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When you use least squares via orthogonal projection, is it common to not necessarily have an exactly 90 degree angle but close? I have attached an image.

For example, I calculated a dot product between p (a projection) and one of the vectors that intersects the projection (x) and I get -1 for the dot product , not 0.

Projection

or am I doing the wrong calculation?

  • Based on the problem I'm doing..p is column vector [2, 3] and b is [column vector [5,1] . Doesn't seem to add up to 0? –  Jul 03 '21 at 14:04
  • [I'm rewriting my previous comment due to an autocorrect error.] If $p$ is the projection of $b$ onto the column space of a matrix $A$, then the vector from $p$ to $b$ is exactly orthogonal to the column space of $A$. – littleO Jul 03 '21 at 20:54
  • Is there an app or program where I can visualize this? I have the span of the column space (A) and I have the orthogonal vector component. Their dot product is 0, so that makes sense algebraically. However, I can't visualize in my head how it is that they are orthogonal. C(A) span is is column vector [4, 6] and ort vector component is [3, -2). projection is [2, 3) and vector b is [5, 1]. Any help appreciated. –  Jul 03 '21 at 22:05
  • Do you have a textbook that you're reading? If $A$ is $m \times n$ then the column space of $A$ is a subspace of $\mathbb R^m$. I picture a plane through the origin in $\mathbb R^3$ or a line through the origin in $\mathbb R^2$. And $b$ is a point in $\mathbb R^m$ [I picture a point off the plane in $\mathbb R^3$], and $p$ is the closest point in $R(A)$ to $b$. Visually you can see that $b - p$ [the vector from $p$ to $b$] is orthogonal to $R(A)$. – littleO Jul 03 '21 at 23:00
  • It is Chad's answer from here where the example is coming from: https://math.stackexchange.com/questions/1298261/difference-between-orthogonal-projection-and-least-squares-solution?newreg=40314d2455994946a1430fbe7eec1557 –  Jul 03 '21 at 23:03

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