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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?)

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  1. zzr0ck3r
    • one year ago
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    I am thinking that it means that one is a scalar multiple of the other?

  2. zzr0ck3r
    • one year ago
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    \(<2,3,4>\iff<8,12,16>\)?

  3. anonymous
    • one year ago
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    if we have that \(T\) represents an orthogonal projection on a line L, then we have that \(L\) is a one-dimensional 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 \(n-1\) and its rank is \(1\). so in that case the columns of the matrix must span just a one-dimensional 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)

  4. anonymous
    • one year ago
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    so 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 one-dimensional subspace)

  5. anonymous
    • one year ago
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    a 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

  6. sh3lsh
    • one year ago
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    You're great! This helps a great deal! Thanks so much!

  7. anonymous
    • one year ago
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    basically, 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

  8. anonymous
    • one year ago
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    oops, I meant n x n matrix

  9. anonymous
    • one year ago
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    and 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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