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Chapter 4: Problem 12

A matrix \(Q \in \mathrm{M}_{n \times n}(R)\) is called orthogonal if $Q Q^{t}=I\(. Prove that if \)Q\( is orthogonal, then \)\operatorname{det}(Q)=\pm 1$.

Short Answer

Expert verified
Given an orthogonal matrix \(Q\), meaning \(QQ^t = I\), we compute the determinant on both sides: \(\det(QQ^t) = \det(Q) \det(Q^t) = \det(I)\). Using the property \(\det(A^t) = \det(A)\) and the fact that \(\det(I) = 1\), we have \(\det(Q)^2 = 1\). Thus, if Q is orthogonal, then \( \operatorname{det}(Q)=\pm 1\).

Step by step solution

01

Write down the definition of an orthogonal matrix

An orthogonal matrix Q is defined by the property \( QQ^t = I \), where \(I\) is the identity matrix and \(Q^t\) is the transpose of Q.
02

Find the determinant of both sides of the equation

Now, let's compute the determinant on both sides of the equation, using the property that \( \det(AB) = \det(A) \det(B) \): \[ \det(QQ^t) = \det(Q) \det(Q^t) = \det(I) \]
03

Use the property of determinants of transposes

We know that the determinant of a matrix's transpose is equal to the determinant of the matrix itself: \( \det(A^t) = \det(A) \). Applying this, we get: \[ \det(Q) \det(Q^t) = \det(Q) \det(Q) \]
04

Evaluate the determinant of the identity matrix

The determinant of the identity matrix I is always 1, since the identity matrix is a diagonal matrix with all diagonal elements equal to 1: \[ \det(I) = 1 \]
05

Combine previous steps to reach the conclusion

Now let's combine all the steps above into one equation: \[\det(Q) \det(Q) = 1\] \[ \Rightarrow \det(Q)^2 = 1 \] The only possible values for a determinant squared equal to 1 are 1 and -1. Therefore, if Q is an orthogonal matrix, then \( \operatorname{det}(Q)=\pm 1 \).

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Consider the vector space \(\mathbf{P}_{3}(t)\) of polynomials of degree \(\leq 3\). (a) Show that \(S=\left\\{(t-1)^{3},(t-1)^{2}, t-1,1\right\\}\) is a basis of \(\mathbf{P}_{3}(t)\), (b) Find the coordinate vector \([v]\) of \(v=3 t^{3}-4 t^{2}+2 t-5\) relative to \(S\).

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