There is a theorem that says if $U$ is an orthogonal matrix, i.e., its columns (or rows) form an orthonormal basis, then the action of $UU^T$ represents orthogonal projection of the vector space onto the space spanned by the columns of $U$.

But, why does this theorem also apply when my "$U$" matrix is a $3\times1$ column vector and hence not an orthogonal matrix (it's not even square)?

I had to compute the matrix that represents orthogonal projection onto the line spanned by $(1, 2, -1)$. So, following the Gram-Schmidt process for this simple case, just set $v_1= (1,2,-1)$. Then I normalized this vector to make it of unit length. Finally, I compute $VV^T$ to get a $3\times3$ matrix. I applied the theorem stated above and concluded that this $3\times3$ matrix represents orthogonal projection onto the line spanned by the *one* column in $V$, which has the same span as the span of $(1, 2,-1)$, since the Gram-Schmidt process produces a set of orthogonal vectors with the same span as the original set of linearly independent vectors.

Have I applied the theorem incorrectly? I got the correct matrix.

Thanks,

## Best Answer

The theorem you have quoted is true but only tells part of the story. An improved version is as follows.

Commentsif it is square. I think this is the situation you are envisaging in your question. But in this case the result is trivial because $W$ is equal to ${\Bbb R}^m$, and $UU^T=I$, and the projection transformation is simply $P({\bf x})={\bf x}$.