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Chapter 4: Problem 11

Show that if \(A\) and \(B\) are similar matrices, then \(\operatorname{det}(A)=\operatorname{det}(B)\)

Short Answer

Expert verified
If A and B are similar matrices, there exists an invertible matrix P such that \(B = P^{-1}AP\). Using determinant properties, we can compute \(\operatorname{det}(P^{-1}AP)\) and simplify the expression to get \(\operatorname{det}(B) = \operatorname{det}(A)\). Thus, if A and B are similar matrices, their determinants are equal.

Step by step solution

01

Recall the definition of similar matrices

Two matrices A and B are similar if there exists an invertible matrix P such that \(B = P^{-1}AP\).
02

Calculate the determinant of both sides of the similarity equation

Using the fact that A and B are similar, we have the equation: \(B = P^{-1}AP\). We will now calculate the determinant of both sides of this equation. That is, we will compute \(\operatorname{det}(B)\) and \(\operatorname{det}(P^{-1}AP)\).
03

Apply determinant properties

The determinant has the following properties: 1. \(\operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B)\) for any two square matrices A and B. 2. \(\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}\) for any invertible matrix A. Using these properties, we can compute \(\operatorname{det}(P^{-1}AP)\) as follows: \[ \operatorname{det}(P^{-1}AP) = \operatorname{det}(P^{-1})\operatorname{det}(A)\operatorname{det}(P) = \frac{1}{\operatorname{det}(P)}\operatorname{det}(A)\operatorname{det}(P) \]
04

Simplify the expression

Now, we can simplify the expression from Step 3: \[ \frac{1}{\operatorname{det}(P)}\operatorname{det}(A)\operatorname{det}(P) = \operatorname{det}(A) \]
05

Conclude

From Steps 2 and 4, we have \(\operatorname{det}(B) = \operatorname{det}(A)\). Therefore, if A and B are similar matrices, their determinants are equal.

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Most popular questions from this chapter

Let \(L\) be the linear operator mapping \(\mathbb{R}^{3}\) into \(\mathbb{R}^{3}\) defined by \(L(\mathbf{x})=A \mathbf{x},\) where $$A=\left(\begin{array}{rrr} 3 & -1 & -2 \\ 2 & 0 & -2 \\ 2 & -1 & -1 \end{array}\right)$$ and let $$\mathbf{v}_{1}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), \quad \mathbf{v}_{2}=\left(\begin{array}{l} 1 \\ 2 \\ 0 \end{array}\right), \quad \mathbf{v}_{3}=\left(\begin{array}{r} 0 \\ -2 \\ 1 \end{array}\right)$$ Find the transition matrix \(V\) corresponding to a change of basis from \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) to \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\\},\) and use it to determine the matrix \(B\) representing \(L\) with respect to \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\)

Show that if \(A\) is similar to \(B\) and \(A\) is nonsingular then \(B\) must also be nonsingular and \(A^{-1}\) and \(B^{-1}\) are similar.

Determine whether the following are linear transformations from \(C[0,1]\) into \(\mathbb{R}^{1}\) : (a) \(L(f)=f(0)\) (b) \(L(f)=|f(0)|\) (c) \(L(f)=[f(0)+f(1)] / 2\) (d) \(L(f)=\left\\{\int_{0}^{1}[f(x)]^{2} d x\right\\}^{1 / 2}\)

The trace of an \(n \times n\) matrix \(A,\) denoted \(\operatorname{tr}(A),\) is the sum of its diagonal entries; that is, $$\operatorname{tr}(A)=a_{11}+a_{22}+\cdots+a_{n n}$$ Show that (a) \(\operatorname{tr}(A B)=\operatorname{tr}(B A)\) (b) if \(A\) is similar to \(B,\) then \(\operatorname{tr}(A)=\operatorname{tr}(B)\)

Find the standard matrix representation for each of the following linear operators: (a) \(L\) is the linear operator that rotates each \(\mathbf{x}\) in \(\mathbb{R}^{2}\) by \(45^{\circ}\) in the clockwise direction. (b) \(L\) is the linear operator that reflects each vector \(\mathbf{x}\) in \(\mathbb{R}^{2}\) about the \(x_{1}\) -axis and then rotates it \(90^{\circ}\) in the counterclockwise direction. (c) \(L\) doubles the length of \(\mathbf{x}\) and then rotates it \(30^{\circ}\) in the counterclockwise direction. (d) \(L\) reflects each vector \(\mathbf{x}\) about the line \(x_{2}=x_{1}\) and then projects it onto the \(x_{1}\) -axis.
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