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sh3lsh
 one year ago
Linear Algebra Question.
Is it true that if a matrix T is the orthogonal projection onto a line T, the columns of the matrix T will be pairwise parallel?
What does pairwise parallel mean? (Is it any two columns will have a dot product to be the same?)
Will a reflection be pairwise orthogonal? Again what does this mean? (Is it any two columns will have a dot product to be 0?)
sh3lsh
 one year ago
Linear Algebra Question. Is it true that if a matrix T is the orthogonal projection onto a line T, the columns of the matrix T will be pairwise parallel? What does pairwise parallel mean? (Is it any two columns will have a dot product to be the same?) Will a reflection be pairwise orthogonal? Again what does this mean? (Is it any two columns will have a dot product to be 0?)

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I am thinking that it means that one is a scalar multiple of the other?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0\(<2,3,4>\iff<8,12,16>\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if we have that \(T\) represents an orthogonal projection on a line L, then we have that \(L\) is a onedimensional subspace of our vector space, the kernel or null space is the orthogonal complement of \(L\), \(L^\bot\), and the image or range is \(L\) itself. recall that the columns of a matrix span its image; this means the nullity of \(T\) is \(n1\) and its rank is \(1\). so in that case the columns of the matrix must span just a onedimensional subspace, which is the case if and only if they are all parallel to one another (i.e. pairwise parallel  which means all possible pairs of columns are themselves parallel vectors)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the point is that if T maps only to vectors on the line, then its columns must all lie on said line as well  in other words, they must be pairwise parallel (all parallel with one another since they live in the same onedimensional subspace)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a reflection on the other hand maps the entire space to a 'flipped' copy of itself, preserving the angles between vectors. since the angles are preserved, and the columns correspond to the matrix's 'output' for the ordered orthonormal basis vector inputs, the fact that the basis vectors are all mutually orthogonal means that their image under the transformation represented by the matrix (i.e. the columns of the matrix) must still be mutually orthogonal

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0You're great! This helps a great deal! Thanks so much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0basically, if we denote the \(i\)th column of a \(m\times n\) matrix \(A\) by \(A_i\), and the \(i\)th orthonormal basis vector \(e_i\) so $$e_1=\begin{bmatrix}1\\0\\0\\\vdots\\0\end{bmatrix},e_2=\begin{bmatrix}0\\1\\0\\\vdots\\0\end{bmatrix},\dots,e_n=\begin{bmatrix}0\\0\\0\\\vdots\\1\end{bmatrix}$$ the column \(A_i\) represents the action of \(A\) on \(e_i\); in particular, \(A_i=Ae_i\). so if \(A\) preserves angles, then \({A_1,\dots,A_n}\) must be orthogonal since \(\{e_1,\dots,e_n\}\) clearly are

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, I meant n x n matrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and similarly in the first case we're told that \(T\) is a map to some line \(L\), so if the vector \(v\) gives the direction of our line, then \(T\) sends every vector to something parallel to \(v\), i.e. \(kv\) for some scalar \(k\). so it follows that \(A\) sends the basis vectors \(e_i\) to such vectors and since \(A_i=Ae_i\) it follows that the columns \(A_i\) are all scalar multiples of some vector \(v\), i.e. they're all mutually parallel
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