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`A`  square matrix `A` is called orthogonal if `A^TA = I_n` . Let `v_1,v_2,ldots,v_n`...

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`A`  square matrix `A` is called orthogonal if `A^TA = I_n` .

Let `v_1,v_2,ldots,v_n` be the columns of an orthogonal matrix `A` . Show that the `v_i"'s"` are mutually perpendicular and unit vectors.

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tiburtius's profile pic

Posted (Answer #1)

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Let's look at the element in `i`-th row and `j`-th column `delta_(ij)` of matrix `A^TA`. That element is scalar product of vectors `v_i` and `v_j` that is 

`delta_(ij)=v_i cdot v_j`.                                                                      (1)

Also since `A` is orthogonal it follows that

`delta_(ij)={(0 if i ne j),(1 if i=j):}`

Hence from (1) we get

`v_i cdot v_j={(0 if i ne j),(1 if i=j):}`

So if we multiply two different vectors `v_i` and `v_j` we get 0 which means they are perpendicular (scalar product is equal to 0 only if vectors are perpendicular or if one of them is zero-vector), and if we multiply vector by itself `v_i cdot v_i` we get 1 which means `v_i` is vector with length 1 that is a unit vector.

Hence all vectors `v_1,v_2,ldots,v_n` are mutually perpendicular unit vectors.

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Posted (Answer #2)

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Multiplyng  on the rigt both side of the  realtion: 

 (1)         A'A= I             (where A' is the transposed matrix of A)

by unity vector u`^k`  we get:

 (2)      A' v`^k` = u`^k`

Now, multiplyng  on the again  both side b u`^h`   h `!=`  k

       v`^h` x v`^k` = u`^k` X u`^h` = 0

Therfore the vectors v`^h` and v`^k`  are orthogonal for h`!=`  k  

  Multiplyng the relation (2) by v`^h` on the right:

   A' v`^k` X v `^h` = u`^k` X v`^h`

since the left side is zero thus:

             u`^k` x v`^h` = 0                 for  h `!=`   k



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