \cdot v = u^T v$, where this time the product on the right is a (well-defined) matrix product. $\endgroup$ – Alex Provost May 1 '16 at 13:50
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Let's start from the definition of orthogonal matrix you feel more comfortable with: A square matrix $Q \in \mathbb{R}^{n\times n}$ is said to be orthogonal iff its rows and its columns form an orthonormal base of $\mathbb{R}^n$, that is if, calling $r_{Q_i}$ and $c_{Q_i}$ the $i$-th row and column of $Q$ respectively, represented as column vectors, the following relations hold: $$r_{Q_i}^T r_{Q_j} = \delta_{ij}$$ $$c_{Q_i}^T c_{Q_j} = \delta_{ij}$$ where: $$\delta_{ij} = \begin{cases} 1,\; i=j \\ 0,\; i\ne j \end{cases}$$ and, of course, $a^T b := \sum\limits_{i=1}^{n}a_ib_i$ is the inner product between the two vectors of $\mathbb{R}^n$ $a$ and $b$, written as they were column vectors.

Then the definition you have found in the textbook is plain to see, since: $$Q = \begin{pmatrix} r_{Q_1}^T \\ r_{Q_2}^T \\ \vdots \\ r_{Q_n}^T \end{pmatrix} \quad Q^T = \begin{pmatrix} r_{Q_1} & r_{Q_2} & \cdots & r_{Q_n} \end{pmatrix}$$

$$QQ^T = \begin{pmatrix} r_{Q_1}^T r_{Q_1} & r_{Q_1}^T r_{Q_2} & \cdots & r_{Q_1}^T r_{Q_n} \\ r_{Q_2}^T r_{Q_1} & r_{Q_2}^T r_{Q_2} & \cdots & r_{Q_2}^T r_{Q_n} \\ \vdots & \vdots & \ddots & \vdots \\ r_{Q_n}^T r_{Q_1} & r_{Q_n}^T r_{Q_2} & \cdots & r_{Q_n}^Tr_{Q_n} \end{pmatrix} = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} = I_n$$

Of course, $Q^TQ = I$ can be proved by considering the columns instead of the rows: $$Q = \begin{pmatrix} c_{Q_1} & c_{Q_2} & \cdots & c_{Q_n} \end{pmatrix} \quad Q^T = \begin{pmatrix} c_{Q_1}^T \\ c_{Q_2}^T \\ \vdots \\ c_{Q_n}^T \end{pmatrix} $$

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