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Chapter 2: Problem 5

Let \(A\) be invertible. Prove that \(A^{t}\) is invertible and \(\left(A^{t}\right)^{-1}=\left(A^{-1}\right)^{t}\). Visit goo.gl/suFm6V for a solution.

Short Answer

Expert verified
We are given that matrix A is invertible. To prove that \(A^t\) is invertible and \((A^t)^{-1} = (A^{-1})^t\), we proceed with the following steps: Step 1: Show that \(A^t\) is invertible. Let B = \((A^{-1})^t\). Since \(A^{-1}\) is the inverse of A, we have \(A A^{-1} = I\). Therefore, transpose both sides: \((A A^{-1})^t = I^t\). Using the transpose property \((AB)^t = B^t A^t\), we obtain \((A^{-1})^t A^t = I\). Thus, \(A^t\) is invertible. Step 2: Show that \((A^t)^{-1} = (A^{-1})^t\). Since B = \((A^{-1})^t\) satisfies the conditions to be the inverse of \(A^t\), we have \((A^t)^{-1} = (A^{-1})^t\). Hence, we have proved that \(A^t\) is invertible, and its inverse is equal to the transpose of the inverse of A.

Step by step solution

01

Show that A^t is invertible

To prove that A^t is invertible, we need to find a matrix B such that A^t times B gives the identity matrix. We will try using the transpose of A^{-1} as our matrix B. Let B = (A^{-1})^t. We can now multiply A^t and B to see if we get the identity matrix. (A^t)(A^{-1})^t = I Recall the property of matrix transpose: (AB)^t = B^t A^t. Using this property, we will reverse transpose both sides of the equation: [(A^t)(A^{-1})^t]^t = I^t Applying the property, we get: [(A^{-1})^t]^t (A^t)^t = I^t Now, using the property that (A^t)^t = A and the fact that the transpose of the identity matrix is still the identity matrix, we can simplify the expression further: A^{-1} A = I This is true since A^{-1} is the inverse of A by definition. Therefore, A^t is invertible.
02

Show that (A^t)^{-1} = (A^{-1})^t

Now that we have shown that A^t is invertible, we need to show that its inverse is equal to the transpose of A^{-1}. From Step 1, we have found a matrix B = (A^{-1})^t, such that: A^t B = I and B A^t = I Since B satisfies the conditions to be the inverse of A^t, we have: (A^t)^{-1} = (A^{-1})^t We have now shown that (A^t)^{-1} = (A^{-1})^t as required. With the two steps above, we have proved that A^t is invertible and its inverse is equal to the transpose of the inverse of A.

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Most popular questions from this chapter

For each of the following vector spaces \(V\) and bases \(\beta\), find explicit formulas for vectors of the dual basis \(\beta^{*}\) for \(\mathrm{V}^{*}\), as in Example 4 . (a) \(\mathrm{V}=\mathrm{R}^{3} ; \beta=\\{(1,0,1),(1,2,1),(0,0,1)\\}\) (b) \(\mathrm{V}=\mathrm{P}_{2}(R) ; \beta=\left\\{1, x, x^{2}\right\\}\)

Suppose \(A\) and \(B\) are symmetric. Show that the following are also symmetric: (a) \(A+B\) (b) \(k A,\) for any scalar \(k\) \((\mathrm{c}) \quad A^{2}\) (d) \(A^{n},\) for \(n>0\) (e) \(f(A),\) for any polynomial \(f(x)\)

Let \(V\) and \(W\) be finite-dimensional vector spaces with ordered bases \(\beta=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) and \(\gamma=\left\\{w_{1}, w_{2}, \ldots, w_{m}\right\\}\), respectively. By Theorem \(2.6\) (p. 73), there exist linear transformations $\mathrm{T}_{i j}: \mathrm{V} \rightarrow \mathrm{W}$ such that $$ \mathrm{T}_{i j}\left(v_{k}\right)= \begin{cases}w_{i} & \text { if } k=j \\\ 0 & \text { if } k \neq j\end{cases} $$ First prove that $\left\\{\mathbf{T}_{i j}: 1 \leq i \leq m, 1 \leq j \leq n\right\\}\( is a basis for \)\mathcal{L}(\mathbf{V}, \mathbf{W})$. Then let \(M^{i j}\) be the \(m \times n\) matrix with 1 in the \(i\) th row and \(j\) th column and 0 elsewhere, and prove that $\left[\mathrm{T}_{i j}\right]_{\beta}^{\gamma}=M^{i j}$. Again by Theorem 2.6, there exists a linear transformation $\Phi_{\beta}^{\gamma}: \mathcal{L}(\mathrm{V}, \mathrm{W}) \rightarrow \mathrm{M}_{m \times n}(F)$ such that \(\Phi_{\beta}^{\gamma}\left(T_{i j}\right)=M^{i j}\). Prove that \(\Phi_{\beta}^{\gamma}\) is an isomorphism.

Let \(V\) be a finite-dimensional vector space with the ordered basis \(\beta\). Prove that \(\psi(\beta)=\beta^{* *}\), where \(\psi\) is defined in Theorem \(2.26\).

Assume the notation in Theorem \(2.13 .\) (a) Suppose that \(z\) is a (column) vector in \(\mathrm{F}^{p}\). Use Theorem 2.13(b) to prove that \(B z\) is a linear combination of the columns of \(B\). In particular, if \(z=\left(a_{1}, a_{2}, \ldots, a_{p}\right)^{t}\), then show that $$ B z=\sum_{j=1}^{p} a_{j} v_{j} . $$ (b) Extend (a) to prove that column \(j\) of \(A B\) is a linear combination of the columns of \(A\) with the coefficients in the linear combination being the entries of column \(j\) of \(B\). (c) For any row vector \(w \in \mathrm{F}^{m}\), prove that \(w A\) is a linear combination of the rows of \(A\) with the coefficients in the linear combination being the coordinates of \(w\). Hint: Use properties of the transpose operation applied to (a). (d) Prove the analogous result to (b) about rows: Row \(i\) of \(A B\) is a linear combination of the rows of \(B\) with the coefficients in the linear combination being the entries of row \(i\) of \(A\).
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