Question:
In an inner product space \(V(F)\), for \(a,b\in F\) and \(\alpha,\beta\in V\), then \((a\alpha,b\beta)=\)
In an inner product space \(V(F)\), for \(a,b\in F\) and \(\alpha,\beta\in V\), then \((a\alpha,b\beta)=\)
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In complex inner product spaces, one scalar comes out conjugated depending on the convention used.
Updated On: Jul 3, 2026
- \(ab(\alpha,\beta)\)
- \(\overline{a}b(\alpha,\beta)\)
- \(\overline{ab}(\alpha,\beta)\)
- \(a\overline{b}(\alpha,\beta)\)
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The Correct Option is B
Solution and Explanation
Concept:
In a complex inner product space, the inner product is conjugate linear in one component and linear in the other component. According to the convention used here, \[ (a\alpha,b\beta)=\overline{a}b(\alpha,\beta) \]
Step 1: Scalar from first component. When scalar \(a\) comes from the first component, it comes out as conjugate: \[ (a\alpha,\beta)=\overline{a}(\alpha,\beta) \]
Step 2: Scalar from second component. When scalar \(b\) comes from the second component, it comes out directly: \[ (\alpha,b\beta)=b(\alpha,\beta) \]
Step 3: Combine both. \[ (a\alpha,b\beta)=\overline{a}b(\alpha,\beta) \]
Step 4: Final answer. \[ \boxed{\overline{a}b(\alpha,\beta)} \]
In a complex inner product space, the inner product is conjugate linear in one component and linear in the other component. According to the convention used here, \[ (a\alpha,b\beta)=\overline{a}b(\alpha,\beta) \]
Step 1: Scalar from first component. When scalar \(a\) comes from the first component, it comes out as conjugate: \[ (a\alpha,\beta)=\overline{a}(\alpha,\beta) \]
Step 2: Scalar from second component. When scalar \(b\) comes from the second component, it comes out directly: \[ (\alpha,b\beta)=b(\alpha,\beta) \]
Step 3: Combine both. \[ (a\alpha,b\beta)=\overline{a}b(\alpha,\beta) \]
Step 4: Final answer. \[ \boxed{\overline{a}b(\alpha,\beta)} \]
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