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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

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

Prove the following generalization of Theorem 2.23. Let $\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}$ be a linear transformation from a finite- dimensional vector space \(V\) to a finite-dimensional vector space W. Let \(\beta\) and \(\beta^{\prime}\) be ordered bases for \(\mathrm{V}\), and let \(\gamma\) and \(\gamma^{\prime}\) be ordered bases for \(\mathrm{W}\). Then $[\mathrm{T}]_{\beta}^{\gamma}^{\prime}=P^{-1}[\mathrm{~T}]_{\beta}^{\gamma} Q\(, where \)Q\( is the matrix that changes \)\beta^{\prime}$-coordinates into \(\beta\)-coordinates and \(P\) is the matrix that changes \(\gamma^{\prime}\)-coordinates into \(\gamma\)-coordinates.

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 \(f(x)\)

Suppose \(A\) is a square matrix. Show (a) \(A+A^{T}\) is symmetric, (b) \(A-A^{T}\) is skew-symmetric, (c) \(A=B+C,\) where \(B\) is symmetric and \(C\) is skew- symmetric.

Compute \(A B\) using block multiplication, where $$A=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 4 & 0 \\ 0 & 0 & 2 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cccc} 1 & 2 & 3 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right].$$
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