1
$\begingroup$

For the vectors: $$v_1 = \frac {1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix},\ v_2 =\frac {1}{\sqrt2} \begin{pmatrix} -1\\ 1 \\ 0 \\ 0 \end{pmatrix}$$

Find an orthogonal projection for $(-1,1,2,2)^T$ onto Span$(v_1,v_2)$, which is a subspace of $V$.

I am unsure how to find the orthogonal projection when I have to do it onto the Span of vectors, I hope somebody can show a method to calculate it or give a hint.

$\endgroup$
1
  • $\begingroup$ Note that $v_1, v_2$ are a pair of orthogonal unit vectors. How do you find the orthogonal projection of a vector onto the subspace spanned by two of the natural basis vectors? What is the orthogonal projection of $(1,2,3,4)$ onto $\langle \mathbf {e_1},\mathbf {e_2}\rangle$? $\endgroup$
    – saulspatz
    Apr 10 '18 at 13:27
1
$\begingroup$

HINT

  • consider the matrix $A=[v_1\quad v_2]$
  • the projection matrix is $P=A(A^TA)^{-1}A^T$
$\endgroup$
1
  • $\begingroup$ I would like to add that if $v_1, v_2$ are orthonormal which they are in this case, the $A^T A$ term cancels out because of the property $A^T A = I$ and you'll end up with $P = A A^T$ which is an orthogonal projector (or projection matrix) onto a span of orthonormal vectors and the longer formula works for any span. $\endgroup$ Mar 24 at 14:03
0
$\begingroup$

The spanning set is orthonormal, so you simply add up the projections onto them: $$(v\cdot v_1)v_1+(v\cdot v_2)v_2.$$ As saulspatz hinted, this is exactly what you do to find the coordinates of $v$ relative to the standard basis, i.e., $v=x_v\mathbf e_1+y_v\mathbf e_2=(v\cdot\mathbf e_1)\mathbf e_1+(v\cdot\mathbf e_2)\mathbf e_2$.

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .