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Chapter 2: Problem 4

Let \(A\) and \(B\) be \(n \times n\) invertible matrices. Prove that \(A B\) is invertible and \((A B)^{-1}=B^{-1} A^{-1}\).

Short Answer

Expert verified
In order to prove that the product of two invertible matrices \(A\) and \(B\) is also invertible, we propose the inverse of \(AB\) as \(B^{-1}A^{-1}\). We then show that \((AB)(B^{-1}A^{-1}) = AA^{-1} = I\), which satisfies the conditions for invertibility. Therefore, the product of two invertible matrices is invertible with inverse given by \((AB)^{-1} = B^{-1}A^{-1}\).

Step by step solution

01

Recall the definition of an invertible matrix

A matrix is called invertible if there exists another matrix, called its inverse, such that their product is equal to the identity matrix. In other words, if \(A\) is invertible, then there exists a matrix \(A^{-1}\) such that \[AA^{-1}=A^{-1}A=I\] Similarly, if \(B\) is invertible, then there exists a matrix \(B^{-1}\) such that \[BB^{-1}=B^{-1}B=I\]
02

Show that the product \(AB\) is invertible

To prove that \(AB\) is invertible, we need to find another matrix that, when multiplied with \(AB\), results in the identity matrix. We will propose that the inverse of \(AB\) is given by the product \(B^{-1}A^{-1}\), and then demonstrate that this product satisfies the conditions for invertibility. Consider the product \((AB)(B^{-1}A^{-1})\). Using the associative property of matrix multiplication, we can rewrite this as: \[A(BB^{-1})A^{-1}\] Since \(BB^{-1}=I\), this simplifies to: \[AI A^{-1}\] We know that \(A\) is invertible, so \(AI=A\), which means we have: \[AA^{-1}\] This is equal to the identity matrix \(I\) by the definition of the inverse of \(A\). Therefore, \(AB\) is invertible with inverse \(B^{-1}A^{-1}\).
03

Conclusion

We have shown that the product of two invertible matrices \(A\) and \(B\) is also invertible, and that the inverse of the product is given by \((AB)^{-1}=B^{-1}A^{-1}\). This is an important property of invertible matrices that can be utilized in various applications, such as solving systems of linear equations.

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Most popular questions from this chapter

Refer to the following matrices: $$A=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{rrr} 4 & 0 & -3 \\ -1 & -2 & 3 \end{array}\right], \quad C=\left[\begin{array}{rrrr} 2 & -3 & 0 & 1 \\ 5 & -1 & -4 & 2 \\ -1 & 0 & 0 & 3 \end{array}\right], \quad D=\left[\begin{array}{rr} 2 \\ -1 \\ 3 \end{array}\right].$$ Find (a) \(3 A-4 B,\) (b) \(A C,\) (c) \(B C,\) (d) \(A D,\) (e) \(B D,\) (f) \(C D.\)

Let \(\mathrm{V}=\mathrm{P}_{n}(F)\), and let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars in \(F\). (a) For \(0 \leq i \leq n\), define \(\mathrm{f}_{i} \in \mathrm{V}^{*}\) by \(\mathrm{f}_{i}(p(x))=p\left(c_{i}\right)\). Prove that \(\left\\{\mathrm{f}_{0}, \mathrm{f}_{1}, \ldots, \mathrm{f}_{n}\right\\}\) is a basis for \(\mathrm{V}^{*}\). Hint: Apply any linear combination of this set that equals the zero transformation to \(p(x)=\) \(\left(x-c_{1}\right)\left(x-c_{2}\right) \cdots\left(x-c_{n}\right)\), and deduce that the first coefficient is zero. (b) Use the corollary to Theorem \(2.26\) and (a) to show that there exist unique polynomials \(p_{0}(x), p_{1}(x), \ldots, p_{n}(x)\) such that \(p_{i}\left(c_{j}\right)=\delta_{i j}\) for \(0 \leq i \leq n\). These polynomials are the Lagrange polynomials defined in Section 1.6. (c) For any scalars \(a_{0}, a_{1}, \ldots, a_{n}\) (not necessarily distinct), deduce that there exists a unique polynomial \(q(x)\) of degree at most \(n\) such that \(q\left(c_{i}\right)=a_{i}\) for \(0 \leq i \leq n .\) In fact, $$ q(x)=\sum_{i=0}^{n} a_{i} p_{i}(x) $$ (d) Deduce the Lagrange interpolation formula: $$ p(x)=\sum_{i=0}^{n} p\left(c_{i}\right) p_{i}(x) $$ for any \(p(x) \in \mathrm{V}\). (e) Prove that $$ \int_{a}^{b} p(t) d t=\sum_{i=0}^{n} p\left(c_{i}\right) d_{i} $$ where $$ d_{i}=\int_{a}^{b} p_{i}(t) d t $$ Suppose now that $$ c_{i}=a+\frac{i(b-a)}{n} \text { for } i=0,1, \ldots, n \text {. } $$ For \(n=1\), the preceding result yields the trapezoidal rule for evaluating the definite integral of a polynomial. For \(n=2\), this result yields Simpson's rule for evaluating the definite integral of a polynomial.

Label the following statements as true or false. In each part, \(V\) and \(W\) are finite-dimensional vector spaces (over \(F\) ), and \(\mathrm{T}\) is a function from \(\mathrm{V}\) to \(\mathrm{W}\). (a) If \(\mathrm{T}\) is linear, then \(\mathrm{T}\) preserves sums and scalar products. (b) If \(\mathrm{T}(x+y)=\mathrm{T}(x)+\mathrm{T}(y)\), then \(\mathrm{T}\) is linear. (c) \(\mathrm{T}\) is one-to-one if and only if the only vector \(x\) such that \(\mathrm{T}(x)=0\) is \(x=0\). (d) If \(T\) is linear, then \(T\left(0_{v}\right)=0_{w}\). (e) If \(T\) is linear, then nullity \((T)+\operatorname{rank}(T)=\operatorname{dim}(W)\). (f) If \(\mathrm{T}\) is linear, then \(\mathrm{T}\) carries linearly independent subsets of \(\mathrm{V}\) onto linearly independent subsets of W. (g) If \(\mathrm{T}, \mathrm{U}: \mathrm{V} \rightarrow \mathrm{W}\) are both linear and agree on a basis for \(\mathrm{V}\), then \(T=U\). (h) Given \(x_{1}, x_{2} \in \mathrm{V}\) and \(y_{1}, y_{2} \in \mathrm{W}\), there exists a linear transformation $\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\( such that \)\mathrm{T}\left(x_{1}\right)=y_{1}$ and \(\mathrm{T}\left(x_{2}\right)=y_{2}\). For Exercises 2 through 6, prove that \(\mathrm{T}\) is a linear transformation, and find bases for both \(N(T)\) and \(R(T)\). Then compute the nullity and rank of \(T\), and verify the dimension theorem. Finally, use the appropriate theorems in this section to determine whether \(\mathrm{T}\) is one-to-one or onto.

Refer to \(A=\left[\begin{array}{rr}2 & -5 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}4 & -2 \\ 1 & -6\end{array}\right], C=\left[\begin{array}{rr}6 & -4 \\ 3 & -2\end{array}\right].\) Find \((a) B^{2}\) and \(B^{3},(b) f(B)\) and \(g(B),\) where $$f(x)=x^{2}+2 x-22, \quad g(x)=x^{2}-3 x-6.$$

Find a basis for each of the following subspaces of \(C^{\infty}\). (a) \(N\left(D^{2}-D-1\right)\) (b) \(N\left(D^{3}-3 D^{2}+3 D-1\right)\) (c) \(\mathrm{N}\left(\mathrm{D}^{3}+6 \mathrm{D}^{2}+8 \mathrm{D}\right)\)
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