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Chapter 1: Problem 5

Prove that \(A+A^{t}\) is symmetric for any square matrix \(A\).

Short Answer

Expert verified
We need to prove that \(A + A^t\) is symmetric for any square matrix \(A\). To do this, we find the transpose of \(A + A^t\) and show it is equal to itself: \((A + A^t)^t = A^t + (A^t)^t\) Using the property that \((A^t)^t = A\), we get: \((A + A^t)^t = A^t + A\) Since matrix addition is commutative, we have \(A^t + A = A + A^t\). Thus, \((A + A^t)^t = A + A^t\) and we conclude that \(A + A^t\) is symmetric.

Step by step solution

01

Find the transpose of the sum of the given matrix and its transpose

Let's begin by finding the transpose of \(A + A^t\), which we will denote as \((A + A^t)^t\).
02

Apply the properties of transpose

Now, we will apply the properties of transpose. One of the properties states that the transpose of the sum of two matrices is equal to the sum of the transposes of those matrices. That is, \((A + B)^t = A^t + B^t\). In our case, this becomes: \[(A + A^t)^t = A^t + (A^t)^t\]
03

Calculate the transpose of the transpose

Now, we need to calculate \((A^t)^t\). Remember that calculating the transpose of a transpose of a matrix returns the original matrix. In other words, \((A^t)^t = A\). Therefore: \[(A + A^t)^t = A^t + A\]
04

Compare the transpose with the original expression

We now need to compare our result with the original expression \(A + A^t\). We have: \[(A + A^t)^t = A^t + A\] Note that the addition of matrices is commutative, which means that the order in which we add the matrices does not matter. That is, \(A^t + A = A + A^t\).
05

Conclude that the matrix is symmetric

Since \((A + A^t)^t = A + A^t\), we can conclude that the matrix resulting from the addition of a given square matrix \(A\) and its transpose is symmetric.

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Most popular questions from this chapter

Prove: For any complex numbers \(z, w \in \mathbf{C},(\text { i) } \overline{z+w}=\bar{z}+\bar{w}\) (ii) \(\overline{z w}=\bar{z} \bar{w}\) (iii) \(\bar{z}=z\)

Let \(u=(1,-2,4), v=(3,5,1), w=(2,1,-3) .\) Find: (a) \(3 u-2 v\) (b) \(5 u+3 v-4 w\) \(\begin{array}{llll}\text { (c) } u \cdot v, & u \cdot w, & v \cdot w ;\end{array}\) (d) \(\|u\|,\|v\|\) (e) \(\cos \theta,\) where \(\theta\) is the angle between \(u\) and \(v\) \((\mathrm{f}) \quad d(u, v)\) \((g) \quad \operatorname{proj}(u, v)\)

Solve the following systems of linear equations by the method introduced in this section. (a) \(2 x_{1}-2 x_{2}-3 x_{3}=-2\) (a) $\begin{aligned} 3 x_{1}-3 x_{2}-2 x_{3}+5 x_{4} &=7 \\ x_{1}-x_{2}-2 x_{3}-x_{4} &=-3 \end{aligned}$ \(3 x_{1}-7 x_{2}+4 x_{3}=10\) (b) \(x_{1}-2 x_{2}+x_{3}=3\) \(2 x_{1}-x_{2}-2 x_{3}=6\) \(x_{1}+2 x_{2}-x_{3}+x_{4}=5\) (c) \(\quad x_{1}+4 x_{2}-3 x_{3}-3 x_{4}=6\) \(2 x_{1}+3 x_{2}-x_{3}+4 x_{4}=8\)Solve the following systems of linear equations by the method introduced in this section. (a) \(2 x_{1}-2 x_{2}-3 x_{3}=-2\) (a) $\begin{aligned} 3 x_{1}-3 x_{2}-2 x_{3}+5 x_{4} &=7 \\ x_{1}-x_{2}-2 x_{3}-x_{4} &=-3 \end{aligned}$ \(3 x_{1}-7 x_{2}+4 x_{3}=10\) (b) \(x_{1}-2 x_{2}+x_{3}=3\) \(2 x_{1}-x_{2}-2 x_{3}=6\) \(x_{1}+2 x_{2}-x_{3}+x_{4}=5\) (c) \(\quad x_{1}+4 x_{2}-3 x_{3}-3 x_{4}=6\) \(2 x_{1}+3 x_{2}-x_{3}+4 x_{4}=8\)

Let \(V\) denote the set of all differentiable real-valued functions defined on the real line. Prove that \(V\) is a vector space with the operations of addition and scalar multiplication defined in Example \(3 .\)

Prove that a subset \(W\) of a vector space \(V\) is a subspace of \(V\) if and only if \(0 \in \mathrm{W}\) and \(a x+y \in \mathrm{W}\) whenever \(a \in F\) and $x, y \in W$.
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