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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 on \(\mathbb{R}^{2}\) defined by \\[ L(\mathbf{x})=\left(x_{1} \cos \alpha-x_{2} \sin \alpha, x_{1} \sin \alpha+x_{2} \cos \alpha\right)^{T} \\] Express \(x_{1}, x_{2},\) and \(L(\mathbf{x})\) in terms of polar coor dinates. Describe geometrically the effect of the linear transformation.

Let \(L\) be the operator on \(P_{3}\) defined by $$L(p(x))=x p^{\prime}(x)+p^{\prime \prime}(x)$$ (a) Find the matrix \(A\) representing \(L\) with respect to \(\left[1, x, x^{2}\right]\) (b) Find the matrix \(B\) representing \(L\) with respect to \(\left[1, x, 1+x^{2}\right]\) (c) Find the matrix \(S\) such that \(B=S^{-1} A S\). (d) If \(p(x)=a_{0}+a_{1} x+a_{2}\left(1+x^{2}\right),\) calculate \(L^{n}(p(x))\)

A linear transformation \(L: V \rightarrow W\) is said to be one-to-one if \(L\left(\mathbf{v}_{1}\right)=L\left(\mathbf{v}_{2}\right)\) implies that \(\mathbf{v}_{1}=\mathbf{v}_{2}\) (i.e., no two distinct vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}\) in \(V\) get mapped into the same vector \(\mathbf{w} \in W\) ). Show that \(L\) is oneto-one if and only if \(\operatorname{ker}(L)=\left\\{\mathbf{0}_{V}\right\\}\)

Suppose that \(A=S T,\) where \(S\) is nonsingular. Let \(B=T S .\) Show that \(B\) is similar to \(A\)

Let \\[ \mathbf{y}_{1}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), \quad \mathbf{y}_{2}=\left(\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right), \quad \mathbf{y}_{3}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) \\] and let \(\mathcal{I}\) be the identity operator on \(\mathbb{R}^{3}\). (a) Find the coordinates of \(\mathcal{I}\left(\mathbf{e}_{1}\right), \mathcal{I}\left(\mathbf{e}_{2}\right),\) and \(\mathcal{I}\left(\mathbf{e}_{3}\right)\) with respect to \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) (b) Find a matrix \(A\) such that \(A \mathbf{x}\) is the coordinate vector of \(\mathbf{x}\) with respect to \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\)
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