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Let $H =\operatorname{Col}(A)$, where $$A =\begin{pmatrix} 1&2\\ 2&4\\ 3& 1\end{pmatrix}$$

Find $H^\perp$, the orthogonal complement of $H$.

$H$ is the same thing as $A$, and as I understand it, Orthogonal complement means the span of vectors that are orthogonal to the matrix, but I don't understand how to solve for this. Nor am I completely clear on how a vector can be orthogonal to a matrix, as I only no how to find the dot product between vectors with only $1$ row.

Thanks.

Luke
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  • $H$ is not the same thing as $A$. $A$ is a matrix; $H$ is a vector space that consists of all linear combinations of the two columns of $A$ treated as vectors. Elements of $H^\bot$ aren’t orthogonal to the matrix $A$, they’re orthogonal to every vector in $H$. – amd Dec 04 '15 at 08:05
  • The $row(A) \perp col(A)$, so $H^\perp$ must be the image of the rowspace of A...@JaneDoe – makansij Jul 02 '17 at 15:51

1 Answers1

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In general, for any matrix $A \in \mathbb{C}^{m\times n}$, the answer may be obtained, using those relations:

$$ \text{im}(A^*) =\ker(A)^{\perp} ~~\text{and}~ ~\ker(A^*) = \text{im}(A)^{\perp}$$ furthermore $$\mathbb{C}^{m} = \ker(A^*) \oplus \text{im}(A) ~~\text{and}~ ~ \mathbb{C}^{n} = \ker(A) \oplus \text{im}(A^*)$$ (this is sometimes called Fredholm alternative)

Where

  • $\ker(\cdot)$ is the kernel of a matrix.
  • $\text{im}(\cdot)$ is the image of a matrix.
  • $A^*$ is the conjugate transpose. When dealing with real matrices only, this becomes the usual transpose.
  • $\oplus$ - direct sum.
them
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  • I think that $H^\perp$ must be the image of the rowspace of A, right? Because $row(A) \perp col(A)$ as far as I understand it, right @them? – makansij Jul 02 '17 at 15:49