/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Show that a matrix \(B\) has a l... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 33

Show that a matrix \(B\) has a left inverse if and only if \(B^{T}\) has a right inverse.

Short Answer

Expert verified
Assuming matrix B has a left inverse A, we have \(AB = I\). Taking the transpose of both sides, we get \((AB)^T = I^T\), which simplifies to \(B^TA^T = I\). This shows that A^T is a right inverse of B^T. On the other hand, assuming that matrix B^T has a right inverse C, we have \(B^TC = I\). Taking the transpose of both sides, we get \((B^TC)^T = I^T\), which simplifies to \(C^TB = I\). This demonstrates that C^T is the left inverse of B. In conclusion, matrix B has a left inverse if and only if its transpose, B^T, has a right inverse.

Step by step solution

01

Assume B has a left inverse

Let's assume that matrix B has a left inverse, which we will denote as A. This means that AB = I, where I is the identity matrix.
02

Find the transpose of both sides of AB = I

In order to find the transpose of both sides, we need to find the transpose of the product AB. Recall that (AB)^T = B^T * A^T, so we have: \( (AB)^T = B^T * A^T \) And since (I)^T = I: \( (AB)^T = I^T \) Putting the two above equations together, we get: \( B^T * A^T = I \)
03

Show that A^T is the right inverse of B^T

From the above equation, we can see that A^T is the right inverse of B^T because their product equals the identity matrix I. So, we have shown that if B has a left inverse, then B^T has a right inverse.
04

Assume B^T has a right inverse

Now, let's assume that B^T has a right inverse C, such that B^T * C = I.
05

Find transpose of both sides of B^T * C = I

To find the transpose of both sides, we need to find the transpose of the product B^T * C. Recall that (B^T * C)^T = C^T * B, so we have: \( (B^T * C)^T = C^T * B \) And since (I)^T = I: \( (B^T * C)^T = I^T \) Putting the two above equations together, we get: \( C^T * B = I \)
06

Show that C^T is the left inverse of B

From the equation in step 5, we can see that C^T is the left inverse of B because their product equals the identity matrix I. So, we have shown that if B^T has a right inverse, then B has a left inverse. In conclusion, we have shown that a matrix B has a left inverse if and only if its transpose, B^T, has a right inverse.

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Most popular questions from this chapter

Let \(\mathrm{x}\) and \(\mathrm{y}\) be nonzero vectors in \(\mathbb{R}^{m}\) and \(\mathbb{R}^{n},\) respectively, and let \(A=\mathbf{x y}^{T}\) (a) Show that \(\\{x\\}\) is a basis for the column space of \(A\) and that \(\left\\{\mathbf{y}^{T}\right\\}\) is a basis for the row space of \(A\). (b) What is the dimension of \(N(A) ?\)

Let \(A \in \mathbb{R}^{m \times n}, B \in \mathbb{R}^{n \times r},\) and \(C=A B .\) Show that (a) if \(A\) and \(B\) both have linearly independent column vectors, then the column vectors of \(C\) will also be linearly independent. (b) if \(A\) and \(B\) both have linearly independent row vectors, then the row vectors of \(C\) will also be linearly independent. [Hint: Apply part (a) to \(C^{T}\).]

We can define a one-to-one correspondence between the elements of \(P_{n}\) and \(\mathbb{R}^{n}\) by $$\begin{array}{l} p(x)=a_{1}+a_{2} x+\cdots+a_{n} x^{n-1} \\ \leftrightarrow\left(a_{1}, \ldots, a_{n}\right)^{T}=\mathbf{a} \end{array}$$ Show that if \(p \leftrightarrow \mathbf{a}\) and \(q \leftrightarrow \mathbf{b},\) then (a) \(\alpha p \leftrightarrow \alpha\) a for any scalar \(\alpha\) (b) \(p+q \leftrightarrow \mathbf{a}+\mathbf{b}\) [In general, two vector spaces are said to be isomorphic if their elements can be put into a one-to-one correspondence that is preserved under scalar multiplication and addition as in (a) and (b).]

Let \(\mathbf{u}_{1}=(1,1,1)^{T}, \mathbf{u}_{2}=(1,2,2)^{T}\) \(\mathbf{u}_{3}=(2,3,4)^{T}\) (a) Find the transition matrix corresponding to the change of basis from \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\\}\) to \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) (b) Find the coordinates of each of the following vectors with respect to \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) (i) \((3,2,5)^{T}\) (ii) \(\quad(1,1,2)^{T}\) (iii) \((2,3,2)^{T}\)

Let \(S\) be the subspace of \(P_{3}\) consisting of all polynomials \(p(x)\) such that \(p(0)=0,\) and let \(T\) be the subspace of all polynomials \(q(x)\) such that \(q(1)=\) 0\. Find bases for (a) \(S\) (b) \(T\) (c) \(S \cap T\)
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