This is kind of a stupid question and I am taking some risk of getting some down-votes here, but, I can't resist posting it. Suppose ( u 1 , u 2 ) is an orthonormal basis for R 2 , and let x be an arbitrary vector in R 2 , then we can decompose x by projecting on u 1 , u 2 , i.e. ( u 1 T x ) u 1 ( u 2 T x ) u 2 Certainly we have x = ( u 1 T x ) u 1 + ( u 2 T x ) u 2 Looks like a simple problem, but how can I prove this?

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This is kind of a stupid question and I am taking some risk of getting some down-votes here, but, I can&#039;t resist posting it. Suppose   (      u    1    ,      u    2    ) is an orthonormal basis for       R    2  , and let x be an arbitrary vector in       R    2  , then we can decompose x by projecting on       u    1    ,      u    2  , i.e.                    (                  u          1          T                x        )                  u          1                                    (                  u          2          T                x        )                  u          2                    Certainly we have                    x        =        (                  u          1          T                x        )                  u          1                +        (                  u          2          T                x        )                  u          2                    Looks like a simple problem, but how can I prove this?

Mackenzie Zimmerman · Accepted Answer

Consider a more general situation. Suppose   {      v    1    ,      v    2    ,  …  ,      v    k    } is an orthogonal basis for a vector space V, and w is any vector in V. So there are unique scalars       c    1    ,      c    2    ,  …  ,      c    k   such that  w  =      c    1        v    1    +      c    2        v    2    +  ⋯  +      c    k        v    k    .Consider the inner product  (  w  ,      v    i    )  =  (      c    1        v    1    +      c    2        v    2    +  ⋯  +      c    k        v    k    ,      v    i    )  =  (      c    i        v    i    ,      v    i    )  =      c    i    (      v    i    ,      v    i    )since   (      v    i    ,      v    j    )  =  0  ,     ∀  i  ≠  j. So the coefficient       c    i   has the form      c    i    =            (      w      ,              v        i            )              (              v        i            ,              v        i            )        .In your question, we know i=1,2 and   (      v    i    ,      v    i    )  =  1 and   (  x  ,      v    i    )  =      x    T        v    i

Jaiden Bowman · Accepted Answer

By definition of a base   ∃    a  ,  b  ∈      R    :    x  =  a      u    1    +  b      u    2  Now   (  x  ,      u    1    )  =  (  a      u    1    +  b      u    2    ,      u    1    )  =  (  a      u    1    ,      u    1    )  +  (  b      u    2    ,      u    1    )  =  a  (      u    1    ,      u    1    )  +  b  (      u    2    ,      u    1    )  =  aSame for b.

This is kind of a stupid question and I am taking some risk of getting some down-votes here, but, I

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