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Chapter 2: Problem 7

Let \(A\) be an \(n \times n\) matrix. (a) Suppose that \(A^{2}=O .\) Prove that \(A\) is not invertible. (b) Suppose that \(A B=O\) for some nonzero \(n \times n\) matrix \(B\). Could \(A\) be invertible? Explain.

Short Answer

Expert verified
(a) If \(A^2=O\), assume \(A\) has an inverse \(A^{-1}\). Pre-multiply both sides by \(A^{-1}\): \((A^{-1}A)A=A^{-1}O \Rightarrow IA =O\). Since \(AA^{-1}=A^{-1}A=I\) but we obtain \(IA=O\), we find a contradiction, and thus, \(A\) is not invertible. (b) If \(AB=O\) for a nonzero matrix \(B\), assume \(A\) has an inverse \(A^{-1}\). Pre-multiply both sides by \(A^{-1}\): \((A^{-1}A)B=A^{-1}O \Rightarrow IB=O\). However, since \(B\) is nonzero, \(IB=B\neq O\), and we find a contradiction. Thus, \(A\) cannot be invertible in this case either.

Step by step solution

01

(a) Assume A has an inverse

Assume that \(A\) has an inverse, \(A^{-1}\). We will show that this assumption leads to a contradiction.
02

(a) Pre-multiply both sides by A^{-1}

Pre-multiply both sides of the equation \(A^2 = O\) by \(A^{-1}\) to obtain: \(A^{-1}(A^2) = A^{-1}O\)
03

(a) Simplify and obtain contradiction

Using the associativity of matrix multiplication, we have: \((A^{-1}A)A = O\) Since \(A^{-1}A = I\), the identity matrix, we get: \(IA = O\) But for a matrix to have an inverse, we must have \(AA^{-1} = A^{-1}A = I\). Since we have found that \(IA = O\), we arrive at a contradiction (as \(O \neq I\)), and thus, \(A\) cannot be invertible.
04

(b) Consider the case where A * B = O

Now, let's consider the case where \(AB = O\) for some nonzero matrix \(B\).
05

(b) Suppose A has an inverse

Assume that \(A\) has an inverse, \(A^{-1}\).
06

(b) Pre-multiply both sides by A^{-1}

Pre-multiply both sides of the equation \(AB = O\) by \(A^{-1}\) to get: \(A^{-1}(AB) = A^{-1}O\)
07

(b) Simplify and obtain contradiction

Using the associativity of matrix multiplication, we have: \((A^{-1}A)B = O\) Since \(A^{-1}A = I\), the identity matrix, we get: \(IB = O\) However, because \(B\) is assumed to be non-zero, and \(IB = B \neq O\), we arrive at a contradiction. Therefore, \(A\) cannot be invertible in this case as well.

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Most popular questions from this chapter

Let \(V\) be a nonzero vector space, and let \(W\) be a proper subspace of \(V\) (i.e., \(W \neq V\) ). (a) Let \(\mathrm{g} \in \mathrm{W}^{*}\) and \(v \in \mathrm{V}\) with $v \notin \mathrm{W}\(. Prove that for any scalar \)a\( there exists a function \)\mathrm{f} \in \mathrm{V}^{*}\( such that \)\mathrm{f}(v)=a$ and \(\mathrm{f}(x)=\mathrm{g}(x)\) for all \(x\) in W. Hint: For the infinite- dimensional case, use Exercise 4 of Section \(1.7\) and Exercise 35 of Section 2.1. (b) Use (a) to prove there exists a nonzero linear functional $f \in \mathrm{V}^{*}\( such that \)\mathrm{f}(x)=0\( for all \)x \in \mathrm{W}$.

Define \(\mathrm{T}: \mathrm{M}_{2 \times 2}(R) \rightarrow \mathrm{P}_{2}(R) \quad\) by $\quad \mathrm{\top}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=(a+b)+(2 d) x+b x^{2} .$ Let $$ \beta=\left\\{\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right),\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right),\left(\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right),\left(\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right)\right\\} \quad \text { and } \quad \gamma=\left\\{1, x, x^{2}\right\\} $$ Compute \([T]_{\beta}^{\gamma}\).

Find \(2 \times 2\) invertible matrices \(A\) and \(B\) such that \(A+B \neq 0\) and \(A+B\) is not invertible.

For each matrix \(A\) and ordered basis \(\beta\), find \(\left[\mathrm{L}_{A}\right]_{\beta}\). Also, find an invertible matrix \(Q\) such that \(\left[\mathrm{L}_{A}\right]_{\beta}=Q^{-1} A Q\). (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 1 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 2\end{array}\right)\right\\}$ (b) \(A=\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{r}1 \\ -1\end{array}\right)\right\\}$ (c) $A=\left(\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 1 \\ 1 & 1 & 0\end{array}\right) \quad\( and \)\quad \beta=\left\\{\left(\begin{array}{l}1 \\\ 1 \\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 2\end{array}\right)\right\\}$ (d) $A=\left(\begin{array}{rrr}13 & 1 & 4 \\ 1 & 13 & 4 \\ 4 & 4 & 10\end{array}\right)\( and \)\beta=\left\\{\left(\begin{array}{r}1 \\ 1 \\\ -2\end{array}\right),\left(\begin{array}{r}1 \\ -1 \\\ 0\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 1\end{array}\right)\right\\}$

Let \(A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right],\) (a) \(\quad\) Find \(A^{n} \cdot(b) \quad\) Find \(B^{n}\).
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