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Let $W=$span$\{\begin{bmatrix}1&1\\0&0\end{bmatrix}\begin{bmatrix}0&0\\1&1\end{bmatrix}\}$ and suppose the span is orthogonal under certain Hermitian inner product space (just suppose). If we are asked to find a basis for $W^{\perp}$, is it ok that I set up a general matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and write $\langle\begin{bmatrix}1&1\\0&0\end{bmatrix},\begin{bmatrix}a&b\\c&d\end{bmatrix}\rangle=0$ and $\langle\begin{bmatrix}0&0\\1&1\end{bmatrix},\begin{bmatrix}a&b\\c&d\end{bmatrix}\rangle=0$ and solve the $2$ row linear system of equation? Then we find the kernel vectors as the basis in $W^{\perp}$. Could someone help?

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    $\begingroup$ yes, that is a right way. $\endgroup$ – Solumilkyu Apr 19 '16 at 4:10
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I see $\left[ \begin{array}{cc} 1 & -1 \\ 0 & 0 \end{array}\right]$ and $\left[ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array}\right]$ by inspection. I'm hopeful these are what you derive from the equations you mention.

To be honest, I think of $\left[ \begin{array}{cc} 1 & -1 \\ 0 & 0 \end{array}\right]$ as $(1,-1,0,0)$ etc... so I just see the problem in $\mathbb{R}^4$ and there is nothing terribly complex about your given data.

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  • $\begingroup$ my answer assumes $\langle A, B \rangle = \text{trace}(AB^T)$ aka the usual inner product. $\endgroup$ – James S. Cook Apr 19 '16 at 4:23

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