Question

Show for a square matrix, the followings are equivalent.
The columns of A are all vectors of length 1, and are all at right angles to each other.
A^T = A^−1

Answer 1

Answer to Question #84219, Math / Linear Algebra

Question: Show for a square matrix, the followings are equivalent.

The columns of $A$ are all vectors of length 1, and are all at right angles to each other.

$$A^T = A^{-1}.$$

Solution:

$$\models$$

Given $A$ is a square matrix of order $n$.

$A = \left[ \begin{array}{cccc} A_1 & A_2 & A_3 & \ldots & A_n \end{array} \right]$, where each $A_i = \left[ \begin{array}{cccc} a_{1i} & a_{2i} & a_{3i} & \ldots & a_{ni} \end{array} \right]^T$ is a vector of length 1.

i.e., $\sqrt{a_{1i}^2 + a_{2i}^2 + \ldots + a_{ni}^2} = 1 \Rightarrow a_{1i}^2 + a_{2i}^2 + \ldots + a_{ni}^2 = 1$.

$A_i^2 = 1$.

Also the columns of $A$ are all at right angles to each other. i.e., $A_i \cdot A_j = 0$.

i.e., $A_i A_j^T = 0, \forall i \neq j \Rightarrow a_{1i}a_{j1} + a_{2i}a_{j2} + \ldots + a_{ni}a_{jn} = 0, \forall i \neq j$.

Now $AA^T = \begin{bmatrix} A_1^2 & A_1 A_2^T & A_1 A_3^T & A_1 A_n^T \\ A_2 A_1^T & A_2^2 & A_2 A_3^T & \ldots & A_2 A_n^T \\ A_3 A_1^T & A_3 A_2^T & A_3^2 & A_3 A_n^T \\ \vdots & & & \\ A_n A_1^T & A_n A_2^T & A_n A_3^T & \ldots & A_n^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 \\ \vdots & & & \\ 0 & 0 & 0 & 1 \end{bmatrix} = I$.

Therefore, $A^T = A^{-1}$.

$\Leftarrow$

Given $A^T = A^{-1}$.

$$\begin{array}{l} \text{i.e., } AA^T = I \quad \Rightarrow \left[ \begin{array}{cccc} A_1^2 & A_1 A_2^T & A_1 A_3^T & A_1 A_n^T \\ A_2 A_1^T & A_2^2 & A_2 A_3^T & \dots A_2 A_n^T \\ A_3 A_1^T & A_3 A_2^T & A_3^2 & A_3 A_n^T \\ \vdots & & & \\ A_n A_1^T & A_n A_2^T & A_n A_3^T & \dots A_n^2 \end{array} \right] = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & 0 \\ \vdots & & & \\ 0 & 0 & 0 & 1 \end{array} \right] \\ \end{array}$$

This implies $A_i^2 = 1, \forall i = 1, 2, \dots, n$.

Thus each vector $A_i$ is of length 1.

And $A_i A_j^T = 0, \forall i \neq j$, which implies that $A_i \cdot A_j = 0, \forall i \neq j$.

Thus the columns of $A$ are all at right angles to each other.

Thus the equivalence of the two statements is proved.

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